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I would like to evaluate this weird looking integral a friend showed me:

$\displaystyle \lim_{n \to \infty} n^2\int_0^{1/n} x^{x+1}\, dx$

I thought that Frullani Integral formula could help, I simply do not know where to begin.

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    $\begingroup$ Looks like it tends towards $\frac{1}{2}$. $\endgroup$ – Benedict W. J. Irwin Aug 9 '18 at 15:54
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I`ll propose a different approach using functions sequences:

We can substitute $t=xn$,$dt=ndx$ and we have : $$\lim_{n\to \infty}n^2\int_0^{1/n}x^{x+1}dx=\lim_{n\to \infty}\int_0^{1/n}n^2x^{x+1}dx=\lim_{n\to \infty}\int_0^{1}n^2(\frac t n)^{t/n+1}dx=$$ $$=\lim_{n\to \infty}\int_0^{1}n(\frac t n)^{t/n+1}dt=\lim_{n\to \infty}\int_0^{1}t(\frac t n)^{t/n}dt$$

Looking at the functions sequence $f_n(x)=x(\frac x n)^{x/n} \to x$ pointwise in $[0,1]$.Also, the sequence $f_n$ is monotonically decreasing(with $n$) , continous, and converges pointwise to a continous function in $[0,1]$ so it converges uniformly to $x$,according to Dini`s theorem.

This means, we can move the limit inside the integral:

$$\lim_{n\to \infty}\int_0^{1}t(\frac t n)^{t/n}dt=\int_0^{1}\lim_{n\to \infty}t(\frac t n)^{t/n}dt=\int_0^{1}t dt = \frac 1 2$$

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  • $\begingroup$ This is pretty nice as well, using dominated convergence theorems. Thanks! $\endgroup$ – Hypergeometry Aug 10 '18 at 3:29
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We have: $$L=\lim_{n \to \infty} \frac{\int_0^{1/n} x^{x+1}\, dx}{\frac1{n^2}}=\lim_{k\to 0} \frac{\int_0^k x^{x+1}dx} {k^2}=\frac00$$ By L'H rule we get: $$L=\lim_{k\to 0} \frac{k\cdot k^{k}}{2k}=\frac12$$

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  • $\begingroup$ Thanks! Didn't even think about L'H-rule. The key is to transform the limits to zero I see. $\endgroup$ – Hypergeometry Aug 10 '18 at 3:27
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You can write $$ x^{x+1} = x e^{x \log (x)}= x + x^2 \log(x) + \frac{x^3}{2} \log^2(x) + \cdots $$ Integrating will give you a series $$ I = \int_0^{1/n} x^{x+1} \; dx = \frac{1}{2n^2} + \frac{-1-3 \log(n)}{9n^3} + \frac{1 +4 \log(n) + 8 \log^2(n)}{64n^4} + \cdots $$ for large $n$ the first term dominates. Then $$ n^2 I = \frac{1}{2} + \frac{-1-3 \log(n)}{9n} + \frac{1 +4 \log(n) + 8 \log^2(n)}{64n^2} + \cdots $$ which goes to $1/2$ as $n\to \infty$. Perhaps this is a fluke but it gives more evidence that the answer is $1/2$, which it appears to be from numerical integration and plotting the function.

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    $\begingroup$ I doubt this a fluke. All it needs is a little formalization for interchanging limit and integration, which we get with work (as the other answer shows). As a first heuristic to posit the correct answer, this is the way I would go. $\endgroup$ – Brevan Ellefsen Aug 9 '18 at 22:40

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