Evaluate the indefinite integral of multiplication of two functionsThe limit of the derivative of an increasing and bounded function is always $0$?Cauchy Criterion for Sequences as opposed to SeriesHow to show these two definitions of the Riemann integral are equivalent?Proof that $lima_n=L$ when $n$ goes to infinity, then $a_n$ its a Cauchy SequenceEvaluate the limit using only the following resultsUniform convergence of improper integralsProof of the Cauchy Criterion for SeriesProving the equivalence between two definitions of Reiman integral.Why this property of the integral doesn't work in this case?Let $gin C[a,b]$, $f_n$ is a sequence of continuous functions, and $f_nto f$ uniformly. Prove $lim_ntoinftyint_a^bf_ng = int_a^bfg$The two Riemann Stieltjes integrals are close in the tails
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Evaluate the indefinite integral of multiplication of two functions
The limit of the derivative of an increasing and bounded function is always $0$?Cauchy Criterion for Sequences as opposed to SeriesHow to show these two definitions of the Riemann integral are equivalent?Proof that $lima_n=L$ when $n$ goes to infinity, then $a_n$ its a Cauchy SequenceEvaluate the limit using only the following resultsUniform convergence of improper integralsProof of the Cauchy Criterion for SeriesProving the equivalence between two definitions of Reiman integral.Why this property of the integral doesn't work in this case?Let $gin C[a,b]$, $f_n$ is a sequence of continuous functions, and $f_nto f$ uniformly. Prove $lim_ntoinftyint_a^bf_ng = int_a^bfg$The two Riemann Stieltjes integrals are close in the tails
$begingroup$
Let $f,g:[a,+infty]tomathbbR$ be continuous and $K>0$ s.t. $$
left|int_c^df(x),dxright|leq K quad forall c,din [a,infty).
$$
If $gin C^1$ is decreasing with $displaystylelim_xto inftyg(x)=0,$ prove that the limit $$int_a^infty f(x)g(x),dx=lim_xtoinfty int_a^x f(t)g(t),dt$$ exists.
I tried to use the Cauchy criterion, that is $forall epsilon>0, exists A>a$ s.t. $A<c<d$ implies that $left|int_c^d f(t)g(t),dtright|<epsilon.$ However I could not get anywhere.
I appreciate any suggestions.
P.S. I already posted this question and as its expression was not completely right, I deleted it and asked it again here.
real-analysis integration riemann-integration
$endgroup$
add a comment |
$begingroup$
Let $f,g:[a,+infty]tomathbbR$ be continuous and $K>0$ s.t. $$
left|int_c^df(x),dxright|leq K quad forall c,din [a,infty).
$$
If $gin C^1$ is decreasing with $displaystylelim_xto inftyg(x)=0,$ prove that the limit $$int_a^infty f(x)g(x),dx=lim_xtoinfty int_a^x f(t)g(t),dt$$ exists.
I tried to use the Cauchy criterion, that is $forall epsilon>0, exists A>a$ s.t. $A<c<d$ implies that $left|int_c^d f(t)g(t),dtright|<epsilon.$ However I could not get anywhere.
I appreciate any suggestions.
P.S. I already posted this question and as its expression was not completely right, I deleted it and asked it again here.
real-analysis integration riemann-integration
$endgroup$
$begingroup$
Are you working in the extended real numbers? Otherwise, I'd say your notation $f, g:[a,+infty]tomathbbR$ is off. Wouldn't it be $f, g:[a,infty)tomathbbR?$
$endgroup$
– Adrian Keister
May 17 at 14:35
$begingroup$
yes ut is indeed extended numbers!
$endgroup$
– Majid
May 17 at 19:12
add a comment |
$begingroup$
Let $f,g:[a,+infty]tomathbbR$ be continuous and $K>0$ s.t. $$
left|int_c^df(x),dxright|leq K quad forall c,din [a,infty).
$$
If $gin C^1$ is decreasing with $displaystylelim_xto inftyg(x)=0,$ prove that the limit $$int_a^infty f(x)g(x),dx=lim_xtoinfty int_a^x f(t)g(t),dt$$ exists.
I tried to use the Cauchy criterion, that is $forall epsilon>0, exists A>a$ s.t. $A<c<d$ implies that $left|int_c^d f(t)g(t),dtright|<epsilon.$ However I could not get anywhere.
