Computing $int_-π/2^π/2 frac28cos^2(θ)+10cos(θ)sin(θ)-28sin^2(θ)2cos^4(θ)+3cos^2(θ)sin^2(θ)+msin^4(θ) dθ$Evaluating $frac1pi int_0^pi e^2costheta dtheta$Complex integration $int_-pi^pi fracsin^2 t3+cos tdt$How to reverse the integration order of the double integral $int_theta=0^2piint_r=0^1+costhetar^2(sintheta+costheta)drdtheta$.Evaluate $int_0^piln(cos(x)+1)cos(nx),dx$Prove that $int_-infty^inftyfracsinomega cos2omegaomegadomega=0$Integrate $int_0^pifracsin(x)(sin^2 (x) + kcos^2(x))^frac12$ dx$ int_0^pi fraccostheta (costheta - costheta_0)(1-costheta_0costheta)^2-sin^2theta_0sin^2varphi_0sin^2theta dtheta $Evaluate $int_0^pi/2fracdthetaleft(cos^3theta+sin^3thetaright)^2/3$Prove $int_0^pifraccos(ntheta)costheta - costheta_0dtheta = fracpisin(ntheta_0)sintheta_0$Evaluating $int_0^2 pi fracsin(phi)(sin(theta)-sin(phi))^2+2a(1-cos(theta -phi)) dphi$
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Computing $int_-π/2^π/2 frac28cos^2(θ)+10cos(θ)sin(θ)-28sin^2(θ)2cos^4(θ)+3cos^2(θ)sin^2(θ)+msin^4(θ) dθ$
Evaluating $frac1pi int_0^pi e^2costheta dtheta$Complex integration $int_-pi^pi fracsin^2 t3+cos tdt$How to reverse the integration order of the double integral $int_theta=0^2piint_r=0^1+costhetar^2(sintheta+costheta)drdtheta$.Evaluate $int_0^piln(cos(x)+1)cos(nx),dx$Prove that $int_-infty^inftyfracsinomega cos2omegaomegadomega=0$Integrate $int_0^pifracsin(x)(sin^2 (x) + kcos^2(x))^frac12$ dx$ int_0^pi fraccostheta (costheta - costheta_0)(1-costheta_0costheta)^2-sin^2theta_0sin^2varphi_0sin^2theta dtheta $Evaluate $int_0^pi/2fracdthetaleft(cos^3theta+sin^3thetaright)^2/3$Prove $int_0^pifraccos(ntheta)costheta - costheta_0dtheta = fracpisin(ntheta_0)sintheta_0$Evaluating $int_0^2 pi fracsin(phi)(sin(theta)-sin(phi))^2+2a(1-cos(theta -phi)) dphi$
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I am studying the integral
$$I=int_-pi/2^pi/2
frac28cos^2(theta)+10cos(theta)sin(theta)-28sin^2(theta)2cos^4(theta)+3cos^2(theta)sin^2(theta)+msin^4(theta)dtheta,$$
where $m>0$. In the problem I am working, it is very important to know what value of $m$ makes this integral positive, negative or equal to zero. I found that if $m=2$, then the integral is zero (introducing it in Wolfram Alpha). However, I don't know how to prove it formally.
On the other hand, I tried some values of $m$ in Wolfram and it seems that if $m<2$, then the integral is negative, and if $m>2$, the integral is positive. But again, I have no proof of this.
Any ideas of how to approach this problem?
Just in case, I found this alternative representation of the integral
$$I=int_-pi/2^pi/2
frac8[5sin(2theta)+28cos(2theta)]cos(4theta)+15+8(m-2)sin^4(theta)dtheta.$$
Any help would be appreciated.
integration definite-integrals trigonometric-integrals
edited Jun 21 at 8:01
user21820
41.4k5 gold badges46 silver badges168 bronze badges
asked Jun 20 at 19:34
user326159user326159
1,4961 gold badge9 silver badges22 bronze badges
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add a comment |
$begingroup$
I am studying the integral
$$I=int_-pi/2^pi/2
frac28cos^2(theta)+10cos(theta)sin(theta)-28sin^2(theta)2cos^4(theta)+3cos^2(theta)sin^2(theta)+msin^4(theta)dtheta,$$
where $m>0$. In the problem I am working, it is very important to know what value of $m$ makes this integral positive, negative or equal to zero. I found that if $m=2$, then the integral is zero (introducing it in Wolfram Alpha). However, I don't know how to prove it formally.
On the other hand, I tried some values of $m$ in Wolfram and it seems that if $m<2$, then the integral is negative, and if $m>2$, the integral is positive. But again, I have no proof of this.
Any ideas of how to approach this problem?
Just in case, I found this alternative representation of the integral
$$I=int_-pi/2^pi/2
frac8[5sin(2theta)+28cos(2theta)]cos(4theta)+15+8(m-2)sin^4(theta)dtheta.$$
Any help would be appreciated.
integration definite-integrals trigonometric-integrals
edited Jun 21 at 8:01
user21820
41.4k5 gold badges46 silver badges168 bronze badges
asked Jun 20 at 19:34
user326159user326159
1,4961 gold badge9 silver badges22 bronze badges
$endgroup$
$begingroup$
Surely it is $8(m-2)sin^4theta$ not $(m-2)sin^4theta$ in the denominator of the last equation?
$endgroup$
– user10354138
Jun 20 at 19:57
$begingroup$
Yes. I've corrected it
$endgroup$
– user326159
Jun 20 at 20:04
3
$begingroup$
Writing $alpha^2 = m/2$, we get $$ I(alpha) = 14 pi left( 1-frac1alpharight)sqrtfrac24alpha+3. $$ From this, we easily deduce the sign of $I$ as function of $alpha$, i.e., $operatornamesign(I(alpha)) = operatornamesign(alpha - 1)$. Now if you ask me how I arrived this expression, it is basically from an ugly residue computation applied to $$I(alpha) = int_-infty^infty frac28(1-t^2)2+3t^2+2alpha^2 t^4, mathrmdt, $$ but I have not enough energy to write up all the intermediate steps...
$endgroup$
– Sangchul Lee
Jun 20 at 22:27
$begingroup$
@SangchulLee Thank you for your answer. It's amazing that you find a closed form for this integral. I've tried to compute it by the residue theorem but I cannot get this formula. Could you be more explicit on how did you do it? To know how to calculate this integral is very important for me, and I've been working many hours on it without success...
