Mixing
$int_{-infty}^{infty} frac{sin(x)}{x} dx $ with integration by parts?
$begingroup$
The integral of the Sinc function over $mathbb{R}$ is well-known, $$int_{-infty}^{infty} frac{sin(x)}{x} dx = pi$$But if I try to evaluate this using integration by parts, I get $$int_{-infty}^{infty} frac{sin(x)}{x} dx = left.-frac{cos(x)}{x} right|_{-infty}^{infty} - int_{-infty}^{infty} frac{cos(x)}{x^2} dx$$ The first part is $0$, and the second part diverges.
What's going on? Is integration by parts just not kosher here? If so, why not?
improper-integrals fake-proofs
asked Jan 28 '13 at 5:59
AndrewGAndrewG
1,6551033
$endgroup$
add a comment |
$begingroup$
The integral of the Sinc function over $mathbb{R}$ is well-known, $$int_{-infty}^{infty} frac{sin(x)}{x} dx = pi$$But if I try to evaluate this using integration by parts, I get $$int_{-infty}^{infty} frac{sin(x)}{x} dx = left.-frac{cos(x)}{x} right|_{-infty}^{infty} - int_{-infty}^{infty} frac{cos(x)}{x^2} dx$$ The first part is $0$, and the second part diverges.
What's going on? Is integration by parts just not kosher here? If so, why not?
improper-integrals fake-proofs
asked Jan 28 '13 at 5:59
AndrewGAndrewG
1,6551033
$endgroup$
add a comment |
$begingroup$
The integral of the Sinc function over $mathbb{R}$ is well-known, $$int_{-infty}^{infty} frac{sin(x)}{x} dx = pi$$But if I try to evaluate this using integration by parts, I get $$int_{-infty}^{infty} frac{sin(x)}{x} dx = left.-frac{cos(x)}{x} right|_{-infty}^{infty} - int_{-infty}^{infty} frac{cos(x)}{x^2} dx$$ The first part is $0$, and the second part diverges.
What's going on? Is integration by parts just not kosher here? If so, why not?
improper-integrals fake-proofs
asked Jan 28 '13 at 5:59
AndrewGAndrewG
1,6551033
$endgroup$
The integral of the Sinc function over $mathbb{R}$ is well-known, $$int_{-infty}^{infty} frac{sin(x)}{x} dx = pi$$But if I try to evaluate this using integration by parts, I get $$int_{-infty}^{infty} frac{sin(x)}{x} dx = left.-frac{cos(x)}{x} right|_{-infty}^{infty} - int_{-infty}^{infty} frac{cos(x)}{x^2} dx$$ The first part is $0$, and the second part diverges.
What's going on? Is integration by parts just not kosher here? If so, why not?
improper-integrals fake-proofs
improper-integrals fake-proofs
asked Jan 28 '13 at 5:59
AndrewGAndrewG
1,6551033
asked Jan 28 '13 at 5:59
AndrewGAndrewG
1,6551033
edited Jan 28 '13 at 6:08
AndrewG
asked Jan 28 '13 at 5:59
AndrewGAndrewG
1,6551033
asked Jan 28 '13 at 5:59
AndrewGAndrewG
1,6551033
asked Jan 28 '13 at 5:59
AndrewGAndrewG
1,6551033
1,6551033
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The definite integration by parts formula
$$int_a^b f(x) g(x) dx = F(x) g(x)bigg|_a^b - int_a^b F(x) g'(x) dx$$
is justified by the product rule for derivative and the fundamental theorem of calculus
$$int_a^b (F g)'(x) dx = F(x) g(x) bigg|_a^b $$
But the "fine print" there is that $F(x) g(x)$ is assumed to be continuously differentiable on the interval
$(a,b)$ and continuous on $[a,b]$. In this case with $F(x) = -cos(x)$ and $g(x) = 1/x$,
$F(x) g(x)$ is undefined at $x=0$, and no definition of it there will make it continuous at $x=0$, let alone differentiable. So you can't use the integration by
parts formula for this integral on any interval containing $0$.
