Since we are summing up the \(\displaystyle I_{n}'s\), we can write:
\(\displaystyle \sum_{k=0}^{\infty}\int_{0}^{1}x^{n}sin(\pi x)dx\)
Now, switch the integral and sum (justified through a convergence theorem that I will not get into here).
\(\displaystyle \int_{0}^{1}\sum_{k=0}^{\infty}x^{k}sin(\pi x)dx\)
Take the sum using the closed form for the sum of a geometric series :
\(\displaystyle \int_{0}^{1}\frac{sin(\pi x)}{1-x}\)
This is equivalent to:
\(\displaystyle \int_{0}^{\pi}\frac{sin(x)}{x}dx\)
As an aside:
We could also evaluate the integral by finding the recurrence in terms of n.
\(\displaystyle \int_{0}^{1}x^{n}sin(\pi x)dx\)
By using parts twice, we get:
\(\displaystyle I_{n}=\frac{1}{\pi}-\frac{n(n-1)}{{\pi}^{2}}\cdot I_{n-2}\)
This gives the value of the integral for any value of n, in terms of the previous.
i.e. say we want \(\displaystyle I_{3}=\int_{0}^{1}x^{3}sin(\pi x)dx\)
This would be \(\displaystyle I_{3}=\frac{1}{\pi}-\frac{3(3-1)}{{\pi}^{2}}\cdot I_{1}\)
where \(\displaystyle I_{1}=\int_{0}^{1}xsin(\pi x)dx=\frac{1}{\pi}\)
