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Integrate using integration by parts:

$F(y) = (y+1)e^{-y}$

Find:

Evaluate the $\int_{a=0}^{b=\infty}F(y)\;dy$ using integration by parts.

Thus far, I've distributed the $e^y$ term and split this into two integrals. One of these integrals becomes trivially easy to solve. The Other integral, the integral of $ye^{-y}$, I solved using integration by parts. I think. However, it's possible I'm making some mistakes somewhere.

My answer follows:

$\left[-ye^{-y} - 2e^{-y}\right]_{a=0}^{b=\infty}$

My concern is that when evaluating with the infinity term I encounter an indeterminate form, do I not? Also, I'm curious if I can say that $e^{-y}$ where $y=\infty$ is defined at all. Wouldn't we only be able to talk about what happens in the limit?

Am I thinking correctly about this problem or have I made some fundamental mistake?

Thank you for any and all help!

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    $\begingroup$ Try taking the derivative of the answer that you got. $\endgroup$ – graydad Aug 27 '14 at 22:20
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    $\begingroup$ If you an infinity, then it is an improper integral and you have to take its limit. $\endgroup$ – OGC Aug 27 '14 at 22:21
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    $\begingroup$ Distributing the exponential is unnecessary and just making extra work. Let $u=x+1,dv=e^{-x}dx,du=dx,v=-e^{-x}$. $\endgroup$ – bof Aug 27 '14 at 22:27
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$\int_0^{\infty}(x+1)e^{-x}dx=-(x+1)e^{-x}\Big|_0^{\infty}+\int_0^{\infty}e^{-x}dx=1+1=2$

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Your answer is right. As for evaluating the integral, the limit at $\infty$ does exist, because $e^{-y}$ converges to zero as $y \rightarrow \infty.$ I think you should get an answer of $2$ after evaluating the integral, but you should check!

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