how to find this type of definite double integral?

Question

could any one tell me how to find this type of definite double integral?

$$\int_{0}^{\infty}\int_{x}^{\infty}{e^{{-y\over2}}\over y}dydx$$

Thank you.

Answer 1

An idea:

$$\int\limits_0^\infty\int\limits_0^y\frac{e^{-y/2}}ydxdy=\int\limits_0^\infty e^{-y/2} dy=2\ldots$$

Answer 2

Gah! Don beat me with the same idea while I was writing this, but I'm posting anyway. :)

We can re-arrange the order of integration: $$\begin{align} \int_0^\infty\int_0^y\frac{e^{-y/2}}{y}\;dx\;dy &= \int_0^\infty x\frac{e^{-y/2}}{y}\Bigg|_{x=0}^{x=y} \;dy\\ &=\int_0^\infty y\frac{e^{-y/2}}{y}\;dy \\ &=\int_0^\infty e^{-y/2}\;dy \\ \end{align}$$

This integral can now be computed in a standard Calc II way.