$endgroup$
– user326159
Jun 24 at 15:08
add a comment |
$begingroup$
I am studying the integral
$$I=int_-pi/2^pi/2
frac28cos^2(theta)+10cos(theta)sin(theta)-28sin^2(theta)2cos^4(theta)+3cos^2(theta)sin^2(theta)+msin^4(theta)dtheta,$$
where $m>0$. In the problem I am working, it is very important to know what value of $m$ makes this integral positive, negative or equal to zero. I found that if $m=2$, then the integral is zero (introducing it in Wolfram Alpha). However, I don't know how to prove it formally.
On the other hand, I tried some values of $m$ in Wolfram and it seems that if $m<2$, then the integral is negative, and if $m>2$, the integral is positive. But again, I have no proof of this.
Any ideas of how to approach this problem?
Just in case, I found this alternative representation of the integral
$$I=int_-pi/2^pi/2
frac8[5sin(2theta)+28cos(2theta)]cos(4theta)+15+8(m-2)sin^4(theta)dtheta.$$
Any help would be appreciated.
integration definite-integrals trigonometric-integrals
edited Jun 21 at 8:01
user21820
41.4k5 gold badges46 silver badges168 bronze badges
asked Jun 20 at 19:34
user326159user326159
1,4961 gold badge9 silver badges22 bronze badges
$endgroup$
I am studying the integral
$$I=int_-pi/2^pi/2
frac28cos^2(theta)+10cos(theta)sin(theta)-28sin^2(theta)2cos^4(theta)+3cos^2(theta)sin^2(theta)+msin^4(theta)dtheta,$$
where $m>0$. In the problem I am working, it is very important to know what value of $m$ makes this integral positive, negative or equal to zero. I found that if $m=2$, then the integral is zero (introducing it in Wolfram Alpha). However, I don't know how to prove it formally.
On the other hand, I tried some values of $m$ in Wolfram and it seems that if $m<2$, then the integral is negative, and if $m>2$, the integral is positive. But again, I have no proof of this.
Any ideas of how to approach this problem?
Just in case, I found this alternative representation of the integral
$$I=int_-pi/2^pi/2
frac8[5sin(2theta)+28cos(2theta)]cos(4theta)+15+8(m-2)sin^4(theta)dtheta.$$
Any help would be appreciated.
integration definite-integrals trigonometric-integrals
integration definite-integrals trigonometric-integrals
edited Jun 21 at 8:01
user21820
41.4k5 gold badges46 silver badges168 bronze badges
asked Jun 20 at 19:34
user326159user326159
1,4961 gold badge9 silver badges22 bronze badges
edited Jun 21 at 8:01
user21820
41.4k5 gold badges46 silver badges168 bronze badges
asked Jun 20 at 19:34
user326159user326159
1,4961 gold badge9 silver badges22 bronze badges
edited Jun 21 at 8:01
user21820
41.4k5 gold badges46 silver badges168 bronze badges
edited Jun 21 at 8:01
user21820
41.4k5 gold badges46 silver badges168 bronze badges
edited Jun 21 at 8:01
user21820
41.4k5 gold badges46 silver badges168 bronze badges
41.4k5 gold badges46 silver badges168 bronze badges
asked Jun 20 at 19:34
user326159user326159
1,4961 gold badge9 silver badges22 bronze badges
asked Jun 20 at 19:34
user326159user326159
1,4961 gold badge9 silver badges22 bronze badges
asked Jun 20 at 19:34
user326159user326159
1,4961 gold badge9 silver badges22 bronze badges
1,4961 gold badge9 silver badges22 bronze badges
$begingroup$
Surely it is $8(m-2)sin^4theta$ not $(m-2)sin^4theta$ in the denominator of the last equation?
$endgroup$
– user10354138
Jun 20 at 19:57
$begingroup$
Yes. I've corrected it
$endgroup$
– user326159
Jun 20 at 20:04
3
$begingroup$
Writing $alpha^2 = m/2$, we get $$ I(alpha) = 14 pi left( 1-frac1alpharight)sqrtfrac24alpha+3. $$ From this, we easily deduce the sign of $I$ as function of $alpha$, i.e., $operatornamesign(I(alpha)) = operatornamesign(alpha - 1)$. Now if you ask me how I arrived this expression, it is basically from an ugly residue computation applied to $$I(alpha) = int_-infty^infty frac28(1-t^2)2+3t^2+2alpha^2 t^4, mathrmdt, $$ but I have not enough energy to write up all the intermediate steps...
$endgroup$
– Sangchul Lee
Jun 20 at 22:27
$begingroup$
@SangchulLee Thank you for your answer. It's amazing that you find a closed form for this integral. I've tried to compute it by the residue theorem but I cannot get this formula. Could you be more explicit on how did you do it? To know how to calculate this integral is very important for me, and I've been working many hours on it without success...
$endgroup$
– user326159
Jun 24 at 15:08
add a comment |
$begingroup$
Surely it is $8(m-2)sin^4theta$ not $(m-2)sin^4theta$ in the denominator of the last equation?
$endgroup$
– user10354138
Jun 20 at 19:57
$begingroup$
Yes. I've corrected it
$endgroup$
– user326159
Jun 20 at 20:04
3
$begingroup$
Writing $alpha^2 = m/2$, we get $$ I(alpha) = 14 pi left( 1-frac1alpharight)sqrtfrac24alpha+3. $$ From this, we easily deduce the sign of $I$ as function of $alpha$, i.e., $operatornamesign(I(alpha)) = operatornamesign(alpha - 1)$. Now if you ask me how I arrived this expression, it is basically from an ugly residue computation applied to $$I(alpha) = int_-infty^infty frac28(1-t^2)2+3t^2+2alpha^2 t^4, mathrmdt, $$ but I have not enough energy to write up all the intermediate steps...
$endgroup$
– Sangchul Lee
Jun 20 at 22:27
$begingroup$
@SangchulLee Thank you for your answer. It's amazing that you find a closed form for this integral. I've tried to compute it by the residue theorem but I cannot get this formula. Could you be more explicit on how did you do it? To know how to calculate this integral is very important for me, and I've been working many hours on it without success...