answered Jan 28 '13 at 6:56
Robert IsraelRobert Israel
330k23219473
$endgroup$
add a comment |
$begingroup$
The issue is the discontinuity at $0$ in the cosine terms when you perform the partial integration, which is overlooked if you integrate over $mathbb{R}$. To get around this, consider:
$$int_{-infty}^{infty}frac{sin x}{x}dx=2int_0^{infty}frac{sin x}{x}dx$$
By parts:
$$int_{0}^{infty} frac{sin(x)}{x} dx = left.-frac{cos(x)}{x} right|_{0}^{infty} - int_{0}^{infty} frac{cos(x)}{x^2} dx$$
And here, both terms on the RHS diverge. No contradiction.
answered Jan 28 '13 at 6:22
L. F.L. F.
7,67032045
$endgroup$
$begingroup$
There is a discontinuity in $cos(x)/x$, though not in $sin(x)/x$.
$endgroup$
– Robert Israel
Jan 28 '13 at 6:26
$begingroup$
@RobertIsrael I thought the OP meant a discontinuity at zero of the function $sin x/x$.
$endgroup$
– user38268
Jan 28 '13 at 6:27
$begingroup$
@BenjaLim Sorry I realise that wasn't absolutely clear in my original post.
$endgroup$
– L. F.
Jan 28 '13 at 6:45
$begingroup$
@L.F., this proof is not accurate. You can not integrate by parts since you have an integrand that is not convergent. If the denominator was $x^{-2}$, than okay, but this is not the case.
$endgroup$
– Jeff Faraci
Dec 15 '13 at 3:44
$begingroup$
@Jeff I haven't proved anything $-$ please re-read the question.
$endgroup$
– L. F.
Dec 15 '13 at 3:48
|
show 3 more comments
$begingroup$
As others have already discussed, integrating by parts with $u=frac1x$ and $v=-cos(x)$ fails due to the singularity at $0$. Here, I thought it would be instructive to present an integration by parts scheme that circumvents the difficulty of the singularity at $0$. To that end, we proceed.
Let $I$ be the integral given by
$$I=int_0^infty frac{sin(x)}{x},dxtag1$$
Enforcing the substitution with $u=frac1x$ and $v=1-cos(x)$ (instead of $displaystyle v=-cos(x)$) in $(1)$ yields
$$begin{align}
I&=left.left(frac{1-cos(x)}{x}right)right|_0^infty+int_0^infty frac{1-cos(x)}{x^2},dx\\
&=2int_0^infty left(frac{sin(x/2)}{x}right)^2,dx\\
&=int_0^infty left(frac{sin(x)}{x}right)^2,dxtag2
end{align}$$
While $(2)$ does not lead immediately to a value for $I$ it does reveal the interesting identity
$$int_0^inftyfrac{sin(x)}{x},dx=int_0^infty left(frac{sin(x)}{x}right)^2,dx$$
answered Jan 30 at 18:36
Mark ViolaMark Viola
134k1278177
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The definite integration by parts formula
$$int_a^b f(x) g(x) dx = F(x) g(x)bigg|_a^b - int_a^b F(x) g'(x) dx$$
is justified by the product rule for derivative and the fundamental theorem of calculus
$$int_a^b (F g)'(x) dx = F(x) g(x) bigg|_a^b $$
But the "fine print" there is that $F(x) g(x)$ is assumed to be continuously differentiable on the interval
$(a,b)$ and continuous on $[a,b]$. In this case with $F(x) = -cos(x)$ and $g(x) = 1/x$,
$F(x) g(x)$ is undefined at $x=0$, and no definition of it there will make it continuous at $x=0$, let alone differentiable. So you can't use the integration by
parts formula for this integral on any interval containing $0$.
answered Jan 28 '13 at 6:56
Robert IsraelRobert Israel
330k23219473
$endgroup$
add a comment |
$begingroup$
The definite integration by parts formula
$$int_a^b f(x) g(x) dx = F(x) g(x)bigg|_a^b - int_a^b F(x) g'(x) dx$$
is justified by the product rule for derivative and the fundamental theorem of calculus
$$int_a^b (F g)'(x) dx = F(x) g(x) bigg|_a^b $$
But the "fine print" there is that $F(x) g(x)$ is assumed to be continuously differentiable on the interval
$(a,b)$ and continuous on $[a,b]$. In this case with $F(x) = -cos(x)$ and $g(x) = 1/x$,
$F(x) g(x)$ is undefined at $x=0$, and no definition of it there will make it continuous at $x=0$, let alone differentiable. So you can't use the integration by
parts formula for this integral on any interval containing $0$.
