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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? –  Bombyx mori 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. –  Bombyx mori Dec 25 '12 at 12:54
    
Check here:en.wikipedia.org/wiki/… –  Bombyx mori Dec 25 '12 at 12:55
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4 Answers

up vote 2 down vote accepted

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
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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
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Assuming $\alpha$ is finite $$\int_{0}^{\infty}\frac{\alpha \sin x}{\alpha^2+x^2} \mathrm{dx},\space \alpha>0$$

Let, $tan\theta=\frac{x}{\alpha}$

$$\int_{0}^{\frac{\pi}{2}}\sin (\alpha\tan\theta) \mathrm{d\theta},\space \alpha>0$$

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It is better if you make drafts on your own and then post full answers or, if you can, provide any hint on how to use what you have done to solve the problem. –  Pedro Tamaroff Dec 25 '12 at 18:22
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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
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