$endgroup$
– user326159
Jun 24 at 15:08
$begingroup$
Surely it is $8(m-2)sin^4theta$ not $(m-2)sin^4theta$ in the denominator of the last equation?
$endgroup$
– user10354138
Jun 20 at 19:57
$begingroup$
Surely it is $8(m-2)sin^4theta$ not $(m-2)sin^4theta$ in the denominator of the last equation?
$endgroup$
– user10354138
Jun 20 at 19:57
$begingroup$
Yes. I've corrected it
$endgroup$
– user326159
Jun 20 at 20:04
$begingroup$
Yes. I've corrected it
$endgroup$
– user326159
Jun 20 at 20:04
3
3
$begingroup$
Writing $alpha^2 = m/2$, we get $$ I(alpha) = 14 pi left( 1-frac1alpharight)sqrtfrac24alpha+3. $$ From this, we easily deduce the sign of $I$ as function of $alpha$, i.e., $operatornamesign(I(alpha)) = operatornamesign(alpha - 1)$. Now if you ask me how I arrived this expression, it is basically from an ugly residue computation applied to $$I(alpha) = int_-infty^infty frac28(1-t^2)2+3t^2+2alpha^2 t^4, mathrmdt, $$ but I have not enough energy to write up all the intermediate steps...
$endgroup$
– Sangchul Lee
Jun 20 at 22:27
$begingroup$
Writing $alpha^2 = m/2$, we get $$ I(alpha) = 14 pi left( 1-frac1alpharight)sqrtfrac24alpha+3. $$ From this, we easily deduce the sign of $I$ as function of $alpha$, i.e., $operatornamesign(I(alpha)) = operatornamesign(alpha - 1)$. Now if you ask me how I arrived this expression, it is basically from an ugly residue computation applied to $$I(alpha) = int_-infty^infty frac28(1-t^2)2+3t^2+2alpha^2 t^4, mathrmdt, $$ but I have not enough energy to write up all the intermediate steps...
$endgroup$
– Sangchul Lee
Jun 20 at 22:27
$begingroup$
@SangchulLee Thank you for your answer. It's amazing that you find a closed form for this integral. I've tried to compute it by the residue theorem but I cannot get this formula. Could you be more explicit on how did you do it? To know how to calculate this integral is very important for me, and I've been working many hours on it without success...
$endgroup$
– user326159
Jun 24 at 15:08
$begingroup$
@SangchulLee Thank you for your answer. It's amazing that you find a closed form for this integral. I've tried to compute it by the residue theorem but I cannot get this formula. Could you be more explicit on how did you do it? To know how to calculate this integral is very important for me, and I've been working many hours on it without success...
$endgroup$
– user326159
Jun 24 at 15:08
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Define the function $mathcalI:mathbbR_>0rightarrowmathbbR$ via the trigonometric integral
$$beginalign
mathcalIleft(muright)
&:=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Let $muinmathbbR_>0$. Since the integral $mathcalI$ has a symmetric interval of integration, the integral of the odd component of the integrand vanishes identically:
$$beginalign
mathcalIleft(muright)
&=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_-fracpi2^0mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)-10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright);~~~smallleft[thetamapsto-thetaright]\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac56cos^2left(thetaright)-56sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Using the double-angle formulas for sine and cosine,
$$beginalign
sinleft(2thetaright)
&=2sinleft(thetaright)cosleft(thetaright),\
cosleft(2thetaright)
&=cos^2left(thetaright)-sin^2left(thetaright)\
&=2cos^2left(thetaright)-1\
&=1-2sin^2left(thetaright),\
endalign$$
we can rewrite the integral as
$$beginalign
mathcalIleft(muright)
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccosleft(2thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)8cos^4left(thetaright)+12cos^2left(thetaright)sin^2left(thetaright)+4musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)2left[1+cosleft(2thetaright)right]^2+3sin^2left(2thetaright)+muleft[1-cosleft(2thetaright)right]^2\
&=56int_0^pimathrmdtheta,frac2cosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2;~~~smallleft[thetamapstofrac12thetaright]\
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2.\
endalign$$
Using the tangent half-angle substitution, the trigonometric integral transforms as
$$beginalign
mathcalIleft(muright)
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2\
&=112int_0^inftymathrmdt,frac21+t^2cdotfracleft(frac1-t^21+t^2right)2left(1+frac1-t^21+t^2right)^2+3left(frac2t1+t^2right)^2+muleft(1-frac1-t^21+t^2right)^2;~~~smallleft[theta=2arctanleft(tright)right]\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+mu,t^4.\
endalign$$
Setting $sqrtfrac2mu=:ainmathbbR_>0$ and $frac34a=:binmathbbR_>0$, we then have
$$beginalign
mathcalIleft(muright)
&=mathcalIleft(frac2a^2right)\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+2a^-2t^4\
&=sqrtaint_0^inftymathrmdu,frac56left(1-au^2right)2+3au^2+2u^4;~~~smallleft[t=usqrtaright]\
&=frac13sqrtaint_0^inftymathrmdu,frac28left(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_1^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3u^2-4bright)1+2bu^2+u^4;~~~smallleft[umapstofrac1uright]\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bright)left(1+u^2right)1+2bu^2+u^4\
&=28left(1-frac43bright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-aright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-sqrtfrac2muright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=frac28left(mu-2right)left(sqrtmu+sqrt2right)sqrtmusqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4.\
endalign$$
Since the integral factor in the last line above is a positive quantity, we don't need to actually solve the integral to determine the sign of $mathcalI$. We have
$$operatornamesgnleft(mathcalIleft(muright)right)=operatornamesgnleft(mu-2right),$$
as you originally conjectured.
answered Jun 21 at 0:28
David HDavid H
22.3k2 gold badges49 silver badges97 bronze badges
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This problem is "nice" in the sense that the integrand is really trig function of $2theta$
$$
I=int_-pi/2^pi/2frac28cos 2theta+5sin2thetafrac12(1+cos2theta)^2+frac34sin^2 2theta+fracm4(cos2theta-1)^2,mathrmdtheta
$$
So
$$
I=frac12int_-pi^pifrac28cosphi+5sinphifrac12(1+cosphi)^2+frac34sin^2 phi+fracm4(cosphi-1)^2,mathrmdphi
$$
which we can rewrite as a contour integral of a rational function over the unit circle $z=e^iphi$, $-pileqphileqpi$ in $mathbbC$, hence it is just a matter of computing residues.
