Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Thanks in advance.

share|improve this question
2  
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
3  
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
add comment

1 Answer

up vote 2 down vote accepted

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...

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.