I am getting undefined as the answer of this integral problem $intlimits_2^3fracmathrm dn(n-2)(3-n)$. Am I doing something wrong? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Improper integral $int^pi/2_0 (operatornamecsc x - frac1x),mathrm dx$an intriguing integral $I=intlimits_0^4 fracdx4+2^x $Computing the integral $intlimits_-infty^infty(t^2-1)delta(t):dt$Help to understand this property: $intlimits_ka^kbsleft(fracxkright)dx = kintlimits_a^bs(x)dx$Prove $intlimits_0^infty mathrmexp(-ax^2-fracbx^2) mathrmd x = frac12sqrtfracpia}mathrme^{-2sqrtab$Computing the integral $int fracdxxsqrt x^2+1$Why am I not getting the right answer for this integral?Show that $intlimits_0^fracpi24cos^2(x)log^2(cos x)~mathrm dx=-pilog 2+pilog^2 2-fracpi2+fracpi^312$Calculating improper integral $int limits_0^inftyfracmathrme^-xsqrtx,mathrmdx$Finding the value of $limlimits_nrightarrow inftysqrtnint^fracpi4_0cos^2n-2(x)mathrm dx$
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I am getting undefined as the answer of this integral problem $intlimits_2^3fracmathrm dn(n-2)(3-n)$. Am I doing something wrong?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Improper integral $int^pi/2_0 (operatornamecsc x - frac1x),mathrm dx$an intriguing integral $I=intlimits_0^4 fracdx4+2^x $Computing the integral $intlimits_-infty^infty(t^2-1)delta(t):dt$Help to understand this property: $intlimits_ka^kbsleft(fracxkright)dx = kintlimits_a^bs(x)dx$Prove $intlimits_0^infty mathrmexp(-ax^2-fracbx^2) mathrmd x = frac12sqrtfracpiamathrme^-2sqrtab$Computing the integral $int fracdxxsqrt x^2+1$Why am I not getting the right answer for this integral?Show that $intlimits_0^fracpi24cos^2(x)log^2(cos x)~mathrm dx=-pilog 2+pilog^2 2-fracpi2+fracpi^312$Calculating improper integral $int limits_0^inftyfracmathrme^-xsqrtx,mathrmdx$Finding the value of $limlimits_nrightarrow inftysqrtnint^fracpi4_0cos^2n-2(x)mathrm dx$
$begingroup$
Find $$intlimits_2^3fracmathrm dn(n-2)(3-n)$$
My Attempt:
Let $$beginalignfrac1(n-2)(3-n)&=fracAn-2+fracB3-n \ &= fracA(3-n)+B(n-2)(n-2)(3-n)\ Rightarrow 1 &= A(3-n)+B(n-2) \ &= 3A - An+Bn-2B \ &= n(B-A)+3A-2B.endalign$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$beginaligntherefore intlimits_2^3fracmathrm dn(n-2)(3-n) &= intlimits_2^3bigg(frac1n-2+frac13-nbigg),mathrm dn \ &= intlimits_2^3frac1n-2,mathrm dn+intlimits_2^3frac13-n,mathrm dn \ &=left[log(n-2)right]_small 2^small 3+left[log(3-n)right]_small 2^small 3 \ &= left[log(3-2)-log(2-2)right]+left[log(3-3)+log(3-2)right] \ &= left[log 1-log 0right]+left[log 0 - log 1right]endalign$$ but $log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?
Thank you in advance.
integration definite-integrals
edited yesterday
YuiTo Cheng
2,43841037
asked yesterday
arandomguyarandomguy
17418
$endgroup$
add a comment |
$begingroup$
Find $$intlimits_2^3fracmathrm dn(n-2)(3-n)$$
My Attempt:
Let $$beginalignfrac1(n-2)(3-n)&=fracAn-2+fracB3-n \ &= fracA(3-n)+B(n-2)(n-2)(3-n)\ Rightarrow 1 &= A(3-n)+B(n-2) \ &= 3A - An+Bn-2B \ &= n(B-A)+3A-2B.endalign$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$beginaligntherefore intlimits_2^3fracmathrm dn(n-2)(3-n) &= intlimits_2^3bigg(frac1n-2+frac13-nbigg),mathrm dn \ &= intlimits_2^3frac1n-2,mathrm dn+intlimits_2^3frac13-n,mathrm dn \ &=left[log(n-2)right]_small 2^small 3+left[log(3-n)right]_small 2^small 3 \ &= left[log(3-2)-log(2-2)right]+left[log(3-3)+log(3-2)right] \ &= left[log 1-log 0right]+left[log 0 - log 1right]endalign$$ but $log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?