So
$$
I=frac12int_mathbbTfrac28cdotfrac12(z+z^-1)+5cdotfrac12i(z-z^-1)frac12(1+frac12(z+z^-1))^2+frac34(-frac14(z-z^-1)^2)+fracm4(frac12(z+z^-1)-1)^2,fracmathrmdziz
$$
which simplifies to
$$
I=frac12iint_mathbbT
frac8 ((28 + 5 i) + (28 - 5 i) z^2),mathrmdz(m-1)(z^4+1)-4(m-2)(z^3+z)+6(m-3)z^2
$$
The poles are at, if $mneq 1$,
$$
z+z^-1=frac2(m-2pmsqrtm)m-1.
$$
so, since $m>0$
$$labeleq:poles
z=
begincases
0,frac12(3pmsqrt5)& m=1\
frac2pmsqrtmpmsqrt3pm 2sqrtm1pmsqrtm&mneq 1.
endcasestag$star$
$$
So all poles are at worst simple unless $m=frac49$ (in which case we get a double pole at $z=4$ which we can ignore), and
$$
I=pisum_substacklvert z_irvert<1\z_iineqrefeq:polesoperatornameres_z_i(dots)+(textcorrection if lvert z_irvert=1).
$$
answered Jun 20 at 20:03
user10354138user10354138
19k3 gold badges13 silver badges34 bronze badges
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Can you give more details of this method?
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– user326159
Jun 20 at 20:30
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@user326159 It's quite a standard method which you can find in any textbook dealing with complex integration. The key point here is the so-called Residue Theorem which I suggest you to look up.
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– mrtaurho
Jun 21 at 7:59
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note that since the function part of the function is odd i.e:
$$f(x)=frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
$$f(-x)=frac28cos^2x-10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
you could notice that the integral can be simplified to:
$$int_-pi/2^pi/2frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=int_-pi/2^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=2int_0^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=56int_0^pi/2fraccos^2x-sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
One route you could try to take is Tangent half-angle substitution, which yields:
$$112int_0^1(1+t^2)frac(1-t^2)^2-(2t)^22(1-t^2)^4+3(1-t^2)^2(2t)^2+m(2t)^4dt$$
the bottom of this fraction can be expanded to:
$$2t^8+4t^6+16t^4m-12t^4+4t^2+2$$
this may be factorisable for certain values of $m$
answered Jun 20 at 20:04
Henry LeeHenry Lee
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$begingroup$
Define the function $mathcalI:mathbbR_>0rightarrowmathbbR$ via the trigonometric integral
$$beginalign
mathcalIleft(muright)
&:=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Let $muinmathbbR_>0$. Since the integral $mathcalI$ has a symmetric interval of integration, the integral of the odd component of the integrand vanishes identically:
$$beginalign
mathcalIleft(muright)
&=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_-fracpi2^0mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)-10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright);~~~smallleft[thetamapsto-thetaright]\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac56cos^2left(thetaright)-56sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Using the double-angle formulas for sine and cosine,
$$beginalign
sinleft(2thetaright)
&=2sinleft(thetaright)cosleft(thetaright),\
cosleft(2thetaright)
&=cos^2left(thetaright)-sin^2left(thetaright)\
&=2cos^2left(thetaright)-1\
&=1-2sin^2left(thetaright),\
endalign$$
we can rewrite the integral as
$$beginalign
mathcalIleft(muright)
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccosleft(2thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)8cos^4left(thetaright)+12cos^2left(thetaright)sin^2left(thetaright)+4musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)2left[1+cosleft(2thetaright)right]^2+3sin^2left(2thetaright)+muleft[1-cosleft(2thetaright)right]^2\
&=56int_0^pimathrmdtheta,frac2cosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2;~~~smallleft[thetamapstofrac12thetaright]\
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2.\
endalign$$
Using the tangent half-angle substitution, the trigonometric integral transforms as
$$beginalign
mathcalIleft(muright)
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2\
&=112int_0^inftymathrmdt,frac21+t^2cdotfracleft(frac1-t^21+t^2right)2left(1+frac1-t^21+t^2right)^2+3left(frac2t1+t^2right)^2+muleft(1-frac1-t^21+t^2right)^2;~~~smallleft[theta=2arctanleft(tright)right]\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+mu,t^4.\
endalign$$
Setting $sqrtfrac2mu=:ainmathbbR_>0$ and $frac34a=:binmathbbR_>0$, we then have
$$beginalign
mathcalIleft(muright)
&=mathcalIleft(frac2a^2right)\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+2a^-2t^4\
&=sqrtaint_0^inftymathrmdu,frac56left(1-au^2right)2+3au^2+2u^4;~~~smallleft[t=usqrtaright]\
&=frac13sqrtaint_0^inftymathrmdu,frac28left(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_1^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3u^2-4bright)1+2bu^2+u^4;~~~smallleft[umapstofrac1uright]\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bright)left(1+u^2right)1+2bu^2+u^4\
&=28left(1-frac43bright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-aright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-sqrtfrac2muright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=frac28left(mu-2right)left(sqrtmu+sqrt2right)sqrtmusqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4.\
endalign$$
Since the integral factor in the last line above is a positive quantity, we don't need to actually solve the integral to determine the sign of $mathcalI$. We have
$$operatornamesgnleft(mathcalIleft(muright)right)=operatornamesgnleft(mu-2right),$$
as you originally conjectured.
