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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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    $\begingroup$ Can you do a standard contour integral? $\endgroup$ Dec 25, 2012 at 12:49
  • $\begingroup$ 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. $\endgroup$ Dec 25, 2012 at 12:54
  • $\begingroup$ Check here:en.wikipedia.org/wiki/… $\endgroup$ Dec 25, 2012 at 12:55

4 Answers 4

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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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  • $\begingroup$ @Mercy I didn't mention that integral. $\endgroup$
    – WimC
    Dec 26, 2012 at 21:45
  • $\begingroup$ @Mercy By integrating only to $2 \pi n / \alpha$, deriving a differential equation and then taking $n \to \infty$. $\endgroup$
    – WimC
    Dec 27, 2012 at 6:13
  • $\begingroup$ @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? $\endgroup$
    – WimC
    Dec 27, 2012 at 10:04
  • $\begingroup$ @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. $\endgroup$
    – WimC
    Dec 27, 2012 at 13:23
  • $\begingroup$ @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. $\endgroup$
    – WimC
    Dec 28, 2012 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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  • $\begingroup$ @Downvoter: Is there something wrong? $\endgroup$ Jan 14, 2013 at 20:51
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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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    $\begingroup$ All of this only shows that $\int_{-\infty}^0\frac{\sin x}{\alpha^2+x^2}\,dx=-\int_0^\infty\frac{\sin x}{\alpha^2+x^2}\,dx$. $\endgroup$
    – Mercy King
    Dec 26, 2012 at 10:32
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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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    $\begingroup$ 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. $\endgroup$
    – Pedro
    Dec 25, 2012 at 18:22

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