I appreciate any suggestions.
P.S. I already posted this question and as its expression was not completely right, I deleted it and asked it again here.
real-analysis integration riemann-integration
$endgroup$
Let $f,g:[a,+infty]tomathbbR$ be continuous and $K>0$ s.t. $$
left|int_c^df(x),dxright|leq K quad forall c,din [a,infty).
$$
If $gin C^1$ is decreasing with $displaystylelim_xto inftyg(x)=0,$ prove that the limit $$int_a^infty f(x)g(x),dx=lim_xtoinfty int_a^x f(t)g(t),dt$$ exists.
I tried to use the Cauchy criterion, that is $forall epsilon>0, exists A>a$ s.t. $A<c<d$ implies that $left|int_c^d f(t)g(t),dtright|<epsilon.$ However I could not get anywhere.
I appreciate any suggestions.
P.S. I already posted this question and as its expression was not completely right, I deleted it and asked it again here.
real-analysis integration riemann-integration
real-analysis integration riemann-integration
edited May 17 at 14:34
Adrian Keister
5,39072133
5,39072133
asked May 17 at 14:23
MajidMajid
2,0092926
2,0092926
$begingroup$
Are you working in the extended real numbers? Otherwise, I'd say your notation $f, g:[a,+infty]tomathbbR$ is off. Wouldn't it be $f, g:[a,infty)tomathbbR?$
$endgroup$
– Adrian Keister
May 17 at 14:35
$begingroup$
yes ut is indeed extended numbers!
$endgroup$
– Majid
May 17 at 19:12
add a comment |
$begingroup$
Are you working in the extended real numbers? Otherwise, I'd say your notation $f, g:[a,+infty]tomathbbR$ is off. Wouldn't it be $f, g:[a,infty)tomathbbR?$
$endgroup$
– Adrian Keister
May 17 at 14:35
$begingroup$
yes ut is indeed extended numbers!
$endgroup$
– Majid
May 17 at 19:12
$begingroup$
Are you working in the extended real numbers? Otherwise, I'd say your notation $f, g:[a,+infty]tomathbbR$ is off. Wouldn't it be $f, g:[a,infty)tomathbbR?$
$endgroup$
– Adrian Keister
May 17 at 14:35
$begingroup$
Are you working in the extended real numbers? Otherwise, I'd say your notation $f, g:[a,+infty]tomathbbR$ is off. Wouldn't it be $f, g:[a,infty)tomathbbR?$
$endgroup$
– Adrian Keister
May 17 at 14:35
$begingroup$
yes ut is indeed extended numbers!
$endgroup$
– Majid
May 17 at 19:12
$begingroup$
yes ut is indeed extended numbers!
$endgroup$
– Majid
May 17 at 19:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For $x in [a,infty)$, define $F(x) := int^x_a f(t),dt$. Then $lvert F(x) rvert le K$, and $F'(x) = f(x)$ by the fundamental theorem of calculus.
Now for $x,y in[a,infty)$, and wlog let $y ge x$. By integrating by parts and using the triangle inequality, we see beginalign*leftlvert int^y_a f(t) g(t) ,dt - int^x_a f(t)g(t),dtrightrvert &= left lvertint^y_x f(t) g(t), dt right rvert\
&= left lvert int^y_x F'(t)g(t) ,dt right rvert\
&= left lvert [F(t)g(t)]^t=y_t=x - int^y_x F(x)g'(x),dx right rvert\
&le lvert F(y)g(y)rvert + lvert F(x)g(x) rvert + int^y_x lvert F(t)g'(t) rvert ,dt\
& le Kleft(lvert g(y) rvert + lvert g(x) rvert + int^y_x lvert g'(t)rvert ,dt right).
endalign* Since $g$ is decreasing, we have $g'(t) le 0$, and so $lvert g'(t)rvert = -g'(t).$ Thus beginalign*leftlvert int^y_a f(t) g(t) ,dt - int^x_a f(t)g(t),dtrightrvert & le Kleft(lvert g(y) rvert + lvert g(x) rvert + int^y_x lvert g'(t)rvert ,dt right)\
&= Kleft(lvert g(y) rvert + lvert g(x) rvert - int^y_x g'(t) ,dt right)\
&= K(lvert g(y) rvert + vert g(x) rvert + g(x) - g(y)) \&le 4K textmax(lvert g(x) rvert, lvert g(y) rvert).
