Ηow to solve this integral $(e^{-y/2})/y$ How to solve this question:
$$\int_{0}^{+\infty}\int_{x}^{+\infty}\frac{e^{-y/2}}y \,dydx$$
One idea that I have is to put $-y/2=\ln t$ then subsequently come to the result as the integration of $1/\ln t$ but what is the integration of $1/\ln t$?
Is there any other way for solving this question?
 A: Since the function $e^{-y/2}/y$ is non-negative for $y\ge 0$, we may apply Tonelli's theorem and interchange the order of integration to obtain 
\begin{align}
\int_{0}^{+\infty}\int_{x}^{+\infty}\frac{e^{-y/2}}y \,dydx &=
\int_{0}^{+\infty}\int_{0}^{y}\frac{e^{-y/2}}y \,dxdy =
\int_{0}^{+\infty}\frac{e^{-y/2}}y[x]_{0}^{y}dy =
\int_{0}^{+\infty}e^{-y/2}dy=2
\end{align}
A: Too long for a comment!
$$\int_{0}^{\infty}\int_{x}^{\infty}\frac{e^{-\frac{y}{2}}}y\space\text{d}y\space\text{d}x=$$
$$\int_{0}^{\infty}\left[\int_{x}^{\infty}\frac{e^{-\frac{y}{2}}}y\space\text{d}y\right]\space\text{d}x=$$
$$\lim_{a\to\infty}\int_{0}^{a}\left[\lim_{b\to\infty}\int_{x}^{b}\frac{e^{-\frac{y}{2}}}y\space\text{d}y\right]\space\text{d}x=$$
$$\lim_{a\to\infty}\int_{0}^{a}\left[\lim_{b\to\infty}\left[\text{Ei}\left(-\frac{y}{2}\right)\right]_{x}^{b}\right]\space\text{d}x=$$
$$\lim_{a\to\infty}\int_{0}^{a}\left[\lim_{b\to\infty}\left(\text{Ei}\left(-\frac{b}{2}\right)-\text{Ei}\left(-\frac{x}{2}\right)\right)\right]\space\text{d}x=$$
$$\lim_{a\to\infty}\int_{0}^{a}\left[0-\text{Ei}\left(-\frac{x}{2}\right)\right]\space\text{d}x=$$
$$\lim_{a\to\infty}\int_{0}^{a}\left[-\text{Ei}\left(-\frac{x}{2}\right)\right]\space\text{d}x=$$
$$-\lim_{a\to\infty}\int_{0}^{a}\left[\text{Ei}\left(-\frac{x}{2}\right)\right]\space\text{d}x=2$$
