# Evaluate $\int_{0}^{\infty}\frac{\alpha \sin x}{\alpha^2+x^2} \mathrm{dx},\space \alpha>0$

Evaluate $$\int_{0}^{\infty}\frac{\alpha \sin x}{\alpha^2+x^2} \mathrm{dx},\space \alpha>0$$ I thought of using Feyman way, but it doesn't seem to help that much.
Some hints, suggestions?
Thanks.

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Can you do a standard contour integral? –  user32240 Dec 25 '12 at 12:49
This is a classical problem using contour integral. I guess once can solve it via other ways, but I believe counter integral might be easier. –  user32240 Dec 25 '12 at 12:54
Check here:en.wikipedia.org/wiki/… –  user32240 Dec 25 '12 at 12:55

One can show that the function $$f(\alpha) = \int_0^\infty\frac{\alpha \sin(x)}{\alpha^2+x^2}dx = \int_0^\infty \frac{\sin(\alpha x)}{1+x^2}dx$$ satisfies the differential equation $$f''(\alpha) = f(\alpha) -\frac{1}{\alpha}$$ which together with $f(0)=0$ and $\lim_{\alpha\to\infty}f(\alpha)=0$ leads to $$f(\alpha) = \frac{e^{-\alpha}\operatorname{Ei}(\alpha)-e^{\alpha}\operatorname{Ei}(-\alpha)}{2}.$$

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@Mercy I didn't mention that integral. –  WimC Dec 26 '12 at 21:45
@Mercy By integrating only to $2 \pi n / \alpha$, deriving a differential equation and then taking $n \to \infty$. –  WimC Dec 27 '12 at 6:13
@Mercy There's an $x$ in there? Anyway, it's not what I got. BTW, do you only question the way to derive the equation or the equation itself? –  WimC Dec 27 '12 at 10:04
@Mercy This is getting a bit annoying but I rechecked my computations anyway. I get $$f_n''(\alpha) = f_n(\alpha) - \frac{1}{\alpha} + \frac{\alpha}{\alpha^2 + (2 \pi n)^2}.$$ I suggest you recheck yours too. –  WimC Dec 27 '12 at 13:23
@Chris'ssister Actually, this one will still take some work to finish properly. See one of the comments for what I did. (Unfortunately, Mercy removed the other side of the conversation.) And no I'm not a teacher but thanks for your kind words, these are rare on math.se. –  WimC Dec 28 '12 at 14:02

Converting $\sin(x)$ in terms of the exponential function and using partial fraction, you can get the answer in terms of the exponential integral

$$\frac{{\rm e}^{-\alpha}}{2}\,{\left( {{\rm e}^{2\, \alpha}}{\it Ei} \left( 1,\alpha \right) -{\it Ei} \left( 1,-\alpha \right) \right) },$$

where

$${\it Ei} \left( a,z \right) =\int _{1}^{\infty }\!{{\rm e}^{-{ \it t}\,z}}{{\it t}}^{-a }{d{\it t}}$$

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@Downvoter: Is there something wrong? –  Mhenni Benghorbal Jan 14 '13 at 20:51

use Residue theory $$\int_{-\infty}^{+\infty} sin(ax)f(x)dx=Im\left[ \oint_{c^+} e^{iaz} f(z)dz \right] = Im \left[ 2 \pi i \sum_{i=1}^n R_i^+ \right]$$ $$\int_{-\infty}^{+\infty} cos(ax)f(x)dx=Re\left[ \oint_{c^+} e^{iaz} f(z)dz \right] = Re \left[ 2 \pi i \sum_{i=1}^n R_i^+ \right]$$ $$Im : Imaginary \space Part \space ; \space Re : Real \space Part$$ $$c^+ : Upper \space half \space plane \space of \space complex \space plane \space (uhp)$$ $$c^- : Lower \space half \space plane \space of \space complex \space plane \space (lhp)$$ $$R_i^+ \rightarrow Residue \space of \space function \space in \space singularities \space ponit \space ( that \space placed \space in \space uhp)$$ $$I=\int_{0}^{\infty}\frac{\alpha \sin x}{\alpha^2+x^2} \mathrm{dx} =\frac{1}{2}\int_{-\infty}^{\infty}\frac{\alpha \sin x}{\alpha^2+x^2} \mathrm{dx}$$ $$I= \frac{1}{2} Im \left[ \oint_{c^+} \frac{\alpha e^{iz}}{\alpha^2+z^2} dz\right] =\frac{1}{2} Im \left[ 2 \pi i \sum_{i=1}^n R_i^+ \right]$$ $$singularities \rightarrow \space z^2+\alpha^2=0 \space \rightarrow \begin{cases} z_1=i \alpha &\mbox{Acceptable (uhp)} \\ z_2=-i \alpha & \mbox{Ineligible (lhp) } \end{cases}$$ $$R_1=Residue( \frac{\alpha e^{iz}}{\alpha^2+z^2} , z=i \alpha)=\lim_{z \rightarrow ia} \frac{\alpha e^{iz} (z-i \alpha)}{(z-i \alpha)(z+i \alpha)} = \frac{e^{-\alpha}}{2i}$$ $$I=\frac{1}{2} Im \left[ 2 \pi i \frac{e^{-\alpha}}{2i} \right] =0$$

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All of this only shows that $\int_{-\infty}^0\frac{\sin x}{\alpha^2+x^2}\,dx=-\int_0^\infty\frac{\sin x}{\alpha^2+x^2}\,dx$. –  Mercy Dec 26 '12 at 10:32
Assuming $\alpha$ is finite $$\int_{0}^{\infty}\frac{\alpha \sin x}{\alpha^2+x^2} \mathrm{dx},\space \alpha>0$$
Let, $tan\theta=\frac{x}{\alpha}$
$$\int_{0}^{\frac{\pi}{2}}\sin (\alpha\tan\theta) \mathrm{d\theta},\space \alpha>0$$