endalign* Since $lvert g(t) rvert to 0$ as $t to infty$, for any $epsilon > 0$, there is $M > 0$, such that $lvert g(t)rvert < epsilon/4K$ for all $t > M$. But then for $x,y > M$, we have shown $$leftlvert int^y_a f(t) g(t), dt - int^x_a f(t)g(t),dtrightrvert < epsilon.$$ Finally, take any sequence $x_n_n=1^infty$ with $x_n to infty$, and this shows that $$leftint^x_n_a f(t)g(t),dt right_n=1^infty$$ is a Cauchy sequence and hence converges. By the sequential criterion theorem, we conclude that $$lim_xtoinfty int^x_a f(t) g(t) ,dt$$ converges.
As an aside, this is completely analogous to the Dirichlet Test for infinite series, which is a generalization of the alternating series test.
$endgroup$
$begingroup$
thanks for the explicit solution :)
$endgroup$
– Majid
May 17 at 19:23
add a comment |
$begingroup$
Since it is given that $g$ is continuously differentiable, so $g'$ exists.
Now the idea is to write $displaystyleint fg$ as $displaystyle gint f-int left(g'int fright)$.
Now using triangle inequality,
$$left|int fgright|leqleft| gint fright|+left|int left(g'int fright)right|,$$
now I think you can complete it, but don't forget to use the fact that $displaystylelim_xtoinfty g'(x)=0.$
$endgroup$
1
$begingroup$
But we are not assuming that $g'(x) to 0$, and indeed, it may not.
$endgroup$
– User8128
May 17 at 14:58
1
$begingroup$
You are right,actually i thought that asymptotically g is becoming constant and hence the derivative must go to $0$,but that's not true.
$endgroup$
– Soumyadip Sarkar
May 17 at 15:04
3
$begingroup$
Exactly right: the prototypical counterexample is something like $g(x) = fracsin(x^10)1+x^2$
$endgroup$
– User8128
May 17 at 15:05
1
$begingroup$
@User8128 Note $g$ is decreasing. That still does not imply $g'to0$, but you will need another counterexample: small jumps with large $|g'|$ such that the sum of the jumps is bounded, for instance, i.e. something like this: math.stackexchange.com/questions/788813/…
$endgroup$
– Jean-Claude Arbaut
May 18 at 11:39
add a comment |
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2 Answers
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2 Answers
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$begingroup$
For $x in [a,infty)$, define $F(x) := int^x_a f(t),dt$. Then $lvert F(x) rvert le K$, and $F'(x) = f(x)$ by the fundamental theorem of calculus.
Now for $x,y in[a,infty)$, and wlog let $y ge x$. By integrating by parts and using the triangle inequality, we see beginalign*leftlvert int^y_a f(t) g(t) ,dt - int^x_a f(t)g(t),dtrightrvert &= left lvertint^y_x f(t) g(t), dt right rvert\
&= left lvert int^y_x F'(t)g(t) ,dt right rvert\
&= left lvert [F(t)g(t)]^t=y_t=x - int^y_x F(x)g'(x),dx right rvert\
&le lvert F(y)g(y)rvert + lvert F(x)g(x) rvert + int^y_x lvert F(t)g'(t) rvert ,dt\
& le Kleft(lvert g(y) rvert + lvert g(x) rvert + int^y_x lvert g'(t)rvert ,dt right).
endalign* Since $g$ is decreasing, we have $g'(t) le 0$, and so $lvert g'(t)rvert = -g'(t).$ Thus beginalign*leftlvert int^y_a f(t) g(t) ,dt - int^x_a f(t)g(t),dtrightrvert & le Kleft(lvert g(y) rvert + lvert g(x) rvert + int^y_x lvert g'(t)rvert ,dt right)\
&= Kleft(lvert g(y) rvert + lvert g(x) rvert - int^y_x g'(t) ,dt right)\
&= K(lvert g(y) rvert + vert g(x) rvert + g(x) - g(y)) \&le 4K textmax(lvert g(x) rvert, lvert g(y) rvert).
