1. ## Integral

Evaluate $\displaystyle \int_0^\infty\frac{e^{-x}\sin x}x\,dx$

2. Fact you should know

$\displaystyle \int_0^\infty e^{-x}\cos(\alpha x)\,dx=\frac1{\alpha^2+1}$

Now, let's introduce the following parameter to construct a double integral as follows:

$\displaystyle \int_0^1\cos(\alpha x)\,d\alpha=\frac{\sin x}x,$ this yields

$\displaystyle \int_0^\infty\frac{e^{-x}\sin x}x\,dx=\int_0^\infty e^{-x}\left(\int_0^1\cos(\alpha x)\,d\alpha\right)\,dx$

Reversing the order of integration

$\displaystyle \int_0^\infty\frac{e^{-x}\sin x}x\,dx=\int_0^1\int_0^\infty e^{-x}\cos(\alpha x)\,dx\,d\alpha=\int_0^1\frac1{\alpha^2+1}\,d\alph a=\frac\pi4\,\blacksquare$

3. Originally Posted by Krizalid
Fact you should know

$\displaystyle \int_0^\infty e^{-x}\cos(\alpha x)\,dx=\frac1{\alpha^2+1}$

Now, let's introduce the following parameter to construct a double integral as follows:

$\displaystyle \int_0^1\cos(\alpha x)\,d\alpha=\frac{\sin x}x,$ this yields

$\displaystyle \int_0^\infty\frac{e^{-x}\sin x}x\,dx=\int_0^\infty e^{-x}\left(\int_0^1\cos(\alpha x)\,d\alpha\right)\,dx$

Reversing the order of integration

$\displaystyle \int_0^\infty\frac{e^{-x}\sin x}x\,dx=\int_0^1\int_0^\infty e^{-x}\cos(\alpha x)\,dx\,d\alpha=\int_0^1\frac1{\alpha^2+1}\,d\alph a=\frac\pi4\,\blacksquare$
Impressive, as always!

-Dan