Agaiepsh

2021-11-19

Consider the integral:
${\int }_{0}^{1}\frac{\mathrm{sin}\left(\pi x\right)}{1-x}dx$
I want to do this via power series and obtain an exact solution.
In power series, I have
${\int }_{0}^{1}\left(\sum _{\left\{n=0\right\}}^{\mathrm{\infty }}{\left(-1\right)}^{n}\frac{{\left(\pi x\right)}^{2n+1}}{\left(2n+1\right)!}\cdot \sum _{\left\{n=0\right\}}^{\mathrm{\infty }}\right)dx$
My question is: how do I multiply these summations together? I have searched online, however, in all cases I found they simply truncated the series and found an approximation.

Drood1980

Let's take a more abstract case, trying to multiply . Note that In the resulting sum, we will have ${a}_{i}{b}_{j}$ for all possibilities of i,j $\in \mathbb{N}$.
One way to make it compact is to sum across diagonals. Think about an integer lattice in the first quadrant of ${\mathbb{R}}^{2}$. Drawing diagonals (origin, then along x+y=1 then along x+y=2, etc), note that the one along the line x+y=n will have length n+1 integer points, and the sum of the indices along all points there will be n - i.e.
(n,0),(n−1,1),…,(k,n−k)…,(0,n). So we can renumber the summation based on these diagonals, getting

Onlaceing