With solving a problem in probability, I find an integral not sure how to do it.
$\int_{|y|}^{\infty}\frac{1}{2x}e^{-x}dx$
Because when I use the theorem \int f(x)g'(x)=f(x)g(x)-\int f'(x)g(x) .i.e.$\int_{|y|}^{\infty}\frac{1}{2x}e^{-x}=-\frac{1}{2x}e^{-x}-\frac{1}{2}\int_{|y|}^{\infty}(-x^{-2}e^{-x})dx$
But how to continue from here