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Prove that

$$\lim_{n\to\infty} n^2 \int _0^{1/n}{x^{x+1}}dx = {1\over2}.$$

My Attempt:

When I saw the integral it was tempting to see the integrand as $x^x x$ and to think of $lim_{x\to0}$ $x^x$ which when evaluated using L-Hopital's gives 1, will it be helpful in reaching a desired conclusion?

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  • $\begingroup$ I think applying l'Hopital to $x^x$ at the start is not helpful. What happens if you apply l'Hopital to the whole thing? $\endgroup$
    – GEdgar
    Jun 10, 2022 at 16:38

5 Answers 5

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$$\lim_{n\to\infty} n^2 \int _0^{1/n}{x^{x+1}}dx = \lim_{n\to\infty}\frac{ \int _0^{1/n}{x^{x+1}}dx}{\frac{1}{n^2}}$$ Using L’Hôpital’s and Leibniz’s rules, $$\lim_{n\to\infty}\frac{(\frac {1}{n})^{\frac{n+1}{n}}(\frac{-1}{n^2})}{\frac{-2}{n^3}}=\frac12 \lim_{n\to\infty}\left(\frac{1}{n}\right)^{\left(\tfrac{1}{n}\right)}=\frac12.$$ , using the limit the OP has already mentioned in the question.

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$\newcommand{\d}{\,\mathrm{d}}$If you enforce $x=t/n$, you get: $$\lim_{n\to\infty}n^2\int_0^{1/n}x^{x+1}\d x=\lim_{n\to\infty}\int_0^1t^{1+t/n}n^{-t/n}\d t$$We have: $$0\le t^{1+t/n}n^{-t/n}\le1$$The dominated convergence theorem is then applicable: It is also the case that $1\ge n^{-t/n}\ge n^{-1/n}\to1$ as $n\to\infty$, so in pointwise evaluation of the integrand (justified by DCT) we find: $$\lim_{n\to\infty}\int_0^1t^{1+t/n}n^{t/n}\d t=\int_0^1 t^1\d t=\frac{1}{2}$$

EDIT: A much cleaner way to prove this directly without appellation to the DCT is as follows: $$t\ge t^{1+t/n}n^{-t/n}\ge t^{1+1/n}n^{-1/n}$$On the desired interval, so we have: $$\begin{align}\frac{1}{2}&=n^{1/n}\int_0^1t\d t\ge\int_0^1t^{1+t/n}n^{-t/n}\d t\ge n^{-1/n}\int_0^1t^{1+1/n}\d t\\&= \frac{1}{2+1/n}\cdot n^{-1/n}\end{align}$$From which the squeeze theorem makes an obvious limit of $1/2$.

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From $\frac{x^{x+1}}{x}=x^x=e^{x\ln(x)}$ and $\lim_{x\to 0^+}x\ln(x)=0$ (prove this yourself), it can be seen that $\lim_{x\to 0^+}\frac{x^{x+1}}{x}=1$. It follows that for $\varepsilon>0$ fixed arbitrarily, we can find a $\delta>0$ satisfying

$$\left|\frac{x^{x+1}}{x}-1\right|<2\varepsilon\text{ whenever }0<x<\delta$$

which says the same thing as

$$\begin{equation}(1-2\varepsilon)x<x^{x+1}<(1+2\varepsilon)x\text{ whenever }0<x<\delta\end{equation}\tag{$\ast$}$$

Pick $N\in\mathbb N$ sufficiently large to get $\frac{1}{N}\leq\delta$. Then for any $n\in\mathbb N$ with $n>N$, we have $0<\frac{1}{n}<\delta$. Integrating the inequality $(\ast)$ over $[0,1/n]$ gives

$$\int_0^{\frac{1}{n}}(1-2\varepsilon)xdx<\int_0^{\frac{1}{n}}x^{x+1}dx<\int_0^{\frac{1}{n}}(1+2\varepsilon)xdx\text{ for every }n>N$$

and consequently, after evaluating the left and rightmost integrals,

$$\left(\frac{1}{2}-\varepsilon\right)\frac{1}{n^2}<\int_0^{\frac{1}{n}}x^{x+1}dx<\left(\frac{1}{2}+\varepsilon\right)\frac{1}{n^2}\text{ for every }n>N$$

Multiplying through by $n^2$ and then subtracting $1/2$ yields

$$-\varepsilon<n^2\int_0^{\frac{1}{n}}x^{x+1}dx-\frac{1}{2}<\varepsilon\text{ for every }n>N$$

or

$$\left|n^2\int_0^{\frac{1}{n}}x^{x+1}dx-\frac{1}{2}\right|<\varepsilon\text{ for every }n>N$$

Since $\varepsilon>0$ was arbitrary, it follows that

$$\lim_{n\to\infty}n^2\int_0^{\frac{1}{n}}x^{x+1}dx=\frac{1}{2}$$

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Since you already recived good answers for the limit, let us try to have asymptotics.

We have, by expansion around $x=0$, $$x^{x+1}=\sum_{k=1}^\infty \frac 1{(k-1)!}x^k\,\big[\log (x)\big]^{k-1} $$ Now, consider the integrals $$I_k=\int_0^{\frac 1{n}}x^k\,\big[\log (x)\big]^{k-1}\,dx$$ $$I_1=\frac{1}{2 n^2}\qquad I_2=-\frac{3 \log (n)+1}{9 n^3}\qquad I_3=\frac{8 \log ^2(n)+4 \log (n)+1}{32 n^4}$$ $$n^2 \big[I_1+ I_2+\frac 12 I_3+\cdots\big]=\frac 12-\frac{3 \log (n)+1}{9 n}+\frac{8 \log ^2(n)+4 \log (n)+1}{64 n^2}+\cdots$$

Trying for $n=10$, the above gives $$\frac{28169-1884 \log (10)+72 \log ^2(10)}{57600}=0.42036$$ while numerical integration would give $0.41985$

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If integral limits gap approaches zero, integral is simply area of trapezoid.

Let $ε = 1/n$

$\displaystyle \lim_{ε\to {0^+}} \frac{1}{ε^2} \int _0^ε{x^{x+1}}dx = \lim_{ε\to {0^+}} \frac{1}{ε^2} \left( \frac{ε^{ε+1}+0^1}{2} \right) ε = \lim_{ε\to {0^+}} \frac{ε^ε}{2} = \frac{1}{2} $

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  • $\begingroup$ If integrand not continuous, we might need more trapezoids. Example, 1/ε^2 * integrate(abs(x)^(x+1), x = -ε/2 .. ε) ≈ 1/ε^2 * (((ε/2)^(ε/2+1)+0)/2*(ε/2) + (ε^(ε+1)+0)/2*ε) ≈ 5/8 $\endgroup$ Jul 5, 2022 at 10:38

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