Using the Fourier cosine integral formula to prove that $$e^{-x}\cos(x)=\frac{2}{\pi}\int_0^\infty \frac{u^2+2}{u^4+4}\cos(ux)\:du$$
Okay, I knew that
$$ \begin{align} f(x)&=\frac{2}{\pi}\int_0^\infty f(t)\:dt\int_0^\infty\cos(ut)\cos(ux)\:du\\ &=\frac{2}{\pi}\int_0^\infty e^{-t}\cos(t)\:dt\int_0^\infty\cos(ut)\cos(ux)\:du \end{align} $$
It seems a tedious integration to work with. How to simplify it further? Or do I need to follow some trick/procedure here?
On
Given $$f(x)=\frac2\pi\int_0^\infty g(u)\cos(ux)du$$you have $$g(u)=\int_0^\infty f(x)\cos(ux)dx$$ So, to prove your problem, you need to show $$\int_0^\infty e^{-x}\cos(x)\cos(ux) dx=\frac{u^2+2}{u^4+4}$$ To do that, use $$\cos x\cos(ux)=\frac12\left(\cos((u+1)x)+\cos((u-1)x)\right)$$ Then $$\int_0^\infty e^{-x}\cos(mx)dx$$ can be solved through integration by parts twice to get $$\frac1{m^2+1}$$ Then $$g(u)=\frac1{(u+1)^2+1}+\frac1{(u-1)^2+1}\\=\frac{u^2-2u+2+u^2+2u+2}{((u^2-2u+2)(u^2+2u+2)}\\=\frac{2(u^2+2)}{2(u^4+4)}$$
Since you are integrating an even function, your integral is equal to$$\frac1\pi\int_{-\infty}^\infty\frac{u^2+2}{u^4+4}\cos(ux)\,\mathrm du=\frac1\pi\operatorname{Re}\left(\int_{-\infty}^\infty\frac{u^2+2}{u^4+4}e^{uxi}\,\mathrm du\right).$$This integral can be computed using residues:\begin{align}\int_{-\infty}^\infty\frac{u^2+2}{u^4+4}e^{uxi}\,\mathrm du&=2\pi i\left(\operatorname{res}_{x=1+i}\left(\frac{u^2+2}{u^4+4}e^{uxi}\right)+\operatorname{res}_{x=-1+i}\left(\frac{u^2+2}{u^4+4}e^{uxi}\right)\right)\\&=2\pi i\left(-\frac{1}{4} i e^{(-1+i) x}-\frac{1}{4} i e^{(-1-i) x}\right)\\&=\frac\pi2\left(e^{(-1+i)x}+e^{(-1-i)x}\right)\\&=\pi e^{-x}\frac{e^{ix}+e^{-ix}}2\\&=\pi e^{-x}\cos(x)\end{align}Therefore, your integral is indeed equal to $e^{-x}\cos(x)$.