Very hard integral limit $$  \lim_{n \to \infty }\int_{0}^{\pi} x^n\sin x \mathrm{d}x$$
I have stumbled across this problem in an old book and havent managed to figure out how to solve it by using basic and intermediate methods. I am a 12th grader. I would appreciate your help!
 A: Over the interval from $\frac{\pi}{2}$ to $\frac{\pi}{2}+\frac{\pi}{3}$, we have $x^n\ge \left(\frac{\pi}{2}\right)^n$, and $\sin x\ge \frac{1}{2}$. Thus our full integral is greater than
$$\frac{\pi}{3}\left(\frac{\pi}{2}\right)^n\frac{1}{2}.$$
Since $\frac{\pi}{2}\gt 1$, it follows that as $n\to\infty$, the integral blows up.
A: Another way, using the symmetry of $\sin$:
$$I_n=\int_{0}^{\pi} x^n\sin x \,dx=\int_{0}^{\pi} (\pi-x)^n\sin x \,dx=\int_{0}^{\pi} \frac{x^n+(\pi-x)^n}{2}\sin x \,dx$$
$$\ge \int_{0}^{\pi} \left(\frac{\pi}{2}\right)^n\sin x \,dx=2\left(\frac{\pi}{2}\right)^n\to\infty$$
A: The antiderivative is
$$\frac 12 i x^{n+1} E_{-n}(-i x)-E_{-n}(i x)+C$$
This makes you definite integral
$$\frac 12 i \pi^{n+1} E_{-n}(-i \pi)-E_{-n}(i \pi)$$
This function is bounded and finite except for $\pi^{n+1}$, which clearly diverges as $n \to \infty$. Thus, we show that the definite integral diverges to infinity. If you want it, here is a WolframAlpha link to the antiderivative
(Note that the antiderivative evaluated at $0$ evaluates to $0$, hence why half of the definite integral disappears)