endalign* Since $lvert g(t) rvert to 0$ as $t to infty$, for any $epsilon > 0$, there is $M > 0$, such that $lvert g(t)rvert < epsilon/4K$ for all $t > M$. But then for $x,y > M$, we have shown $$leftlvert int^y_a f(t) g(t), dt - int^x_a f(t)g(t),dtrightrvert < epsilon.$$ Finally, take any sequence $x_n_n=1^infty$ with $x_n to infty$, and this shows that $$leftint^x_n_a f(t)g(t),dt right_n=1^infty$$ is a Cauchy sequence and hence converges. By the sequential criterion theorem, we conclude that $$lim_xtoinfty int^x_a f(t) g(t) ,dt$$ converges.
As an aside, this is completely analogous to the Dirichlet Test for infinite series, which is a generalization of the alternating series test.
$endgroup$
$begingroup$
thanks for the explicit solution :)
$endgroup$
– Majid
May 17 at 19:23
add a comment |
$begingroup$
For $x in [a,infty)$, define $F(x) := int^x_a f(t),dt$. Then $lvert F(x) rvert le K$, and $F'(x) = f(x)$ by the fundamental theorem of calculus.
Now for $x,y in[a,infty)$, and wlog let $y ge x$. By integrating by parts and using the triangle inequality, we see beginalign*leftlvert int^y_a f(t) g(t) ,dt - int^x_a f(t)g(t),dtrightrvert &= left lvertint^y_x f(t) g(t), dt right rvert\
&= left lvert int^y_x F'(t)g(t) ,dt right rvert\
&= left lvert [F(t)g(t)]^t=y_t=x - int^y_x F(x)g'(x),dx right rvert\
&le lvert F(y)g(y)rvert + lvert F(x)g(x) rvert + int^y_x lvert F(t)g'(t) rvert ,dt\
& le Kleft(lvert g(y) rvert + lvert g(x) rvert + int^y_x lvert g'(t)rvert ,dt right).
endalign* Since $g$ is decreasing, we have $g'(t) le 0$, and so $lvert g'(t)rvert = -g'(t).$ Thus beginalign*leftlvert int^y_a f(t) g(t) ,dt - int^x_a f(t)g(t),dtrightrvert & le Kleft(lvert g(y) rvert + lvert g(x) rvert + int^y_x lvert g'(t)rvert ,dt right)\
&= Kleft(lvert g(y) rvert + lvert g(x) rvert - int^y_x g'(t) ,dt right)\
&= K(lvert g(y) rvert + vert g(x) rvert + g(x) - g(y)) \&le 4K textmax(lvert g(x) rvert, lvert g(y) rvert).
endalign* Since $lvert g(t) rvert to 0$ as $t to infty$, for any $epsilon > 0$, there is $M > 0$, such that $lvert g(t)rvert < epsilon/4K$ for all $t > M$. But then for $x,y > M$, we have shown $$leftlvert int^y_a f(t) g(t), dt - int^x_a f(t)g(t),dtrightrvert < epsilon.$$ Finally, take any sequence $x_n_n=1^infty$ with $x_n to infty$, and this shows that $$leftint^x_n_a f(t)g(t),dt right_n=1^infty$$ is a Cauchy sequence and hence converges. By the sequential criterion theorem, we conclude that $$lim_xtoinfty int^x_a f(t) g(t) ,dt$$ converges.
As an aside, this is completely analogous to the Dirichlet Test for infinite series, which is a generalization of the alternating series test.
$endgroup$
$begingroup$
thanks for the explicit solution :)
$endgroup$
– Majid
May 17 at 19:23
add a comment |
$begingroup$
For $x in [a,infty)$, define $F(x) := int^x_a f(t),dt$. Then $lvert F(x) rvert le K$, and $F'(x) = f(x)$ by the fundamental theorem of calculus.