answered Jun 21 at 0:28
David HDavid H
22.3k2 gold badges49 silver badges97 bronze badges
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Define the function $mathcalI:mathbbR_>0rightarrowmathbbR$ via the trigonometric integral
$$beginalign
mathcalIleft(muright)
&:=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Let $muinmathbbR_>0$. Since the integral $mathcalI$ has a symmetric interval of integration, the integral of the odd component of the integrand vanishes identically:
$$beginalign
mathcalIleft(muright)
&=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_-fracpi2^0mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)-10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright);~~~smallleft[thetamapsto-thetaright]\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac56cos^2left(thetaright)-56sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Using the double-angle formulas for sine and cosine,
$$beginalign
sinleft(2thetaright)
&=2sinleft(thetaright)cosleft(thetaright),\
cosleft(2thetaright)
&=cos^2left(thetaright)-sin^2left(thetaright)\
&=2cos^2left(thetaright)-1\
&=1-2sin^2left(thetaright),\
endalign$$
we can rewrite the integral as
$$beginalign
mathcalIleft(muright)
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccosleft(2thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)8cos^4left(thetaright)+12cos^2left(thetaright)sin^2left(thetaright)+4musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)2left[1+cosleft(2thetaright)right]^2+3sin^2left(2thetaright)+muleft[1-cosleft(2thetaright)right]^2\
&=56int_0^pimathrmdtheta,frac2cosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2;~~~smallleft[thetamapstofrac12thetaright]\
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2.\
endalign$$
Using the tangent half-angle substitution, the trigonometric integral transforms as
$$beginalign
mathcalIleft(muright)
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2\
&=112int_0^inftymathrmdt,frac21+t^2cdotfracleft(frac1-t^21+t^2right)2left(1+frac1-t^21+t^2right)^2+3left(frac2t1+t^2right)^2+muleft(1-frac1-t^21+t^2right)^2;~~~smallleft[theta=2arctanleft(tright)right]\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+mu,t^4.\
endalign$$
Setting $sqrtfrac2mu=:ainmathbbR_>0$ and $frac34a=:binmathbbR_>0$, we then have
$$beginalign
mathcalIleft(muright)
&=mathcalIleft(frac2a^2right)\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+2a^-2t^4\
&=sqrtaint_0^inftymathrmdu,frac56left(1-au^2right)2+3au^2+2u^4;~~~smallleft[t=usqrtaright]\
&=frac13sqrtaint_0^inftymathrmdu,frac28left(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_1^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3u^2-4bright)1+2bu^2+u^4;~~~smallleft[umapstofrac1uright]\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bright)left(1+u^2right)1+2bu^2+u^4\
&=28left(1-frac43bright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-aright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-sqrtfrac2muright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=frac28left(mu-2right)left(sqrtmu+sqrt2right)sqrtmusqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4.\
endalign$$
Since the integral factor in the last line above is a positive quantity, we don't need to actually solve the integral to determine the sign of $mathcalI$. We have
$$operatornamesgnleft(mathcalIleft(muright)right)=operatornamesgnleft(mu-2right),$$
as you originally conjectured.
answered Jun 21 at 0:28
David HDavid H
22.3k2 gold badges49 silver badges97 bronze badges
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$begingroup$
Define the function $mathcalI:mathbbR_>0rightarrowmathbbR$ via the trigonometric integral
$$beginalign
mathcalIleft(muright)
&:=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Let $muinmathbbR_>0$. Since the integral $mathcalI$ has a symmetric interval of integration, the integral of the odd component of the integrand vanishes identically:
$$beginalign
mathcalIleft(muright)
&=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_-fracpi2^0mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)-10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright);~~~smallleft[thetamapsto-thetaright]\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac56cos^2left(thetaright)-56sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Using the double-angle formulas for sine and cosine,
$$beginalign
sinleft(2thetaright)
&=2sinleft(thetaright)cosleft(thetaright),\
cosleft(2thetaright)
&=cos^2left(thetaright)-sin^2left(thetaright)\
&=2cos^2left(thetaright)-1\
&=1-2sin^2left(thetaright),\
endalign$$
we can rewrite the integral as
$$beginalign
mathcalIleft(muright)
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccosleft(2thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)8cos^4left(thetaright)+12cos^2left(thetaright)sin^2left(thetaright)+4musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)2left[1+cosleft(2thetaright)right]^2+3sin^2left(2thetaright)+muleft[1-cosleft(2thetaright)right]^2\
&=56int_0^pimathrmdtheta,frac2cosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2;~~~smallleft[thetamapstofrac12thetaright]\
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2.\
endalign$$
Using the tangent half-angle substitution, the trigonometric integral transforms as
$$beginalign
mathcalIleft(muright)
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2\
&=112int_0^inftymathrmdt,frac21+t^2cdotfracleft(frac1-t^21+t^2right)2left(1+frac1-t^21+t^2right)^2+3left(frac2t1+t^2right)^2+muleft(1-frac1-t^21+t^2right)^2;~~~smallleft[theta=2arctanleft(tright)right]\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+mu,t^4.\
endalign$$
Setting $sqrtfrac2mu=:ainmathbbR_>0$ and $frac34a=:binmathbbR_>0$, we then have
$$beginalign
mathcalIleft(muright)
&=mathcalIleft(frac2a^2right)\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+2a^-2t^4\
&=sqrtaint_0^inftymathrmdu,frac56left(1-au^2right)2+3au^2+2u^4;~~~smallleft[t=usqrtaright]\
&=frac13sqrtaint_0^inftymathrmdu,frac28left(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_1^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3u^2-4bright)1+2bu^2+u^4;~~~smallleft[umapstofrac1uright]\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bright)left(1+u^2right)1+2bu^2+u^4\
&=28left(1-frac43bright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-aright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-sqrtfrac2muright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=frac28left(mu-2right)left(sqrtmu+sqrt2right)sqrtmusqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4.\
endalign$$
Since the integral factor in the last line above is a positive quantity, we don't need to actually solve the integral to determine the sign of $mathcalI$. We have
$$operatornamesgnleft(mathcalIleft(muright)right)=operatornamesgnleft(mu-2right),$$
as you originally conjectured.
