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How can I integrate $r\,e^{-r/a}$ from $0$ to $\infty$?

I integrated by parts, and then was left with something in an indeterminate form. Is there an alternative to solving this without using L'Hospital's Rule?

EDIT: I flipped the exponent.

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You changed the question! –  Mhenni Benghorbal Sep 26 '13 at 6:24
So what should be in the epxonent? $-\frac{a}{r}$ or $-\frac{r}{a}$? I see two different versions in one question. –  Caran-d'Ache Sep 26 '13 at 6:39
Please can you clear it out: the title says $\frac{−a}{r}$ the body says $\frac{−r}{a}$. –  Caran-d'Ache Sep 26 '13 at 6:46

3 Answers 3

Making the change of variables $r=\frac{1}{t}$ gives rise to the integral

$$ \int _{0}^{\infty }\!{\frac {{{\rm e}^{-at}}}{{t}^{3}}}{dt}. $$

So, can you see if it diverges or converges? Note that, the integrand near zero behaves like

$$ \frac{1}{t^3}. $$

Added: For the new integral, using the change of variables $t=r/a$ and the gamma function

$$ \Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,{\rm d}t, $$


$$ a^2\int _{0}^{\infty }\!t{{\rm e}^{-t}}{dt} = a^2 \Gamma(2)=a^2 1!=a^2. $$

Note: You can use integration by parts with $u=r$ to evaluate he integral without using the gamma function.

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I flipped the exponent by accident! –  Anthony Sep 26 '13 at 6:22
@Tony: You can evaluate it the new one using the gamma function. –  Mhenni Benghorbal Sep 26 '13 at 6:24
I've never used a gamma function.... –  Anthony Sep 26 '13 at 6:26
@Tony: I'll add more material. –  Mhenni Benghorbal Sep 26 '13 at 6:27

This problem looks a lot like a standard template, so that you'd instinctively try $u = r$ and $dv = e^{-a/r}dr$. But there's another way to set it up: $u = e^{-a/r}$, $dv = r\cdot dr$. What happens if you use that instead?

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Then an $r^2$ pops up, right? So I'm not sure why that would help. –  Anthony Sep 26 '13 at 6:15
@Tony, The integration by parts is $uv - \int vdu$. What will this explicitly be? –  Hovercouch Sep 26 '13 at 6:18
I flipped the exponent by accident! I edited it. –  Anthony Sep 26 '13 at 6:23

Remark: The original question was about $\int_0^\infty re^{-a/r}\,dr$.

The improper integral does not exist. The integrand blows up.

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