# Thread: physicist needs help solving integral problem

1. ## 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!

2. ## Is this partial fraction rep any help?

$
1/{ (z (z^2 +b ^2)) } =
{1/b^2} { [ 1/z - 1/{(2(z-ib))} - 1/{(2(z+ib))} ]}
$

3. Originally Posted by qmech
$
1/{ (z (z^2 +b ^2)) } =
{1/b^2} { [ 1/z - 1/{(2(z-ib))} - 1/{(2(z+ib))} ]}
$
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!

4. Originally Posted by AndrewKaz
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.