answered Jun 21 at 0:28
David HDavid H
22.3k2 gold badges49 silver badges97 bronze badges
$endgroup$
Define the function $mathcalI:mathbbR_>0rightarrowmathbbR$ via the trigonometric integral
$$beginalign
mathcalIleft(muright)
&:=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Let $muinmathbbR_>0$. Since the integral $mathcalI$ has a symmetric interval of integration, the integral of the odd component of the integrand vanishes identically:
$$beginalign
mathcalIleft(muright)
&=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_-fracpi2^0mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)-10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright);~~~smallleft[thetamapsto-thetaright]\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac56cos^2left(thetaright)-56sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Using the double-angle formulas for sine and cosine,
$$beginalign
sinleft(2thetaright)
&=2sinleft(thetaright)cosleft(thetaright),\
cosleft(2thetaright)
&=cos^2left(thetaright)-sin^2left(thetaright)\
&=2cos^2left(thetaright)-1\
&=1-2sin^2left(thetaright),\
endalign$$
we can rewrite the integral as
$$beginalign
mathcalIleft(muright)
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccosleft(2thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)8cos^4left(thetaright)+12cos^2left(thetaright)sin^2left(thetaright)+4musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)2left[1+cosleft(2thetaright)right]^2+3sin^2left(2thetaright)+muleft[1-cosleft(2thetaright)right]^2\
&=56int_0^pimathrmdtheta,frac2cosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2;~~~smallleft[thetamapstofrac12thetaright]\
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2.\
endalign$$
Using the tangent half-angle substitution, the trigonometric integral transforms as
$$beginalign
mathcalIleft(muright)
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2\
&=112int_0^inftymathrmdt,frac21+t^2cdotfracleft(frac1-t^21+t^2right)2left(1+frac1-t^21+t^2right)^2+3left(frac2t1+t^2right)^2+muleft(1-frac1-t^21+t^2right)^2;~~~smallleft[theta=2arctanleft(tright)right]\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+mu,t^4.\
endalign$$
Setting $sqrtfrac2mu=:ainmathbbR_>0$ and $frac34a=:binmathbbR_>0$, we then have
$$beginalign
mathcalIleft(muright)
&=mathcalIleft(frac2a^2right)\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+2a^-2t^4\
&=sqrtaint_0^inftymathrmdu,frac56left(1-au^2right)2+3au^2+2u^4;~~~smallleft[t=usqrtaright]\
&=frac13sqrtaint_0^inftymathrmdu,frac28left(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_1^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3u^2-4bright)1+2bu^2+u^4;~~~smallleft[umapstofrac1uright]\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bright)left(1+u^2right)1+2bu^2+u^4\
&=28left(1-frac43bright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-aright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-sqrtfrac2muright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=frac28left(mu-2right)left(sqrtmu+sqrt2right)sqrtmusqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4.\
endalign$$
Since the integral factor in the last line above is a positive quantity, we don't need to actually solve the integral to determine the sign of $mathcalI$. We have
$$operatornamesgnleft(mathcalIleft(muright)right)=operatornamesgnleft(mu-2right),$$
as you originally conjectured.
answered Jun 21 at 0:28
David HDavid H
22.3k2 gold badges49 silver badges97 bronze badges
answered Jun 21 at 0:28
David HDavid H
22.3k2 gold badges49 silver badges97 bronze badges
answered Jun 21 at 0:28
David HDavid H
22.3k2 gold badges49 silver badges97 bronze badges
answered Jun 21 at 0:28
David HDavid H
22.3k2 gold badges49 silver badges97 bronze badges
22.3k2 gold badges49 silver badges97 bronze badges
add a comment |
add a comment |
$begingroup$
This problem is "nice" in the sense that the integrand is really trig function of $2theta$
$$
I=int_-pi/2^pi/2frac28cos 2theta+5sin2thetafrac12(1+cos2theta)^2+frac34sin^2 2theta+fracm4(cos2theta-1)^2,mathrmdtheta
$$
So
$$
I=frac12int_-pi^pifrac28cosphi+5sinphifrac12(1+cosphi)^2+frac34sin^2 phi+fracm4(cosphi-1)^2,mathrmdphi
$$
which we can rewrite as a contour integral of a rational function over the unit circle $z=e^iphi$, $-pileqphileqpi$ in $mathbbC$, hence it is just a matter of computing residues.
So
$$
I=frac12int_mathbbTfrac28cdotfrac12(z+z^-1)+5cdotfrac12i(z-z^-1)frac12(1+frac12(z+z^-1))^2+frac34(-frac14(z-z^-1)^2)+fracm4(frac12(z+z^-1)-1)^2,fracmathrmdziz
$$
which simplifies to
$$
I=frac12iint_mathbbT
frac8 ((28 + 5 i) + (28 - 5 i) z^2),mathrmdz(m-1)(z^4+1)-4(m-2)(z^3+z)+6(m-3)z^2
$$
The poles are at, if $mneq 1$,
$$
z+z^-1=frac2(m-2pmsqrtm)m-1.
$$
so, since $m>0$
$$labeleq:poles
z=
begincases
0,frac12(3pmsqrt5)& m=1\
frac2pmsqrtmpmsqrt3pm 2sqrtm1pmsqrtm&mneq 1.
endcasestag$star$
$$
So all poles are at worst simple unless $m=frac49$ (in which case we get a double pole at $z=4$ which we can ignore), and
$$
I=pisum_substacklvert z_irvert<1\z_iineqrefeq:polesoperatornameres_z_i(dots)+(textcorrection if lvert z_irvert=1).
$$
answered Jun 20 at 20:03
user10354138user10354138
19k3 gold badges13 silver badges34 bronze badges
$endgroup$
$begingroup$
Can you give more details of this method?
$endgroup$
– user326159
Jun 20 at 20:30
$begingroup$
@user326159 It's quite a standard method which you can find in any textbook dealing with complex integration. The key point here is the so-called Residue Theorem which I suggest you to look up.
$endgroup$
– mrtaurho
Jun 21 at 7:59
add a comment |
$begingroup$
This problem is "nice" in the sense that the integrand is really trig function of $2theta$
$$
I=int_-pi/2^pi/2frac28cos 2theta+5sin2thetafrac12(1+cos2theta)^2+frac34sin^2 2theta+fracm4(cos2theta-1)^2,mathrmdtheta
$$
So
$$
I=frac12int_-pi^pifrac28cosphi+5sinphifrac12(1+cosphi)^2+frac34sin^2 phi+fracm4(cosphi-1)^2,mathrmdphi
$$
which we can rewrite as a contour integral of a rational function over the unit circle $z=e^iphi$, $-pileqphileqpi$ in $mathbbC$, hence it is just a matter of computing residues.