answered Jan 28 '13 at 6:56
Robert IsraelRobert Israel
330k23219473
$endgroup$
add a comment |
$begingroup$
The definite integration by parts formula
$$int_a^b f(x) g(x) dx = F(x) g(x)bigg|_a^b - int_a^b F(x) g'(x) dx$$
is justified by the product rule for derivative and the fundamental theorem of calculus
$$int_a^b (F g)'(x) dx = F(x) g(x) bigg|_a^b $$
But the "fine print" there is that $F(x) g(x)$ is assumed to be continuously differentiable on the interval
$(a,b)$ and continuous on $[a,b]$. In this case with $F(x) = -cos(x)$ and $g(x) = 1/x$,
$F(x) g(x)$ is undefined at $x=0$, and no definition of it there will make it continuous at $x=0$, let alone differentiable. So you can't use the integration by
parts formula for this integral on any interval containing $0$.
answered Jan 28 '13 at 6:56
Robert IsraelRobert Israel
330k23219473
$endgroup$
The definite integration by parts formula
$$int_a^b f(x) g(x) dx = F(x) g(x)bigg|_a^b - int_a^b F(x) g'(x) dx$$
is justified by the product rule for derivative and the fundamental theorem of calculus
$$int_a^b (F g)'(x) dx = F(x) g(x) bigg|_a^b $$
But the "fine print" there is that $F(x) g(x)$ is assumed to be continuously differentiable on the interval
$(a,b)$ and continuous on $[a,b]$. In this case with $F(x) = -cos(x)$ and $g(x) = 1/x$,
$F(x) g(x)$ is undefined at $x=0$, and no definition of it there will make it continuous at $x=0$, let alone differentiable. So you can't use the integration by
parts formula for this integral on any interval containing $0$.
answered Jan 28 '13 at 6:56
Robert IsraelRobert Israel
330k23219473
answered Jan 28 '13 at 6:56
Robert IsraelRobert Israel
330k23219473
answered Jan 28 '13 at 6:56
Robert IsraelRobert Israel
330k23219473
answered Jan 28 '13 at 6:56
Robert IsraelRobert Israel
330k23219473
330k23219473
add a comment |
add a comment |
$begingroup$
The issue is the discontinuity at $0$ in the cosine terms when you perform the partial integration, which is overlooked if you integrate over $mathbb{R}$. To get around this, consider:
$$int_{-infty}^{infty}frac{sin x}{x}dx=2int_0^{infty}frac{sin x}{x}dx$$
By parts:
$$int_{0}^{infty} frac{sin(x)}{x} dx = left.-frac{cos(x)}{x} right|_{0}^{infty} - int_{0}^{infty} frac{cos(x)}{x^2} dx$$
And here, both terms on the RHS diverge. No contradiction.
answered Jan 28 '13 at 6:22
L. F.L. F.
7,67032045
$endgroup$
$begingroup$
There is a discontinuity in $cos(x)/x$, though not in $sin(x)/x$.
$endgroup$
– Robert Israel
Jan 28 '13 at 6:26
$begingroup$
@RobertIsrael I thought the OP meant a discontinuity at zero of the function $sin x/x$.
$endgroup$
– user38268
Jan 28 '13 at 6:27
$begingroup$
@BenjaLim Sorry I realise that wasn't absolutely clear in my original post.
$endgroup$
– L. F.
Jan 28 '13 at 6:45
$begingroup$
@L.F., this proof is not accurate. You can not integrate by parts since you have an integrand that is not convergent. If the denominator was $x^{-2}$, than okay, but this is not the case.
$endgroup$
– Jeff Faraci
Dec 15 '13 at 3:44
$begingroup$
@Jeff I haven't proved anything $-$ please re-read the question.
