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Math Help - physicist needs help solving integral problem

  1. #1
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    physicist needs help solving integral problem

    Hi, this is the integral and how much I managed to solve
    http://img689.imageshack.us/img689/4719/img8366.jpg
    So at this point I have two problems:
    Firstly finding a limit around zero pole and secondly finding a residue around pole which is inside contour. While I had limits in my high maths course, I still struggle with this one; as for residue, have no idea how to find it and at the moment I don't have any good maths books with myself and even more important - time to learn it, so I'm in a pretty bad situation, any help would more than grateful!
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  2. #2
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    Is this partial fraction rep any help?

    <br />
1/{ (z (z^2 +b ^2)) } =<br />
{1/b^2} { [ 1/z - 1/{(2(z-ib))} - 1/{(2(z+ib))} ]}<br />
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  3. #3
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    Quote Originally Posted by qmech View Post
    <br />
1/{ (z (z^2 +b ^2)) } =<br />
{1/b^2} { [ 1/z - 1/{(2(z-ib))} - 1/{(2(z+ib))} ]}<br />
    Firstly I thought It wasn't useful, but no I think that I can convert limit (D part) into semicircular integral (which equals "-i*pi", "-" because I'm going clockwise) and multiplicate it of function residue at 1, so the limit then equals to "-i*pi/b^2 "which seams to be reasonable.

    I managed to solve E part, which equals to "2i*pi*e^(iaib)/ib2ib = -i*pi*e^(-ab)/b^2" .
    After simple algebra, my integral equals "pi(1-e^(-ab))/2b^2"

    Thanks!
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by AndrewKaz View Post
    Hi, this is the integral and how much I managed to solve
    http://img689.imageshack.us/img689/4719/img8366.jpg
    So at this point I have two problems:
    Firstly finding a limit around zero pole and secondly finding a residue around pole which is inside contour. While I had limits in my high maths course, I still struggle with this one; as for residue, have no idea how to find it and at the moment I don't have any good maths books with myself and even more important - time to learn it, so I'm in a pretty bad situation, any help would more than grateful!
    I think that this is what your integral is (I'm sorry my comp. has bad resolution)

    Problem: Compute \int_0^{\infty}\frac{\sin(x)}{x(x^2+b^2)}dx\quad b>0

    Solution: Think for a second about the complex integral \int_{C_{\delta,R}}\frac{e^{iz}}{z(z^2+b^2)}dz where that C_{\delta,R} is the semi-circular countour of radius R centered at 0, which is oriented counterclockwise. Notice that the integrand of this integral as simple poles at both 0 and ib. Also, notice that \frac{\sin(x)}{x(x^2+b^2(}=\mathfrak{R}\left(\frac  {e^{ix}}{x(x^2+b^2)}\right) \quad x\in\mathbb{R}. By Jordan's lemma (Jordan's Lemma -- from Wolfram MathWorld) we can see that \left|\int_{\Gamma_R}\frac{e^{iz}}{z(z^2+b^2)}\rig  ht|\le\frac{\pi}{R}\int_{\Gamma_R}\frac{dz}{z^2+b^  2}=o(1)\quad |R|\to\infty (where \Gamma_R is the bigger semi-circle). Therefore, at the limit the contribution from the integral with respect to the bigger semi-circle approaches zero. Lastly, using the residue theorem we can see that \int_{C_{\delta,R}}\frac{e^{iz}}{z(z^2+b^2)}=2\pi i\text{Res}\left(\frac{e^{iz}}{z(z^2+b^2)},ib\righ  t)+\pi i\text{Res}\left(\frac{e^{iz}}{z(z^2+b^2)}, 0\right) which reduces to \frac{-\pi i}{e^{b}b^2}+\frac{\pi i}{b^2} and using the fact that the integral is even we may deduce that \int_0^{\infty}\frac{\sin(x)}{x(x^2+b^2)}dx=\frac{  \pi}{2b^2}\left(1-e^{-b}\right)


    Remark: I have no idea how to draw stuff on here, but if the countour I used isn't clear I can try to figure it out. Also, you can do this another way considering the function \frac{e^{iz}-1}{z(z^2+b^2)} which makes the pole at the origin removable.
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