So
$$
I=frac12int_mathbbTfrac28cdotfrac12(z+z^-1)+5cdotfrac12i(z-z^-1)frac12(1+frac12(z+z^-1))^2+frac34(-frac14(z-z^-1)^2)+fracm4(frac12(z+z^-1)-1)^2,fracmathrmdziz
$$
which simplifies to
$$
I=frac12iint_mathbbT
frac8 ((28 + 5 i) + (28 - 5 i) z^2),mathrmdz(m-1)(z^4+1)-4(m-2)(z^3+z)+6(m-3)z^2
$$
The poles are at, if $mneq 1$,
$$
z+z^-1=frac2(m-2pmsqrtm)m-1.
$$
so, since $m>0$
$$labeleq:poles
z=
begincases
0,frac12(3pmsqrt5)& m=1\
frac2pmsqrtmpmsqrt3pm 2sqrtm1pmsqrtm&mneq 1.
endcasestag$star$
$$
So all poles are at worst simple unless $m=frac49$ (in which case we get a double pole at $z=4$ which we can ignore), and
$$
I=pisum_substacklvert z_irvert<1\z_iineqrefeq:polesoperatornameres_z_i(dots)+(textcorrection if lvert z_irvert=1).
$$
answered Jun 20 at 20:03
user10354138user10354138
19k3 gold badges13 silver badges34 bronze badges
$endgroup$
$begingroup$
Can you give more details of this method?
$endgroup$
– user326159
Jun 20 at 20:30
$begingroup$
@user326159 It's quite a standard method which you can find in any textbook dealing with complex integration. The key point here is the so-called Residue Theorem which I suggest you to look up.
$endgroup$
– mrtaurho
Jun 21 at 7:59
add a comment |
$begingroup$
This problem is "nice" in the sense that the integrand is really trig function of $2theta$
$$
I=int_-pi/2^pi/2frac28cos 2theta+5sin2thetafrac12(1+cos2theta)^2+frac34sin^2 2theta+fracm4(cos2theta-1)^2,mathrmdtheta
$$
So
$$
I=frac12int_-pi^pifrac28cosphi+5sinphifrac12(1+cosphi)^2+frac34sin^2 phi+fracm4(cosphi-1)^2,mathrmdphi
$$
which we can rewrite as a contour integral of a rational function over the unit circle $z=e^iphi$, $-pileqphileqpi$ in $mathbbC$, hence it is just a matter of computing residues.
So
$$
I=frac12int_mathbbTfrac28cdotfrac12(z+z^-1)+5cdotfrac12i(z-z^-1)frac12(1+frac12(z+z^-1))^2+frac34(-frac14(z-z^-1)^2)+fracm4(frac12(z+z^-1)-1)^2,fracmathrmdziz
$$
which simplifies to
$$
I=frac12iint_mathbbT
frac8 ((28 + 5 i) + (28 - 5 i) z^2),mathrmdz(m-1)(z^4+1)-4(m-2)(z^3+z)+6(m-3)z^2
$$
The poles are at, if $mneq 1$,
$$
z+z^-1=frac2(m-2pmsqrtm)m-1.
$$
so, since $m>0$
$$labeleq:poles
z=
begincases
0,frac12(3pmsqrt5)& m=1\
frac2pmsqrtmpmsqrt3pm 2sqrtm1pmsqrtm&mneq 1.
endcasestag$star$
$$
So all poles are at worst simple unless $m=frac49$ (in which case we get a double pole at $z=4$ which we can ignore), and
$$
I=pisum_substacklvert z_irvert<1\z_iineqrefeq:polesoperatornameres_z_i(dots)+(textcorrection if lvert z_irvert=1).
$$
answered Jun 20 at 20:03
user10354138user10354138
19k3 gold badges13 silver badges34 bronze badges
$endgroup$
This problem is "nice" in the sense that the integrand is really trig function of $2theta$
$$
I=int_-pi/2^pi/2frac28cos 2theta+5sin2thetafrac12(1+cos2theta)^2+frac34sin^2 2theta+fracm4(cos2theta-1)^2,mathrmdtheta
$$
So
$$
I=frac12int_-pi^pifrac28cosphi+5sinphifrac12(1+cosphi)^2+frac34sin^2 phi+fracm4(cosphi-1)^2,mathrmdphi
$$
which we can rewrite as a contour integral of a rational function over the unit circle $z=e^iphi$, $-pileqphileqpi$ in $mathbbC$, hence it is just a matter of computing residues.
So
$$
I=frac12int_mathbbTfrac28cdotfrac12(z+z^-1)+5cdotfrac12i(z-z^-1)frac12(1+frac12(z+z^-1))^2+frac34(-frac14(z-z^-1)^2)+fracm4(frac12(z+z^-1)-1)^2,fracmathrmdziz
$$
which simplifies to
$$
I=frac12iint_mathbbT
frac8 ((28 + 5 i) + (28 - 5 i) z^2),mathrmdz(m-1)(z^4+1)-4(m-2)(z^3+z)+6(m-3)z^2
$$
The poles are at, if $mneq 1$,
$$
z+z^-1=frac2(m-2pmsqrtm)m-1.
$$
so, since $m>0$
$$labeleq:poles
z=
begincases
0,frac12(3pmsqrt5)& m=1\
frac2pmsqrtmpmsqrt3pm 2sqrtm1pmsqrtm&mneq 1.
endcasestag$star$
$$
So all poles are at worst simple unless $m=frac49$ (in which case we get a double pole at $z=4$ which we can ignore), and
$$
I=pisum_substacklvert z_irvert<1\z_iineqrefeq:polesoperatornameres_z_i(dots)+(textcorrection if lvert z_irvert=1).
$$
answered Jun 20 at 20:03
user10354138user10354138
19k3 gold badges13 silver badges34 bronze badges
edited Jun 20 at 20:46
answered Jun 20 at 20:03
user10354138user10354138
19k3 gold badges13 silver badges34 bronze badges
answered Jun 20 at 20:03
user10354138user10354138
19k3 gold badges13 silver badges34 bronze badges
answered Jun 20 at 20:03
user10354138user10354138
19k3 gold badges13 silver badges34 bronze badges
19k3 gold badges13 silver badges34 bronze badges
$begingroup$
Can you give more details of this method?
$endgroup$
– user326159
Jun 20 at 20:30
$begingroup$
@user326159 It's quite a standard method which you can find in any textbook dealing with complex integration. The key point here is the so-called Residue Theorem which I suggest you to look up.