$endgroup$
– L. F.
Dec 15 '13 at 3:48
|
show 3 more comments
$begingroup$
The issue is the discontinuity at $0$ in the cosine terms when you perform the partial integration, which is overlooked if you integrate over $mathbb{R}$. To get around this, consider:
$$int_{-infty}^{infty}frac{sin x}{x}dx=2int_0^{infty}frac{sin x}{x}dx$$
By parts:
$$int_{0}^{infty} frac{sin(x)}{x} dx = left.-frac{cos(x)}{x} right|_{0}^{infty} - int_{0}^{infty} frac{cos(x)}{x^2} dx$$
And here, both terms on the RHS diverge. No contradiction.
answered Jan 28 '13 at 6:22
L. F.L. F.
7,67032045
$endgroup$
$begingroup$
There is a discontinuity in $cos(x)/x$, though not in $sin(x)/x$.
$endgroup$
– Robert Israel
Jan 28 '13 at 6:26
$begingroup$
@RobertIsrael I thought the OP meant a discontinuity at zero of the function $sin x/x$.
$endgroup$
– user38268
Jan 28 '13 at 6:27
$begingroup$
@BenjaLim Sorry I realise that wasn't absolutely clear in my original post.
$endgroup$
– L. F.
Jan 28 '13 at 6:45
$begingroup$
@L.F., this proof is not accurate. You can not integrate by parts since you have an integrand that is not convergent. If the denominator was $x^{-2}$, than okay, but this is not the case.
$endgroup$
– Jeff Faraci
Dec 15 '13 at 3:44
$begingroup$
@Jeff I haven't proved anything $-$ please re-read the question.
$endgroup$
– L. F.
Dec 15 '13 at 3:48
|
show 3 more comments
$begingroup$
The issue is the discontinuity at $0$ in the cosine terms when you perform the partial integration, which is overlooked if you integrate over $mathbb{R}$. To get around this, consider:
$$int_{-infty}^{infty}frac{sin x}{x}dx=2int_0^{infty}frac{sin x}{x}dx$$
By parts:
$$int_{0}^{infty} frac{sin(x)}{x} dx = left.-frac{cos(x)}{x} right|_{0}^{infty} - int_{0}^{infty} frac{cos(x)}{x^2} dx$$
And here, both terms on the RHS diverge. No contradiction.
answered Jan 28 '13 at 6:22
L. F.L. F.
7,67032045
$endgroup$
The issue is the discontinuity at $0$ in the cosine terms when you perform the partial integration, which is overlooked if you integrate over $mathbb{R}$. To get around this, consider:
$$int_{-infty}^{infty}frac{sin x}{x}dx=2int_0^{infty}frac{sin x}{x}dx$$
By parts:
$$int_{0}^{infty} frac{sin(x)}{x} dx = left.-frac{cos(x)}{x} right|_{0}^{infty} - int_{0}^{infty} frac{cos(x)}{x^2} dx$$
And here, both terms on the RHS diverge. No contradiction.
answered Jan 28 '13 at 6:22
L. F.L. F.
7,67032045
edited Jan 28 '13 at 6:33
answered Jan 28 '13 at 6:22
L. F.L. F.
7,67032045
answered Jan 28 '13 at 6:22
L. F.L. F.
7,67032045
answered Jan 28 '13 at 6:22
L. F.L. F.
7,67032045
7,67032045
$begingroup$
There is a discontinuity in $cos(x)/x$, though not in $sin(x)/x$.
$endgroup$
– Robert Israel
Jan 28 '13 at 6:26
$begingroup$
@RobertIsrael I thought the OP meant a discontinuity at zero of the function $sin x/x$.
$endgroup$
– user38268
Jan 28 '13 at 6:27
$begingroup$
@BenjaLim Sorry I realise that wasn't absolutely clear in my original post.
$endgroup$
– L. F.
Jan 28 '13 at 6:45
$begingroup$
@L.F., this proof is not accurate. You can not integrate by parts since you have an integrand that is not convergent. If the denominator was $x^{-2}$, than okay, but this is not the case.
