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The value of the Laplace integral $$\int_0^{+\infty} \frac{\cos ax}{x^2+1}dx=\frac{\pi}{2}e^{-|a|}$$ is well-known. Perhaps a little less known is the value of complementary integral where $\cos$ is replaced with $\sin$: $$\int_0^{+\infty} \frac{\sin ax}{x^2+1}dx=\frac{1}{2}(e^{-a}\mathrm{Ei}(a)-e^{a}\mathrm{Ei}(-a)),$$ where $a\neq0$ and $$\mathrm{Ei}(x)=\mathrm{v.p.}\int_{-\infty}^x \frac{e^t}{t}dt$$ is exponential integral (v.p. is only needed for $x>0)$. I was able to deduce this formula using the representation $\frac{1}{1+x^2}=\int_0^{+\infty} e^{-t}\cos xt\,dt$ and exchanging the order of integration, but some problems with convergence arise, and justification here is rather complicated. My question: is there a simpler way to prove it?

Also, if you know where one can find it, the reference would be nice.

I'm only interested in a solution using real calculus methods, without contour integration.

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    Does expressing $sin ax$ as Im($e^{iax}$) and splitting $\frac{1}{x^2+1}$ as partial fractions help? – 2017-02-10
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    What about sin(ax) = cos (a(x-pi/2)) and adding a small finite part to the known integral? – 2017-02-10

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