$endgroup$
– mrtaurho
Jun 21 at 7:59
add a comment |
$begingroup$
Can you give more details of this method?
$endgroup$
– user326159
Jun 20 at 20:30
$begingroup$
@user326159 It's quite a standard method which you can find in any textbook dealing with complex integration. The key point here is the so-called Residue Theorem which I suggest you to look up.
$endgroup$
– mrtaurho
Jun 21 at 7:59
$begingroup$
Can you give more details of this method?
$endgroup$
– user326159
Jun 20 at 20:30
$begingroup$
Can you give more details of this method?
$endgroup$
– user326159
Jun 20 at 20:30
$begingroup$
@user326159 It's quite a standard method which you can find in any textbook dealing with complex integration. The key point here is the so-called Residue Theorem which I suggest you to look up.
$endgroup$
– mrtaurho
Jun 21 at 7:59
$begingroup$
@user326159 It's quite a standard method which you can find in any textbook dealing with complex integration. The key point here is the so-called Residue Theorem which I suggest you to look up.
$endgroup$
– mrtaurho
Jun 21 at 7:59
add a comment |
$begingroup$
note that since the function part of the function is odd i.e:
$$f(x)=frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
$$f(-x)=frac28cos^2x-10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
you could notice that the integral can be simplified to:
$$int_-pi/2^pi/2frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=int_-pi/2^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=2int_0^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=56int_0^pi/2fraccos^2x-sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
One route you could try to take is Tangent half-angle substitution, which yields:
$$112int_0^1(1+t^2)frac(1-t^2)^2-(2t)^22(1-t^2)^4+3(1-t^2)^2(2t)^2+m(2t)^4dt$$
the bottom of this fraction can be expanded to:
$$2t^8+4t^6+16t^4m-12t^4+4t^2+2$$
this may be factorisable for certain values of $m$
answered Jun 20 at 20:04
Henry LeeHenry Lee
2,9211 gold badge4 silver badges19 bronze badges
$endgroup$
add a comment |
$begingroup$
note that since the function part of the function is odd i.e:
$$f(x)=frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
$$f(-x)=frac28cos^2x-10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
you could notice that the integral can be simplified to:
$$int_-pi/2^pi/2frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=int_-pi/2^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=2int_0^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=56int_0^pi/2fraccos^2x-sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
One route you could try to take is Tangent half-angle substitution, which yields:
$$112int_0^1(1+t^2)frac(1-t^2)^2-(2t)^22(1-t^2)^4+3(1-t^2)^2(2t)^2+m(2t)^4dt$$
the bottom of this fraction can be expanded to:
$$2t^8+4t^6+16t^4m-12t^4+4t^2+2$$
this may be factorisable for certain values of $m$
answered Jun 20 at 20:04
Henry LeeHenry Lee
2,9211 gold badge4 silver badges19 bronze badges
$endgroup$
add a comment |
$begingroup$
note that since the function part of the function is odd i.e:
$$f(x)=frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
$$f(-x)=frac28cos^2x-10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
you could notice that the integral can be simplified to:
$$int_-pi/2^pi/2frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=int_-pi/2^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=2int_0^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=56int_0^pi/2fraccos^2x-sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
One route you could try to take is Tangent half-angle substitution, which yields:
$$112int_0^1(1+t^2)frac(1-t^2)^2-(2t)^22(1-t^2)^4+3(1-t^2)^2(2t)^2+m(2t)^4dt$$
the bottom of this fraction can be expanded to:
$$2t^8+4t^6+16t^4m-12t^4+4t^2+2$$
this may be factorisable for certain values of $m$
answered Jun 20 at 20:04
Henry LeeHenry Lee
2,9211 gold badge4 silver badges19 bronze badges
$endgroup$
note that since the function part of the function is odd i.e:
$$f(x)=frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
$$f(-x)=frac28cos^2x-10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
you could notice that the integral can be simplified to:
$$int_-pi/2^pi/2frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=int_-pi/2^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=2int_0^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=56int_0^pi/2fraccos^2x-sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
One route you could try to take is Tangent half-angle substitution, which yields:
$$112int_0^1(1+t^2)frac(1-t^2)^2-(2t)^22(1-t^2)^4+3(1-t^2)^2(2t)^2+m(2t)^4dt$$
the bottom of this fraction can be expanded to:
$$2t^8+4t^6+16t^4m-12t^4+4t^2+2$$
this may be factorisable for certain values of $m$
answered Jun 20 at 20:04
Henry LeeHenry Lee
2,9211 gold badge4 silver badges19 bronze badges
edited Jun 20 at 20:16
answered Jun 20 at 20:04
Henry LeeHenry Lee
2,9211 gold badge4 silver badges19 bronze badges
answered Jun 20 at 20:04
Henry LeeHenry Lee
2,9211 gold badge4 silver badges19 bronze badges
answered Jun 20 at 20:04
Henry LeeHenry Lee
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$begingroup$
Surely it is $8(m-2)sin^4theta$ not $(m-2)sin^4theta$ in the denominator of the last equation?
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– user10354138
Jun 20 at 19:57
$begingroup$
Yes. I've corrected it
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– user326159
Jun 20 at 20:04
3
$begingroup$
Writing $alpha^2 = m/2$, we get $$ I(alpha) = 14 pi left( 1-frac1alpharight)sqrtfrac24alpha+3. $$ From this, we easily deduce the sign of $I$ as function of $alpha$, i.e., $operatornamesign(I(alpha)) = operatornamesign(alpha - 1)$. Now if you ask me how I arrived this expression, it is basically from an ugly residue computation applied to $$I(alpha) = int_-infty^infty frac28(1-t^2)2+3t^2+2alpha^2 t^4, mathrmdt, $$ but I have not enough energy to write up all the intermediate steps...
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– Sangchul Lee
Jun 20 at 22:27
$begingroup$
@SangchulLee Thank you for your answer. It's amazing that you find a closed form for this integral. I've tried to compute it by the residue theorem but I cannot get this formula. Could you be more explicit on how did you do it? To know how to calculate this integral is very important for me, and I've been working many hours on it without success...
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– user326159
Jun 24 at 15:08