Thank you in advance.
integration definite-integrals
edited yesterday
YuiTo Cheng
2,43841037
asked yesterday
arandomguyarandomguy
17418
$endgroup$
$begingroup$
Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
$endgroup$
– Dbchatto67
yesterday
add a comment |
$begingroup$
Find $$intlimits_2^3fracmathrm dn(n-2)(3-n)$$
My Attempt:
Let $$beginalignfrac1(n-2)(3-n)&=fracAn-2+fracB3-n \ &= fracA(3-n)+B(n-2)(n-2)(3-n)\ Rightarrow 1 &= A(3-n)+B(n-2) \ &= 3A - An+Bn-2B \ &= n(B-A)+3A-2B.endalign$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$beginaligntherefore intlimits_2^3fracmathrm dn(n-2)(3-n) &= intlimits_2^3bigg(frac1n-2+frac13-nbigg),mathrm dn \ &= intlimits_2^3frac1n-2,mathrm dn+intlimits_2^3frac13-n,mathrm dn \ &=left[log(n-2)right]_small 2^small 3+left[log(3-n)right]_small 2^small 3 \ &= left[log(3-2)-log(2-2)right]+left[log(3-3)+log(3-2)right] \ &= left[log 1-log 0right]+left[log 0 - log 1right]endalign$$ but $log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?
Thank you in advance.
integration definite-integrals
edited yesterday
YuiTo Cheng
2,43841037
asked yesterday
arandomguyarandomguy
17418
$endgroup$
Find $$intlimits_2^3fracmathrm dn(n-2)(3-n)$$
My Attempt:
Let $$beginalignfrac1(n-2)(3-n)&=fracAn-2+fracB3-n \ &= fracA(3-n)+B(n-2)(n-2)(3-n)\ Rightarrow 1 &= A(3-n)+B(n-2) \ &= 3A - An+Bn-2B \ &= n(B-A)+3A-2B.endalign$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$beginaligntherefore intlimits_2^3fracmathrm dn(n-2)(3-n) &= intlimits_2^3bigg(frac1n-2+frac13-nbigg),mathrm dn \ &= intlimits_2^3frac1n-2,mathrm dn+intlimits_2^3frac13-n,mathrm dn \ &=left[log(n-2)right]_small 2^small 3+left[log(3-n)right]_small 2^small 3 \ &= left[log(3-2)-log(2-2)right]+left[log(3-3)+log(3-2)right] \ &= left[log 1-log 0right]+left[log 0 - log 1right]endalign$$ but $log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?
Thank you in advance.
integration definite-integrals
integration definite-integrals
edited yesterday
YuiTo Cheng
2,43841037
asked yesterday
arandomguyarandomguy
17418
edited yesterday
YuiTo Cheng
2,43841037
asked yesterday
arandomguyarandomguy
17418
edited yesterday
YuiTo Cheng
2,43841037
edited yesterday
YuiTo Cheng
2,43841037
edited yesterday
YuiTo Cheng
2,43841037
2,43841037
asked yesterday
arandomguyarandomguy
17418
asked yesterday
arandomguyarandomguy
17418
asked yesterday
arandomguyarandomguy
17418
17418
$begingroup$
Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
$endgroup$
– Dbchatto67
yesterday
add a comment |
$begingroup$
Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
$endgroup$
– Dbchatto67
yesterday
$begingroup$
Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
$endgroup$
– Dbchatto67
yesterday
$begingroup$
Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
$endgroup$
– Dbchatto67
yesterday
add a comment |
2 Answers
2
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$begingroup$
No.
There are two singularities at $n = 2$ and $n = 3$, as can be determined by inspecting the denominator, and they are quite strong. The integral does not converge.
answered yesterday
The_SympathizerThe_Sympathizer
7,8852246
$endgroup$
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
yesterday
$begingroup$
@MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
$endgroup$
– The_Sympathizer
yesterday
$begingroup$
More quantitatively, a singular term $x^-alpha$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
$endgroup$
– The_Sympathizer
yesterday
add a comment |
$begingroup$
No, there is nothing wrong. You havebeginalignint_2^3frac1(x-2)(x-3),mathrm dx&=int_2^5/2frac1(x-2)(x-3),mathrm dx+int_5/2^3frac1(x-2)(x-3),mathrm dx\&=lim_tto2^+int_t^5/2frac1(x-2)(x-3),mathrm dx+lim_tto3^-int_5/2^tfrac1(x-2)(x-3),mathrm dx.endalignNone of these limits exist, since the function that's being integrated behaves as $frac1x-3$ near $3$ and as $frac1x-2$ near $2$.
answered yesterday
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
No.
There are two singularities at $n = 2$ and $n = 3$, as can be determined by inspecting the denominator, and they are quite strong. The integral does not converge.
answered yesterday
The_SympathizerThe_Sympathizer
7,8852246
$endgroup$
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
yesterday
$begingroup$
@MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
$endgroup$
– The_Sympathizer
yesterday
$begingroup$
More quantitatively, a singular term $x^-alpha$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
$endgroup$
– The_Sympathizer
yesterday
add a comment |
$begingroup$
No.
