# Evaluating $\int_0^\infty \! \!\int_x^\infty \frac{1}{2}e^{-x-\frac{y}{2}} \, \mathrm{d}y\,\mathrm{d}x$

$$\int_0^\infty \! \!\int_x^\infty \frac{1}{2}e^{-x-\frac{y}{2}} \, \mathrm{d}y \, \mathrm{d}x$$ I can't remember how to solve it, mainly because of $-x$ part.

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Does $$\frac12\int_0^\infty \exp(-x)\left(\int_x^\infty \exp\left(-\frac{y}{2}\right)\mathrm dy\right)\mathrm dx$$ look to be easier to evaluate? – J. M. May 1 '12 at 22:18
It seems almost universal among non-mathematicians to speak of "solving" an integral. One solves problems; one solves equations; one evaluates expressions, including integrals. One may solve the problem of evaluating the integral, but if it's phrase that way, why one does with the integral is still called "evaluating", not "solving". – Michael Hardy May 1 '12 at 23:47

First off, by linearity we can pull out a factor (recall that $e^{a+b}=e^ae^b$):
$$\int_0^\infty \int_x^\infty \frac{1}{2}e^{-x-y/2}dydx= \frac{1}{2}\int_0^\infty e^{-x}\left(\int_x^\infty e^{-y/2}dy\right)dx.$$
What's the antiderivative of $e^{-y/2}$ with respect to $y$? Apply the fundamental theorem of calculus...