$endgroup$
– Jeff Faraci
Dec 15 '13 at 3:44
$begingroup$
@Jeff I haven't proved anything $-$ please re-read the question.
$endgroup$
– L. F.
Dec 15 '13 at 3:48
|
show 3 more comments
$begingroup$
There is a discontinuity in $cos(x)/x$, though not in $sin(x)/x$.
$endgroup$
– Robert Israel
Jan 28 '13 at 6:26
$begingroup$
@RobertIsrael I thought the OP meant a discontinuity at zero of the function $sin x/x$.
$endgroup$
– user38268
Jan 28 '13 at 6:27
$begingroup$
@BenjaLim Sorry I realise that wasn't absolutely clear in my original post.
$endgroup$
– L. F.
Jan 28 '13 at 6:45
$begingroup$
@L.F., this proof is not accurate. You can not integrate by parts since you have an integrand that is not convergent. If the denominator was $x^{-2}$, than okay, but this is not the case.
$endgroup$
– Jeff Faraci
Dec 15 '13 at 3:44
$begingroup$
@Jeff I haven't proved anything $-$ please re-read the question.
$endgroup$
– L. F.
Dec 15 '13 at 3:48
$begingroup$
There is a discontinuity in $cos(x)/x$, though not in $sin(x)/x$.
$endgroup$
– Robert Israel
Jan 28 '13 at 6:26
$begingroup$
There is a discontinuity in $cos(x)/x$, though not in $sin(x)/x$.
$endgroup$
– Robert Israel
Jan 28 '13 at 6:26
$begingroup$
@RobertIsrael I thought the OP meant a discontinuity at zero of the function $sin x/x$.
$endgroup$
– user38268
Jan 28 '13 at 6:27
$begingroup$
@RobertIsrael I thought the OP meant a discontinuity at zero of the function $sin x/x$.
$endgroup$
– user38268
Jan 28 '13 at 6:27
$begingroup$
@BenjaLim Sorry I realise that wasn't absolutely clear in my original post.
$endgroup$
– L. F.
Jan 28 '13 at 6:45
$begingroup$
@BenjaLim Sorry I realise that wasn't absolutely clear in my original post.
$endgroup$
– L. F.
Jan 28 '13 at 6:45
$begingroup$
@L.F., this proof is not accurate. You can not integrate by parts since you have an integrand that is not convergent. If the denominator was $x^{-2}$, than okay, but this is not the case.
$endgroup$
– Jeff Faraci
Dec 15 '13 at 3:44
$begingroup$
@L.F., this proof is not accurate. You can not integrate by parts since you have an integrand that is not convergent. If the denominator was $x^{-2}$, than okay, but this is not the case.
$endgroup$
– Jeff Faraci
Dec 15 '13 at 3:44
$begingroup$
@Jeff I haven't proved anything $-$ please re-read the question.
$endgroup$
– L. F.
Dec 15 '13 at 3:48
$begingroup$
@Jeff I haven't proved anything $-$ please re-read the question.
$endgroup$
– L. F.
Dec 15 '13 at 3:48
|
show 3 more comments
$begingroup$
As others have already discussed, integrating by parts with $u=frac1x$ and $v=-cos(x)$ fails due to the singularity at $0$. Here, I thought it would be instructive to present an integration by parts scheme that circumvents the difficulty of the singularity at $0$. To that end, we proceed.
Let $I$ be the integral given by
$$I=int_0^infty frac{sin(x)}{x},dxtag1$$
Enforcing the substitution with $u=frac1x$ and $v=1-cos(x)$ (instead of $displaystyle v=-cos(x)$) in $(1)$ yields
$$begin{align}
I&=left.left(frac{1-cos(x)}{x}right)right|_0^infty+int_0^infty frac{1-cos(x)}{x^2},dx\\
&=2int_0^infty left(frac{sin(x/2)}{x}right)^2,dx\\
&=int_0^infty left(frac{sin(x)}{x}right)^2,dxtag2
end{align}$$
While $(2)$ does not lead immediately to a value for $I$ it does reveal the interesting identity
$$int_0^inftyfrac{sin(x)}{x},dx=int_0^infty left(frac{sin(x)}{x}right)^2,dx$$
answered Jan 30 at 18:36
Mark ViolaMark Viola
134k1278177
$endgroup$
add a comment |
$begingroup$
As others have already discussed, integrating by parts with $u=frac1x$ and $v=-cos(x)$ fails due to the singularity at $0$. Here, I thought it would be instructive to present an integration by parts scheme that circumvents the difficulty of the singularity at $0$. To that end, we proceed.
