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What is the value of $\displaystyle\lim_{R\to\infty} {\int_R^{\infty}{r^ne^{-r^2/2}dr}\over{R^{n-1}e^{-R^2/2}}}$, where n is a positive integer? I don't know how to integrate the numerator.

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3 Answers 3

up vote 4 down vote accepted

The quotient we are investigating is of the form $0/0$, and so we may apply L'Hopital's rule. We differentiate the numerator using the second fundamental theorem of calculus and the denominator using the product rule to get that our limit is the same as $$ \lim_{R \to \infty} \frac{-R^ne^{-R^2/2}}{(n-1)R^{n-2}e^{-R^2/2} - R^n e^{-R^2/2}}. $$

Now you can factor $R^ne^{-R^2/2}$ out of numerator and denominator to get $$ \lim_{R \to \infty} \frac{-1}{(n-1)R^{-2} - 1} $$

which is clearly 1.

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+1 Nice and elementary! –  Sasha Apr 23 '12 at 15:02

After a change of variables $u = \frac{r^2}{2}$ the numerator is expressible in terms of the upper incomplete $\Gamma$-function: $$ 2^{\frac{n-1}{2}} \frac{\Gamma_{\frac{n+1}{2}}\left(\frac{R^2}{2}\right)}{ R^{n-1} \exp\left(-\frac{R^2}{2} \right)} $$ For odd $n = 2m+1$ this can be expressed in elementary functions: $$ \frac{2^m}{R^{2m}} \sum_{k=0}^m \frac{m!}{k!} \left( \frac{R^2}{2} \right)^k $$ For even $n$ the answer will involve the incomplete error function.

See the section on asymptotic expansion of the incomplete $\Gamma$-function: $$ \Gamma_{\frac{n+1}{2}}\left(\frac{R^2}{2}\right) \sim \exp\left(-\frac{R^2}{2} \right) \left( \frac{R^2}{2} \right)^{\frac{n-1}{2}} $$ Thus the large $R$ limit equals to 1.

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The question is asking for the limit as $R \to \infty$. :) –  cardinal Apr 22 '12 at 20:27
Thanks @cardinal. I expanded my answer and corrected a typo. –  Sasha Apr 22 '12 at 20:41

We assume $n\geq 2$, for $n=1$ the integral is computable. We integrate by parts: \begin{align*} \int_R^{+\infty}r^ne^{-r^2/2}dr&=\left[-r^{n-1}e^{-r^2/2}\right]_R^{+\infty}-\int_R^{+\infty}(n-1)r^{n-2}e^{-r^2/2}dr\\\ &=R^{N-1}e^{-R^2/2}-(n-1)\int_R^{+\infty}r^{n-2}e^{-r^2/2}dr\\\ &=R^{N-1}e^{-R^2/2}-(n-1)\int_1^{+\infty}R(Rs)^{n-2}e^{-R^2s^2/2}ds\\\ &=R^{N-1}e^{-R^2/2}\left(1-(n-1)\int_1^{+\infty}s^{n-2}e^{-R^2s/2}ds\right). \end{align*} Since the map $R\mapsto \int_1^{+\infty}s^{n-2}e^{-R^2(s-1)/2}ds$ is decreasing and $\int_1^{+\infty}s^{n-2}e^{-s/2}ds<\infty$, we have $$\lim_{R\to \infty}\frac{\int_R^{+\infty}r^ne^{-r^2/2}dr}{R^{N-1}e^{-R^2/2}}=1.$$

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Hmmm, Davide: $$\int_R^\infty r^n e^{-r^2/2} \,\mathrm dr \geq R^{n-1} \int_R^\infty r e^{-r^2/2} \,\mathrm dr = R^{n-1} e^{-R^2/2} \>.$$ –  cardinal Apr 22 '12 at 20:14
@cardinal Thanks for pointing this out. Hoping it's correct now. –  Davide Giraudo Apr 22 '12 at 20:51
I think this looks good, except I think you're missing a $< \infty$ on the second integral in the last sentence. –  cardinal Apr 22 '12 at 20:54
@cardinal Thanks, fixed now. –  Davide Giraudo Apr 22 '12 at 20:58
I meant $\int_1^\infty s^{n-2} e^{-s/2} \,\mathrm ds < \infty$; sorry I wasn't very clear. :) (+1 by the way.) –  cardinal Apr 22 '12 at 21:00

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