How to calculate the limit $\lim_{n\rightarrow\infty} n\int^{1}_{0}x^{kn}e^{x^{n}}dx$

Question

Let $k$ be a fixed positive integer. How to calculate the following limit? $$\lim_{n\rightarrow\infty} n\int^{1}_{0}x^{kn}e^{x^{n}}dx$$

Try this technique. Do not forget to up vote the answers if you benefit from them. — Mhenni Benghorbal, May 04 '13 at 20:47

score 2 · Accepted Answer · answered May 04 '13 at 21:03

The change of variable $t=x^n$ yields $$ n\int^{1}_{0}x^{kn}\mathrm e^{x^{n}}\mathrm dx=\int^{1}_{0}t^{k-1}\mathrm e^tt^{1/n}\mathrm dt, $$ and $t^{1/n}\to1$ when $n\to\infty$ hence the LHS converges to $$ I_k=\int^{1}_{0}t^{k-1}\mathrm e^t\mathrm dt. $$ A standard recursion on $k\geqslant1$ then leads to $$ I_k=\mathrm e\,(k-1)!\sum_{i=0}^{+\infty}\frac{(-1)^i}{(k+i)!}. $$ Sanity checks: $I_1=\mathrm e-1$ and $I_k\to0$ when $k\to\infty$.

score 1 · Answer 2 · answered May 04 '13 at 20:14

1

Your integrand is bounded and the terms go to zero pointwise. This is one way to go.

answered May 04 '13 at 20:14

ncmathsadist

49,383

Your integrand is bounded... No, not uniformly, only by $n\mathrm e$. – Did May 04 '13 at 21:04

How to calculate the limit $\lim_{n\rightarrow\infty} n\int^{1}_{0}x^{kn}e^{x^{n}}dx$

2 Answers2