What is the value of $\displaystyle\lim_{R\to\infty} {\int_R^{\infty}{r^ne^{-r^2/2}dr}\over{R^{n-1}e^{-R^2/2}}}$ 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.
 A: 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.
A: 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.
A: 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 $\int_1^{+\infty}s^{n-2}e^{-s/2}ds&lt\infty$, we have
$$\lim_{R\to \infty}\frac{\int_R^{+\infty}r^ne^{-r^2/2}dr}{R^{N-1}e^{-R^2/2}}=1.$$