Let $I$ be the integral given by
$$I=int_0^infty frac{sin(x)}{x},dxtag1$$
Enforcing the substitution with $u=frac1x$ and $v=1-cos(x)$ (instead of $displaystyle v=-cos(x)$) in $(1)$ yields
$$begin{align}
I&=left.left(frac{1-cos(x)}{x}right)right|_0^infty+int_0^infty frac{1-cos(x)}{x^2},dx\\
&=2int_0^infty left(frac{sin(x/2)}{x}right)^2,dx\\
&=int_0^infty left(frac{sin(x)}{x}right)^2,dxtag2
end{align}$$
While $(2)$ does not lead immediately to a value for $I$ it does reveal the interesting identity
$$int_0^inftyfrac{sin(x)}{x},dx=int_0^infty left(frac{sin(x)}{x}right)^2,dx$$
answered Jan 30 at 18:36
Mark ViolaMark Viola
134k1278177
$endgroup$
add a comment |
$begingroup$
As others have already discussed, integrating by parts with $u=frac1x$ and $v=-cos(x)$ fails due to the singularity at $0$. Here, I thought it would be instructive to present an integration by parts scheme that circumvents the difficulty of the singularity at $0$. To that end, we proceed.
Let $I$ be the integral given by
$$I=int_0^infty frac{sin(x)}{x},dxtag1$$
Enforcing the substitution with $u=frac1x$ and $v=1-cos(x)$ (instead of $displaystyle v=-cos(x)$) in $(1)$ yields
$$begin{align}
I&=left.left(frac{1-cos(x)}{x}right)right|_0^infty+int_0^infty frac{1-cos(x)}{x^2},dx\\
&=2int_0^infty left(frac{sin(x/2)}{x}right)^2,dx\\
&=int_0^infty left(frac{sin(x)}{x}right)^2,dxtag2
end{align}$$
While $(2)$ does not lead immediately to a value for $I$ it does reveal the interesting identity
$$int_0^inftyfrac{sin(x)}{x},dx=int_0^infty left(frac{sin(x)}{x}right)^2,dx$$
answered Jan 30 at 18:36
Mark ViolaMark Viola
134k1278177
$endgroup$
As others have already discussed, integrating by parts with $u=frac1x$ and $v=-cos(x)$ fails due to the singularity at $0$. Here, I thought it would be instructive to present an integration by parts scheme that circumvents the difficulty of the singularity at $0$. To that end, we proceed.
Let $I$ be the integral given by
$$I=int_0^infty frac{sin(x)}{x},dxtag1$$
Enforcing the substitution with $u=frac1x$ and $v=1-cos(x)$ (instead of $displaystyle v=-cos(x)$) in $(1)$ yields
$$begin{align}
I&=left.left(frac{1-cos(x)}{x}right)right|_0^infty+int_0^infty frac{1-cos(x)}{x^2},dx\\
&=2int_0^infty left(frac{sin(x/2)}{x}right)^2,dx\\
&=int_0^infty left(frac{sin(x)}{x}right)^2,dxtag2
end{align}$$
While $(2)$ does not lead immediately to a value for $I$ it does reveal the interesting identity
$$int_0^inftyfrac{sin(x)}{x},dx=int_0^infty left(frac{sin(x)}{x}right)^2,dx$$
answered Jan 30 at 18:36
Mark ViolaMark Viola
134k1278177
answered Jan 30 at 18:36
Mark ViolaMark Viola
134k1278177
answered Jan 30 at 18:36
Mark ViolaMark Viola
134k1278177
answered Jan 30 at 18:36
Mark ViolaMark Viola
134k1278177
134k1278177
add a comment |
add a comment |
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