Now for $x,y in[a,infty)$, and wlog let $y ge x$. By integrating by parts and using the triangle inequality, we see beginalign*leftlvert int^y_a f(t) g(t) ,dt - int^x_a f(t)g(t),dtrightrvert &= left lvertint^y_x f(t) g(t), dt right rvert\
&= left lvert int^y_x F'(t)g(t) ,dt right rvert\
&= left lvert [F(t)g(t)]^t=y_t=x - int^y_x F(x)g'(x),dx right rvert\
&le lvert F(y)g(y)rvert + lvert F(x)g(x) rvert + int^y_x lvert F(t)g'(t) rvert ,dt\
& le Kleft(lvert g(y) rvert + lvert g(x) rvert + int^y_x lvert g'(t)rvert ,dt right).
endalign* Since $g$ is decreasing, we have $g'(t) le 0$, and so $lvert g'(t)rvert = -g'(t).$ Thus beginalign*leftlvert int^y_a f(t) g(t) ,dt - int^x_a f(t)g(t),dtrightrvert & le Kleft(lvert g(y) rvert + lvert g(x) rvert + int^y_x lvert g'(t)rvert ,dt right)\
&= Kleft(lvert g(y) rvert + lvert g(x) rvert - int^y_x g'(t) ,dt right)\
&= K(lvert g(y) rvert + vert g(x) rvert + g(x) - g(y)) \&le 4K textmax(lvert g(x) rvert, lvert g(y) rvert).
endalign* Since $lvert g(t) rvert to 0$ as $t to infty$, for any $epsilon > 0$, there is $M > 0$, such that $lvert g(t)rvert < epsilon/4K$ for all $t > M$. But then for $x,y > M$, we have shown $$leftlvert int^y_a f(t) g(t), dt - int^x_a f(t)g(t),dtrightrvert < epsilon.$$ Finally, take any sequence $x_n_n=1^infty$ with $x_n to infty$, and this shows that $$leftint^x_n_a f(t)g(t),dt right_n=1^infty$$ is a Cauchy sequence and hence converges. By the sequential criterion theorem, we conclude that $$lim_xtoinfty int^x_a f(t) g(t) ,dt$$ converges.
As an aside, this is completely analogous to the Dirichlet Test for infinite series, which is a generalization of the alternating series test.
$endgroup$
For $x in [a,infty)$, define $F(x) := int^x_a f(t),dt$. Then $lvert F(x) rvert le K$, and $F'(x) = f(x)$ by the fundamental theorem of calculus.
Now for $x,y in[a,infty)$, and wlog let $y ge x$. By integrating by parts and using the triangle inequality, we see beginalign*leftlvert int^y_a f(t) g(t) ,dt - int^x_a f(t)g(t),dtrightrvert &= left lvertint^y_x f(t) g(t), dt right rvert\
&= left lvert int^y_x F'(t)g(t) ,dt right rvert\
&= left lvert [F(t)g(t)]^t=y_t=x - int^y_x F(x)g'(x),dx right rvert\
&le lvert F(y)g(y)rvert + lvert F(x)g(x) rvert + int^y_x lvert F(t)g'(t) rvert ,dt\
& le Kleft(lvert g(y) rvert + lvert g(x) rvert + int^y_x lvert g'(t)rvert ,dt right).
endalign* Since $g$ is decreasing, we have $g'(t) le 0$, and so $lvert g'(t)rvert = -g'(t).$ Thus beginalign*leftlvert int^y_a f(t) g(t) ,dt - int^x_a f(t)g(t),dtrightrvert & le Kleft(lvert g(y) rvert + lvert g(x) rvert + int^y_x lvert g'(t)rvert ,dt right)\
&= Kleft(lvert g(y) rvert + lvert g(x) rvert - int^y_x g'(t) ,dt right)\
&= K(lvert g(y) rvert + vert g(x) rvert + g(x) - g(y)) \&le 4K textmax(lvert g(x) rvert, lvert g(y) rvert).
endalign* Since $lvert g(t) rvert to 0$ as $t to infty$, for any $epsilon > 0$, there is $M > 0$, such that $lvert g(t)rvert < epsilon/4K$ for all $t > M$. But then for $x,y > M$, we have shown $$leftlvert int^y_a f(t) g(t), dt - int^x_a f(t)g(t),dtrightrvert < epsilon.$$ Finally, take any sequence $x_n_n=1^infty$ with $x_n to infty$, and this shows that $$leftint^x_n_a f(t)g(t),dt right_n=1^infty$$ is a Cauchy sequence and hence converges. By the sequential criterion theorem, we conclude that $$lim_xtoinfty int^x_a f(t) g(t) ,dt$$ converges.
