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Integration, Help!

  1. Aug 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate the following integrals:

    2. Relevant 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]

    3. 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]
     
  2. jcsd
  3. Aug 9, 2012 #2
    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.
     
  4. Aug 9, 2012 #3
    how would u solve the counter integral, I've never done counter integration.
     
  5. Aug 9, 2012 #4
    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.
     
  6. Aug 9, 2012 #5
    If counter integration is the better way, then can u show me how to do it that way? I'll learn it.
     
  7. Aug 9, 2012 #6

    tiny-tim

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    Hi XtremePhysX! :smile:

    (1) looks excessively nasty :yuck:

    are you sure it isn't cos-1x on the top?
     
  8. Aug 9, 2012 #7
    yes I'm sure :) it is cos(x)
     
  9. Aug 9, 2012 #8
    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.
     
  10. Aug 9, 2012 #9

    gabbagabbahey

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    Use the derivative of the Gamma function and a the substitution u=log(1/x).
     
  11. Aug 9, 2012 #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.
     
  12. Aug 9, 2012 #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.
     
  13. Aug 9, 2012 #12

    gabbagabbahey

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    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]
     
  14. Aug 9, 2012 #13
    Where do you go from here?
     
  15. Aug 9, 2012 #14
    Do you use a substitution to integrate [tex]-\frac{1}{x^3} \left (2+\frac{1}{x^2-1}\right)[/tex]
     
  16. Aug 9, 2012 #15

    gabbagabbahey

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

    (As long as the integrals converge)
     
  17. Aug 9, 2012 #16
    Like this:
    [tex]-\frac{1}{x^3}\left ( \int2dx + \int \frac{1}{x^2-1}\right)dx[/tex]
     
  18. Aug 9, 2012 #17

    gabbagabbahey

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    Is [itex]\frac{1}{x^3}[/itex] a constant? If not, how can you pull it out of the integral like that?
     
  19. Aug 9, 2012 #18
    I don't know, then how do u apply that identity?
     
  20. Aug 9, 2012 #19
    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]
     
  21. Aug 10, 2012 #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.
     
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