I wanna calculate: $$\int_0^\infty e^{-tx}\cos(x)\cdot A \, dx$$
where $A=\frac 2 {\sqrt\pi} \int_0^\infty e^{-xu^2}\,du$
What I've done :
Rewrite the integral to $$\int_0^\infty e^{-tx} \cos(x)\cdot\frac{2}{\sqrt\pi} \int_0^\infty e^{-xu^2} \, du \,dx$$
$$=\int_0^\infty \left(\int_0^\infty e^{-tx} \cos(x)\cdot\frac{2}{\sqrt\pi} e^{-xu^2} \, du \right)\, dx$$
Substitute $w=\sqrt xu\Rightarrow dt=\sqrt x \,du$
We get: $$=\int_0^\infty \left(\int_0^\infty e^{-tx} \cos(x)\cdot\frac 2 {\sqrt\pi} e^{w^2}\, dw \right)\, dx$$
Any hint ?
We have $$I=\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}e^{-tx}\cos\left(x\right)\int_{0}^{\infty}e^{-xu^{2}}dudx $$ now take $u\sqrt{x}=v,\, du=dv/\sqrt{x} $. We get $$I=\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}\frac{e^{-tx}\cos\left(x\right)}{\sqrt{x}}\int_{0}^{\infty}e^{-v^{2}}dvdx=\int_{0}^{\infty}\frac{e^{-tx}\cos\left(x\right)}{\sqrt{x}} $$ $$=\textrm{Re}\left(\int_{0}^{\infty}\frac{e^{-x\left(t-i\right)}}{\sqrt{x}}\right)=\textrm{Re}\left(\frac{1}{\sqrt{t-i}}\int_{0}^{\infty}\frac{e^{-y}}{\sqrt{y}}\right)=\textrm{Re}\left(\frac{\sqrt{\pi}}{\sqrt{t-i}}\right)$$ and now you can calculate the integral using the trigonometric representation of complex numbers.