Evaluate $\int_{0}^{\infty}\frac{\alpha \sin x}{\alpha^2+x^2} \mathrm{dx},\space \alpha>0$ Evaluate 
$$\int_{0}^{\infty}\frac{\alpha \sin x}{\alpha^2+x^2} \mathrm{dx},\space \alpha&gt;0$$
I thought of using Feyman way, but it doesn't seem to help that much.
Some hints, suggestions?
Thanks.
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 $$f(\alpha) = \frac{e^{-\alpha}\operatorname{Ei}(\alpha)-e^{\alpha}\operatorname{Ei}(-\alpha)}{2}.$$
Assuming $\alpha$ is finite
$$\int_{0}^{\infty}\frac{\alpha \sin x}{\alpha^2+x^2} \mathrm{dx},\space \alpha&gt;0$$

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

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