Proving $\int_{\Bbb R} \frac{\cos x}{(x^2+t^2)^2}dx=\frac{\pi(t+1)}{2t^3e^t}, t>0$ 
I want to prove that $$\int_{\Bbb R}\frac{\cos x}{(x^2+t^2)^2}dx=\frac{\pi(t+1)}{2t^3e^t},t > 0$$

Integral calculator with steps gives the following answer.
$\displaystyle\int_{\Bbb R}\frac{\cos{(x)}}{(x^2+t^2)^2 }dx=\frac{(t*\cosh{(t)}-\sinh{(t)})*Si(x+it)-i*(t*\sinh{(t)}-\cosh{(t)})*Ci(x+it)+(t*\cosh{(t)}-\sinh{(t)}*Si(x-it)+i*(t*\sinh{(t)}-\cosh{(t)})*Ci(x-it)}{4*t^3}+\frac{x*\cos{(x)}}{2*t^2*(x^2+t^2)}+C$
Now my question is how to evaluate $Si{(x+it)},Ci(x+it),Si(x-it),Ci(x-it)?$
My another question is how to prove this equation using "Differentiation under Integral Sign making  change of variables"
 A: First let's solve an easier integral that doesn't have the denominator squared, namely:$$y(t)=\int_0^\infty \frac{\cos(tx)}{\color{blue}{1+x^2}}dx=\int_0^\infty \cos(tx){\color{blue}{\int_0^\infty \sin y e^{-xy}dy}}dx$$
$$={\int_0^\infty}\sin y{\color{red}{\int_0^\infty \cos(tx)e^{-xy}dx}}dy=\int_0^\infty \frac{{\color{red}y}\sin y}{{\color{red}{t^2+y^2}}}dy $$
$$\overset{y=tx}=\int_0^\infty \frac{x\sin(tx)}{1+x^2}dx=-y'(t)\Rightarrow \frac{y'(t)}{y(t)}=-1 \Rightarrow y(t)=c\cdot e^{-t}$$
$$y(0)=\int_0^\infty \frac{dx}{1+x^2}=\frac{\pi}{2}\Rightarrow c=\frac{\pi}{2}\Rightarrow y(t)=\frac{\pi}{2}e^{-t}$$
$$y(t)\overset{tx=z}=t\int_0^\infty\frac{\cos z}{t^2+z^2}dz\Rightarrow\boxed{\int_{-\infty}^\infty\frac{\cos x}{t^2+x^2}dx=\frac{\pi}{t}e^{-t},\ t>0}$$

Now in order to find the desired integral, we will differentiate under the integral sign, with respect to $t$ then divide by $-2t$, since:
$$\frac{d}{dt}\int_{-\infty}^\infty\frac{\cos x}{t^2+x^2}dx=-2t\int_{-\infty}^\infty\frac{\cos x}{(t^2+x^2)^2}dx=\frac{d}{dt}\left(\frac{\pi}{t}e^{-t}\right)$$
$$\Rightarrow \int_{-\infty}^\infty\frac{\cos x}{(t^2+x^2)^2}dx=-\frac{1}{2t}\frac{d}{dt}\left(\frac{\pi}{t}e^{-t}\right)=\frac{\pi(t+1)}{2t^3}e^{-t},\ t>0$$
A: Another way to do it is to note that$$\frac{1}{x^2+t^2}=\frac{1}{2it}\left(\frac{1}{x-it}-\frac{1}{x+it}\right)\implies\frac{1}{(x^2+t^2)^2}=\frac{-1}{4t^2}\left(\frac{1}{(x-it)^2}+\frac{1}{(x+it)^2}+\frac{i}{t}\left(\frac{1}{x-it}-\frac{1}{x+it}\right)\right).$$The $e^{ix}$ in $\cos x$ doesn't diverge when $x=i\infty$, so the $\Im x\ge0$ semicircle contains a pole $it$ contributing$$\frac{-i\pi}{2t^2}\left.\left(\frac{d}{dx}e^{ix}+\frac{i}{t}e^{ix}\right)\right|_{x=it}=\frac{-i\pi}{2t^2}\left(ie^{-t}+\frac{i}{t}e^{-t}\right)=\frac{\pi(t+1)}{2t^3e^t}.$$
