Integration, Help!

  • #1

Homework Statement



Evaluate the following integrals:

Homework Equations



(1)[tex]\int_{0}^{\infty }\frac{cos(x)}{1+(x)^{2}}dx[/tex]
(2)[tex]\int_{0}^{\infty}e^{-x^2}dx[/tex]
(3)[tex]\int_{0}^{1}\log\log\left(\frac{1}{x}\right)\,dx[/tex]
(4)[tex]\int_{0}^{\infty}\frac{x\,dx}{e^x-1}[/tex]
(5)[tex]\int_{-\infty}^{\infty}\frac{\sin(x)\,dx}{x}[/tex]

The Attempt at a Solution



I know how to do the second one, so please help me with the rest:

(2) Let [tex]I = \int_{0}^{\infty }e^{-x^{2}}dx[/tex] then we have by the square of an integral:
[tex]I^{2}=\left ( \int_{0}^{\infty }e^{-x^{2}}dx \right )\left ( \int_{0}^{\infty }e^{-y^{2}}dy \right ) = \int_{0}^{\infty }\int_{0}^{\infty }e^{-x^{2}-y^{2}}dydx[/tex]
Shifting to polar coordinates:
[tex]\int_{0}^{\frac{\pi }{2}}\int_{0}^{\infty }e^{-r^{2}}rdrd\theta[/tex]
For the interior integral, we use the transformation:
[tex]u=r^{2}, \frac{du}{2}=rdr[/tex] to obtain:
[tex]\frac{1}{2}\int_{0}^{\infty }e^{-u}du = -\frac{1}{2}\left [ e^{-u} \right ]\infty to 0 = \frac{1}{2}[/tex]
Plugging this into our integral, we're left with:
[tex]I^{2}=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}d\theta = \frac{\pi }{4}[/tex]
By taking the square root of both sides, we have the answer:
[tex]I=\int_{0}^{\infty }e^{-x^{2}}dx=\frac{\sqrt{\pi }}{2}[/tex]
 

Answers and Replies

  • #2
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For the first one, I would selfishly resort to contour integration and write:

[tex]\int_0^{\infty}\frac{\cos(x)}{(x+i)(x-i)}dx=1/2 \int_{-\infty}^{\infty} \frac{\cos(x)}{(x+i)(x-i)}dx[/tex]

then consider the contour integral:

[tex]\oint \frac{e^{iz}}{(z+i)(z-i)}dz[/tex]

where the contour is the half-disc in the upper half plane, then just use the Residue Theorem but not sure you want to go this way.
 
  • #3
how would u solve the counter integral, I've never done counter integration.
 
  • #4
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how would u solve the counter integral, I've never done counter integration.

If this is a homework exercise then perhaps you're not expected to know contour integration and need to find another way. I can't think of one at the moment though. Also, really not helpful to you I think for me to try and explain the contour integration if you've not had any Complex Analysis.
 
  • #5
If this is a homework exercise then perhaps you're not expected to know contour integration and need to find another way. I can't think of one at the moment though. Also, really not helpful to you I think for me to try and explain the contour integration if you've not had any Complex Analysis.

If counter integration is the better way, then can u show me how to do it that way? I'll learn it.
 
  • #6
tiny-tim
Science Advisor
Homework Helper
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Hi XtremePhysX! :smile:

(1) looks excessively nasty :yuck:

are you sure it isn't cos-1x on the top?
 
  • #7
yes I'm sure :) it is cos(x)
 
  • #8
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The third one is one definition of the Euler Mascheroni constant so we would need to show:

[tex]\int_0^1 \log\left(\log(1/x)\right)dx=-\gamma[/tex]

but I can't think of a way to show this.
 
  • #9
gabbagabbahey
Homework Helper
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The third one is one definition of the Euler Mascheroni constant so we would need to show:

[tex]\int_0^1 \log\left(\log(1/x)\right)dx=-\gamma[/tex]

but I can't think of a way to show this.

Use the derivative of the Gamma function and a the substitution u=log(1/x).
 
  • #10
Can anyone help me with this integral please:
[tex]\int^{\frac{\sqrt{3}}{2}}_{\frac{1}{\sqrt{2}}} \frac{1 - 2x^2}{x^5 - x^3}dx[/tex] using x=sin(u)

please, please.
 
  • #11
[tex]\int_{\frac{\pi }{3}}^{\frac{\pi }{4}}\frac{1-2sin^{2}u}{sin^{5}u-sin^{3}u}cosudu[/tex]

I got up to here.
 
  • #12
gabbagabbahey
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Can anyone help me with this integral please:
[tex]\int^{\frac{\sqrt{3}}{2}}_{\frac{1}{\sqrt{2}}} \frac{1 - 2x^2}{x^5 - x^3}dx[/tex] using x=sin(u)

please, please.

Why use that substitution?

[tex]\frac{1-2x^2}{x^5-x^3}=-\frac{1}{x^3} \left (2+\frac{1}{x^2-1}\right)[/tex]
 
  • #13
Why use that substitution?

[tex]\frac{1-2x^2}{x^5-x^3}=-\frac{1}{x^3} \left (2+\frac{1}{x^2-1}\right)[/tex]

Where do you go from here?
 
  • #14
Do you use a substitution to integrate [tex]-\frac{1}{x^3} \left (2+\frac{1}{x^2-1}\right)[/tex]
 
  • #15
gabbagabbahey
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Where do you go from here?

[tex]\int \left( f(x) + g(x) \right) dx = \int f(x) dx + \int g(x) dx[/tex]

(As long as the integrals converge)
 
  • #16
Like this:
[tex]-\frac{1}{x^3}\left ( \int2dx + \int \frac{1}{x^2-1}\right)dx[/tex]
 
  • #17
gabbagabbahey
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Like this:
[tex]-\frac{1}{x^3}\left ( \int2dx + \int \frac{1}{x^2-1}\right)dx[/tex]

Is [itex]\frac{1}{x^3}[/itex] a constant? If not, how can you pull it out of the integral like that?
 
  • #18
I don't know, then how do u apply that identity?
 
  • #19
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Just distribute to the terms in parentheses:[tex]-\frac{1}{x^3}\left(2 + \frac{1}{x^2 - 1}\right) = -\frac{2}{x^3} - \frac{1}{x^3(x^2 - 1)}[/tex]
 
  • #20
Can you guys help me with the original integrals please? I kind of know how to do the first, second and third, please help me do the last two.
 
  • #21
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For the fourth, use the identity
[tex]\int_{0}^{\infty}e^{-nx}x^{s-1}dx=\frac{\Gamma(s)}{n^s}[/tex]
along with the monotone convergence theorem.
The fifth can be obtained by considering the complex valued integral
[tex]\int_{C}\frac{e^{iz}}{z}\,dz[/tex]
on a semicircular contour.
 
  • #22
For the fourth, use the identity
[tex]\int_{0}^{\infty}e^{-nx}x^{s-1}dx=\frac{\Gamma(s)}{n^s}[/tex]
along with the monotone convergence theorem.

Thank you mill :)
 
  • #23
gabbagabbahey
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The fifth can also be done by looking at [itex]\frac{d}{da}\int_{-\infty}^{\infty}\frac{e^{iax}}{x}dx[/itex]
 

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