As an aside, this is completely analogous to the Dirichlet Test for infinite series, which is a generalization of the alternating series test.
edited May 17 at 20:20
Adrian Keister
5,39072133
5,39072133
answered May 17 at 14:56
User8128User8128
11.3k1622
11.3k1622
$begingroup$
thanks for the explicit solution :)
$endgroup$
– Majid
May 17 at 19:23
add a comment |
$begingroup$
thanks for the explicit solution :)
$endgroup$
– Majid
May 17 at 19:23
$begingroup$
thanks for the explicit solution :)
$endgroup$
– Majid
May 17 at 19:23
$begingroup$
thanks for the explicit solution :)
$endgroup$
– Majid
May 17 at 19:23
add a comment |
$begingroup$
Since it is given that $g$ is continuously differentiable, so $g'$ exists.
Now the idea is to write $displaystyleint fg$ as $displaystyle gint f-int left(g'int fright)$.
Now using triangle inequality,
$$left|int fgright|leqleft| gint fright|+left|int left(g'int fright)right|,$$
now I think you can complete it, but don't forget to use the fact that $displaystylelim_xtoinfty g'(x)=0.$
$endgroup$
1
$begingroup$
But we are not assuming that $g'(x) to 0$, and indeed, it may not.
$endgroup$
– User8128
May 17 at 14:58
1
$begingroup$
You are right,actually i thought that asymptotically g is becoming constant and hence the derivative must go to $0$,but that's not true.
$endgroup$
– Soumyadip Sarkar
May 17 at 15:04
3
$begingroup$
Exactly right: the prototypical counterexample is something like $g(x) = fracsin(x^10)1+x^2$
$endgroup$
– User8128
May 17 at 15:05
1
$begingroup$
@User8128 Note $g$ is decreasing. That still does not imply $g'to0$, but you will need another counterexample: small jumps with large $|g'|$ such that the sum of the jumps is bounded, for instance, i.e. something like this: math.stackexchange.com/questions/788813/…
$endgroup$
– Jean-Claude Arbaut
May 18 at 11:39
add a comment |
$begingroup$
Since it is given that $g$ is continuously differentiable, so $g'$ exists.
Now the idea is to write $displaystyleint fg$ as $displaystyle gint f-int left(g'int fright)$.
Now using triangle inequality,
$$left|int fgright|leqleft| gint fright|+left|int left(g'int fright)right|,$$
now I think you can complete it, but don't forget to use the fact that $displaystylelim_xtoinfty g'(x)=0.$
$endgroup$
1
$begingroup$
But we are not assuming that $g'(x) to 0$, and indeed, it may not.
$endgroup$
– User8128
May 17 at 14:58
1
$begingroup$
You are right,actually i thought that asymptotically g is becoming constant and hence the derivative must go to $0$,but that's not true.
$endgroup$
– Soumyadip Sarkar
May 17 at 15:04
3
$begingroup$
Exactly right: the prototypical counterexample is something like $g(x) = fracsin(x^10)1+x^2$
$endgroup$
– User8128
May 17 at 15:05
1
$begingroup$
@User8128 Note $g$ is decreasing. That still does not imply $g'to0$, but you will need another counterexample: small jumps with large $|g'|$ such that the sum of the jumps is bounded, for instance, i.e. something like this: math.stackexchange.com/questions/788813/…
$endgroup$
– Jean-Claude Arbaut
May 18 at 11:39
add a comment |
$begingroup$
Since it is given that $g$ is continuously differentiable, so $g'$ exists.
Now the idea is to write $displaystyleint fg$ as $displaystyle gint f-int left(g'int fright)$.
Now using triangle inequality,
$$left|int fgright|leqleft| gint fright|+left|int left(g'int fright)right|,$$
now I think you can complete it, but don't forget to use the fact that $displaystylelim_xtoinfty g'(x)=0.$
$endgroup$
Since it is given that $g$ is continuously differentiable, so $g'$ exists.
Now the idea is to write $displaystyleint fg$ as $displaystyle gint f-int left(g'int fright)$.
