$I_n=\frac{1}{\pi}-\frac{n(n-1)}{\pi^2}I_{n-2}$

Question

I have to calculate $I_n=\int_0^1 x^n \sin{(\pi x)} dx$.

By double partial integration I got $I_n=\frac{1}{\pi}-\frac{n(n-1)}{\pi^2}I_{n-2}$ and $I_0=\frac{2}{\pi}$, $I_1=\frac{1}{\pi}$.

How to solve this recurrence, in order to find $\lim_{n\to\infty} I_n$ and to deduce convergence of $\sum_{n=1}^\infty I_n$?

Any help is welcome. Thanks in advance.

Answer 1

$$(\forall n\in \Bbb N)\;\;\big|I_n\big|=\left|\int_0^1x^n\sin(\pi x)dx\right|$$ $$\le \int_0^1\big|x^n\sin(\pi x)\big|dx$$ $$\le \int_0^1x^ndx=\frac{1}{n+1}$$

$$\implies \lim_{n\to+\infty}I_n=0$$

The recursive relation you got by double partial integration, can also be written as

$$\boxed{I_{n+2}\pi^2=\color{red}{\pi}-(n+2)(n+1)I_n.}$$ $$\lim_{n\to+\infty}I_{n+2}=0\implies$$ $$\implies \lim_{n\to+\infty}(n+2)(n+1)I_n=\color{red}{\pi}$$

$$\implies I_n\sim \frac{\pi}{(n+2)(n+1)}\sim \frac{\pi}{n^2}$$ $$\implies \sum I_n \text{ converges}$$

Answer 2

$\;\sum\limits_{n=1}^\infty I_n=\text{Si}(\pi)-\dfrac2\pi\;$

and as far as I know there is not a simplier way to write the sum of your series.