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

Question: Prove that $$\lim_{n\to\infty}n^2\int_0^{\frac{1}{n}}x^{x+1}dx=\frac{1}{2}.$$

Solution: Let $$I_n:=n^2\int_0^{\frac{1}{n}}x^{x+1}dx, \forall \in\mathbb{N}.$$ Substituting $nx=t$ in $I_n$, we have $$I_n=n\int_0^1\left(\frac{t}{n}\right)^{1+\frac{t}{n}}dt.$$

Now for all $0\le t\le 1$ and for all $n\in\mathbb{N}, n+t\le n+1\implies 1+\frac{t}{n}\le1+\frac{1}{n}.$ This implies that for all $0\le t\le 1$ and for all $n\in\mathbb{N}$, we have $$\left(\frac{t}{n}\right)^{1+\frac{t}{n}}\ge \left(\frac{t}{n}\right)^{1+\frac{1}{n}}.$$

Therefore, for all $n\in\mathbb{N},$ $$\int_0^1\left(\frac{t}{n}\right)^{1+\frac{t}{n}}dt\ge \int_0^1\left(\frac{t}{n}\right)^{1+\frac{1}{n}}dt=n^{-\left(1+\frac{1}{n}\right)}\frac{n}{2n+1}.$$ This implies that $$I_n\ge n^{-\frac{1}{n}}\frac{n}{2n+1},\forall n\in\mathbb{N}.$$

Next note that for all $0\le t\le 1$ and for all $n\in\mathbb{N}$, $1+\frac{t}{n}>1$, which implies that $\left(\frac{t}{n}\right)^{1+\frac{t}{n}}<\frac{t}{n}.$ Therefore, $$\int_0^1\left(\frac{t}{n}\right)^{1+\frac{t}{n}}dt<\int_0^1\left(\frac{t}{n}\right)dt=\frac{1}{2n}.$$ This implies that $$I_n<\frac{1}{2},\forall n\in\mathbb{N}.$$ Thus, for all $n\in\mathbb{N}$, we have $$n^{-\frac{1}{n}}\frac{n}{2n+1}\le I_n<\frac{1}{2}.$$ Now since $$\lim_{n\to\infty}n^{-\frac{1}{n}}\frac{n}{2n+1}=\frac{1}{2},$$ therefore by Sandwich theorem we can conclude that $$\lim_{n\to\infty}I_n=\frac{1}{2}.$$

Is this solution correct and rigorous enough and is there any other way to solve the problem?

Solution 1:

Here is another way. You should be able to verify that $\lim_{x \to 0^+} x^x = 1$. So, given an $\epsilon > 0$ there exists $N$ with the property that $1-\epsilon < x^x < 1 + \epsilon$ whenever $0 < x < \dfrac 1N$.

If $n \ge N$ then $$ \frac{1-\epsilon}{2n^2} = (1 - \epsilon) \int_0^{1/n} x \, dx < \int_0^{1/n} x^{x+1} \, dx < (1+\epsilon) \int_0^{1/n} x \, dx = \frac{1+\epsilon}{2n^2}$$ which rearranges to $$ - \frac \epsilon 2 < n^2 \int_0^{1/n} x^{x+1} \, dx - \frac 12 < \frac \epsilon 2.$$ Thus
$$n \ge N \implies \left| n^2 \int_0^{1/n} x^{x+1} \, dx - \frac 12 \right| < \epsilon$$ giving you the limit you want.

Solution 2:

On the interval $(0, \frac{1}{n}] \log x>-\infty$, so you can rewrite the integrand as $e^{x\log x}x$. The function $\phi(x)=x \log x $ is monotone decreasing on this interval (easy to check), and it achieves its minimum at $x=\frac{1}{n}:\phi(\frac{1}{n})=-\frac{\log n}{n}$ and maximum at $x=0:\phi(0)=0$. At the same time, $e^{x}$ is a monotone function on this interval, so it maintains the order: if $e^x<e^y \implies x<y$. Therefore, we can take the upper and lower bounds on the integrand: $$ \int_{0}^{\frac{1}{n}}xe^{-\frac{\log n}{n}}dx \le \int_{0}^{\frac{1}{n}}xe^{\phi(x)}dx \leq \int_{0}^{\frac{1}{n}}xdx $$ Upper bound converges to $\frac{1}{2}$ and so does the lower bound, because $n^2$ cancels out, and $e^{-\frac{\log n}{n}} \to_n 1$. By squeeze lemma, the integral converges to $\frac{1}{2}$.

Solution 3:

A little tricky way: Règle de L'Hôpital

In fact, the original question can be transformed into: $$\lim_{n\to\infty}2n^2\int_0^{\frac{1}{n}}x^{x+1}dx=1.$$

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

Use L'Hôpital, the question becomes prove: $$\lim_{n\to\infty}\frac{\frac{1}{n}^{\frac{1}{n}+1} \cdot( -1/n^2)}{-1/n^3}=1.$$ Then $$\lim_{n\to\infty}\frac{1}{n}^{\frac{1}{n}}=1$$ Then $$\lim_{n\to\infty}-\frac{1}{n}\ln{n}=0$$

Then use L'Hôpital again, the equation above is obvious.

Solution 4:

$$ \lim_{n\to\infty}n^2\int_0^{\frac{1}{n}}x^{x+1}\mathrm{d}x = \lim_{n \to \infty} n^2 I_n =\frac{1}{2}. $$ My approach is quite similar to @Sanket. Sandwiching is the basic idea.

We have that $$0 \leqslant x \leqslant \frac{1}{n} \implies 1 \leqslant x + 1 \leqslant 1 + \frac{1}{n} \implies x \geqslant x^{x + 1} \geqslant x^{1 + \frac{1}{n}} $$ $$ \implies n^2 \int_0^{\frac{1}{n}} x~\mathrm{d}x \geqslant n^2 I_n \geqslant n^2 \int_{0}^{\frac{1}{n}} x^{1 + \frac{1}{n}}~\mathrm{d}x \implies \frac{1}{2} \geqslant n^2 I_n \geqslant \frac{1}{n^{1/n}\left(2 + \frac{1}{n}\right)} $$ $$ \implies \frac{1}{2} \geqslant \lim_{n \to \infty} n^2 I_n \geqslant \lim_{n \to \infty} \frac{1}{n^{1/n}\left(2 + \frac{1}{n}\right)} $$ We'll basically use the standard limit $n^{1/n} \to 1$ as $n \to \infty$. Thus: $$ \frac{1}{2} \geqslant \lim_{n \to \infty} n^2 I_n \geqslant \frac{1}{2} $$ Thus by sandwich theorem we have that $\lim\limits_{n \to \infty} n^2 I_n $ exists and equals $\frac{1}{2}$.