Now using triangle inequality,
$$left|int fgright|leqleft| gint fright|+left|int left(g'int fright)right|,$$
now I think you can complete it, but don't forget to use the fact that $displaystylelim_xtoinfty g'(x)=0.$
edited May 17 at 20:26
Adrian Keister
5,39072133
5,39072133
answered May 17 at 14:57
Soumyadip SarkarSoumyadip Sarkar
728
728
1
$begingroup$
But we are not assuming that $g'(x) to 0$, and indeed, it may not.
$endgroup$
– User8128
May 17 at 14:58
1
$begingroup$
You are right,actually i thought that asymptotically g is becoming constant and hence the derivative must go to $0$,but that's not true.
$endgroup$
– Soumyadip Sarkar
May 17 at 15:04
3
$begingroup$
Exactly right: the prototypical counterexample is something like $g(x) = fracsin(x^10)1+x^2$
$endgroup$
– User8128
May 17 at 15:05
1
$begingroup$
@User8128 Note $g$ is decreasing. That still does not imply $g'to0$, but you will need another counterexample: small jumps with large $|g'|$ such that the sum of the jumps is bounded, for instance, i.e. something like this: math.stackexchange.com/questions/788813/…
$endgroup$
– Jean-Claude Arbaut
May 18 at 11:39
add a comment |
1
$begingroup$
But we are not assuming that $g'(x) to 0$, and indeed, it may not.
$endgroup$
– User8128
May 17 at 14:58
1
$begingroup$
You are right,actually i thought that asymptotically g is becoming constant and hence the derivative must go to $0$,but that's not true.
$endgroup$
– Soumyadip Sarkar
May 17 at 15:04
3
$begingroup$
Exactly right: the prototypical counterexample is something like $g(x) = fracsin(x^10)1+x^2$
$endgroup$
– User8128
May 17 at 15:05
1
$begingroup$
@User8128 Note $g$ is decreasing. That still does not imply $g'to0$, but you will need another counterexample: small jumps with large $|g'|$ such that the sum of the jumps is bounded, for instance, i.e. something like this: math.stackexchange.com/questions/788813/…
$endgroup$
– Jean-Claude Arbaut
May 18 at 11:39
1
1
$begingroup$
But we are not assuming that $g'(x) to 0$, and indeed, it may not.
$endgroup$
– User8128
May 17 at 14:58
$begingroup$
But we are not assuming that $g'(x) to 0$, and indeed, it may not.
$endgroup$
– User8128
May 17 at 14:58
1
1
$begingroup$
You are right,actually i thought that asymptotically g is becoming constant and hence the derivative must go to $0$,but that's not true.
$endgroup$
– Soumyadip Sarkar
May 17 at 15:04
$begingroup$
You are right,actually i thought that asymptotically g is becoming constant and hence the derivative must go to $0$,but that's not true.
$endgroup$
– Soumyadip Sarkar
May 17 at 15:04
3
3
$begingroup$
Exactly right: the prototypical counterexample is something like $g(x) = fracsin(x^10)1+x^2$
$endgroup$
– User8128
May 17 at 15:05
$begingroup$
Exactly right: the prototypical counterexample is something like $g(x) = fracsin(x^10)1+x^2$
$endgroup$
– User8128
May 17 at 15:05
1
1
$begingroup$
@User8128 Note $g$ is decreasing. That still does not imply $g'to0$, but you will need another counterexample: small jumps with large $|g'|$ such that the sum of the jumps is bounded, for instance, i.e. something like this: math.stackexchange.com/questions/788813/…
$endgroup$
– Jean-Claude Arbaut
May 18 at 11:39
$begingroup$
@User8128 Note $g$ is decreasing. That still does not imply $g'to0$, but you will need another counterexample: small jumps with large $|g'|$ such that the sum of the jumps is bounded, for instance, i.e. something like this: math.stackexchange.com/questions/788813/…
$endgroup$
– Jean-Claude Arbaut
May 18 at 11:39
add a comment |
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$begingroup$
Are you working in the extended real numbers? Otherwise, I'd say your notation $f, g:[a,+infty]tomathbbR$ is off. Wouldn't it be $f, g:[a,infty)tomathbbR?$
$endgroup$
– Adrian Keister
May 17 at 14:35
$begingroup$
yes ut is indeed extended numbers!
$endgroup$
– Majid
May 17 at 19:12