There are two singularities at $n = 2$ and $n = 3$, as can be determined by inspecting the denominator, and they are quite strong. The integral does not converge.
answered yesterday
The_SympathizerThe_Sympathizer
7,8852246
$endgroup$
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
yesterday
$begingroup$
@MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
$endgroup$
– The_Sympathizer
yesterday
$begingroup$
More quantitatively, a singular term $x^-alpha$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
$endgroup$
– The_Sympathizer
yesterday
add a comment |
$begingroup$
No.
There are two singularities at $n = 2$ and $n = 3$, as can be determined by inspecting the denominator, and they are quite strong. The integral does not converge.
answered yesterday
The_SympathizerThe_Sympathizer
7,8852246
$endgroup$
No.
There are two singularities at $n = 2$ and $n = 3$, as can be determined by inspecting the denominator, and they are quite strong. The integral does not converge.
answered yesterday
The_SympathizerThe_Sympathizer
7,8852246
answered yesterday
The_SympathizerThe_Sympathizer
7,8852246
answered yesterday
The_SympathizerThe_Sympathizer
7,8852246
answered yesterday
The_SympathizerThe_Sympathizer
7,8852246
7,8852246
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
yesterday
$begingroup$
@MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
$endgroup$
– The_Sympathizer
yesterday
$begingroup$
More quantitatively, a singular term $x^-alpha$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
$endgroup$
– The_Sympathizer
yesterday
add a comment |
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
yesterday
$begingroup$
@MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
$endgroup$
– The_Sympathizer
yesterday
$begingroup$
More quantitatively, a singular term $x^-alpha$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
$endgroup$
– The_Sympathizer
yesterday
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
yesterday
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
yesterday
$begingroup$
@MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
$endgroup$
– The_Sympathizer
yesterday
$begingroup$
@MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
$endgroup$
– The_Sympathizer
yesterday
$begingroup$
More quantitatively, a singular term $x^-alpha$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
$endgroup$
– The_Sympathizer
yesterday
$begingroup$
More quantitatively, a singular term $x^-alpha$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
$endgroup$
– The_Sympathizer
yesterday
add a comment |
$begingroup$
No, there is nothing wrong. You havebeginalignint_2^3frac1(x-2)(x-3),mathrm dx&=int_2^5/2frac1(x-2)(x-3),mathrm dx+int_5/2^3frac1(x-2)(x-3),mathrm dx\&=lim_tto2^+int_t^5/2frac1(x-2)(x-3),mathrm dx+lim_tto3^-int_5/2^tfrac1(x-2)(x-3),mathrm dx.endalignNone of these limits exist, since the function that's being integrated behaves as $frac1x-3$ near $3$ and as $frac1x-2$ near $2$.
answered yesterday
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
add a comment |
$begingroup$
No, there is nothing wrong. You havebeginalignint_2^3frac1(x-2)(x-3),mathrm dx&=int_2^5/2frac1(x-2)(x-3),mathrm dx+int_5/2^3frac1(x-2)(x-3),mathrm dx\&=lim_tto2^+int_t^5/2frac1(x-2)(x-3),mathrm dx+lim_tto3^-int_5/2^tfrac1(x-2)(x-3),mathrm dx.endalignNone of these limits exist, since the function that's being integrated behaves as $frac1x-3$ near $3$ and as $frac1x-2$ near $2$.
answered yesterday
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
add a comment |
$begingroup$
No, there is nothing wrong. You havebeginalignint_2^3frac1(x-2)(x-3),mathrm dx&=int_2^5/2frac1(x-2)(x-3),mathrm dx+int_5/2^3frac1(x-2)(x-3),mathrm dx\&=lim_tto2^+int_t^5/2frac1(x-2)(x-3),mathrm dx+lim_tto3^-int_5/2^tfrac1(x-2)(x-3),mathrm dx.endalignNone of these limits exist, since the function that's being integrated behaves as $frac1x-3$ near $3$ and as $frac1x-2$ near $2$.
answered yesterday
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
No, there is nothing wrong. You havebeginalignint_2^3frac1(x-2)(x-3),mathrm dx&=int_2^5/2frac1(x-2)(x-3),mathrm dx+int_5/2^3frac1(x-2)(x-3),mathrm dx\&=lim_tto2^+int_t^5/2frac1(x-2)(x-3),mathrm dx+lim_tto3^-int_5/2^tfrac1(x-2)(x-3),mathrm dx.endalignNone of these limits exist, since the function that's being integrated behaves as $frac1x-3$ near $3$ and as $frac1x-2$ near $2$.
answered yesterday
José Carlos SantosJosé Carlos Santos
175k24134243
answered yesterday
José Carlos SantosJosé Carlos Santos
175k24134243
answered yesterday
José Carlos SantosJosé Carlos Santos
175k24134243
answered yesterday
José Carlos SantosJosé Carlos Santos
175k24134243
175k24134243
add a comment |
add a comment |
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$begingroup$
Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
$endgroup$
– Dbchatto67
yesterday