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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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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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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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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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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