Find $\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x$ Evaluate $$\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x$$
My attempt:
Consider
$$I=\int_{0}^{1}x^ne^{x^2}$$
using Taylor's series expansion for $e^{x^2}$ we get
$$I=\int_{0}^{1}x^n\left(1+x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+\cdots\right)$$
So we get
$$I=\int_{0}^{1}x^n+\int_{0}^{1}\frac{x^{n+2}}{1!}+\int_{0}^{1}\frac{x^{n+4}}{2!}+\cdots$$
Integrating we get
$$I=\frac{1}{0!(n+1)}+\frac{1}{1!(n+3)}+\frac{1}{2!(n+5)}+\cdots$$
Now we have,
$$\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x=\lim_{n \to \infty}nI$$
Thus we obtain
$$\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x=\lim_{n \to \infty}\frac{n}{0!(n+1)}+\frac{n}{1!(n+3)}+\frac{n}{2!(n+5)}+\cdots=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+..$$
So we get
$$\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x=e$$
is this a valid approach?
 A: I think there is an easier way around this, just integrate by parts to get,
$$\int_{0}^{1} (x^n)'e^{x^2}x dx = e-\int_{0}^{1} (e^{x^2}x)'x^{n}dx$$
and by the dominated convergence theorem the right side tends to $e$ and we are done.
A: Yes. You still have to justify the last step (i.e., the interchange of $\lim\limits_{n\to\infty}$ and the infinite summation), but that's easy, and can be viewed as an application of (a discrete form of) DCT.
Here are two approaches that look easier for me:

*

*Integrate by parts, and then apply DCT: $$\int_0^1 nx^n e^{x^2}\,dx=n\frac{x^{n+1}}{n+1}e^{x^2}\Big|_0^1-\frac{2n}{n+1}\int_0^1 x^{n+2}e^{x^2}\,dx\underset{n\to\infty}{\longrightarrow}e-0=e.$$

*Substitute $x^{n+1}=y$ and (again) apply DCT: $$\int_0^1 nx^n e^{x^2}\,dx=\frac{n}{n+1}\int_0^1 e^{y^{2/(n+1)}}\,dy\underset{n\to\infty}{\longrightarrow}\int_0^1 e^1\,dy=e.$$
A: Here is yet another approach, based on an observation that the integral is concentrated near $1$.
Fix any $0 < \delta < 1 $ and write
$$
I_n := \int\limits_0^1 n x ^n e^{x^2} dx = \int_0^\delta + \int_{1 - \delta}^1 =: I_1 + I_2.
$$
Clearly
$$
\tag{1} I_1 \leq e^{\delta^2} \int\limits_0^\delta n x^n dx = e^{\delta^2} \frac{n}{n+1} \delta^{n+1} \to 0 , \text{ as } n \to \infty.
$$
For the second part, in view of the monotonicity of $e^{x^2}$ we get
$$
e^{(1-\delta)^2} \frac{n}{n + 1} \left( 1 - (1-\delta)^{n+1}\right) = e^{(1-\delta)^2} \int\limits_{1-\delta}^1 n x^n dx\leq I_2 \leq e \int\limits_0^1 n x^n dx = e \frac{n}{n+1}.
$$
Thanks to $(1)$ and taking $n\to \infty$ in the last inequalities, we get that
$$
e^{(1-\delta)^2} \leq \liminf I_n \leq \limsup I_n \leq e
$$
for any $0 < \delta < 1$. Passing to limit in the last expression as $\delta \to 0$ we prove that the limit of $I_n$ exists as $n\to \infty$ and equals $e$.
