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.
Evaluate \int_{0}^{\infty}\frac{\alpha \sin x}{\alpha^2+x^2} \mathrm{dx},\space \alpha>0
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0Check here:http://en.wikipedia.org/wiki/Methods_of_contour_integration#Example_.28IV.29_.E2.80.93_branch_cuts – 2012-12-25
4 Answers
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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0@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. – 2012-12-28
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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0@Downvoter: Is there something wrong? – 2013-01-14
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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1All of this only shows that $\int_{-\infty}^0\frac{\sin x}{\alpha^2+x^2}\,dx=-\int_0^\infty\frac{\sin x}{\alpha^2+x^2}\,dx$. – 2012-12-26
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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4It 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. – 2012-12-25