Multiplying two summations together exactly.
Emily Wong 
Consider the integral: $$\int_0^1 \frac{\sin(\pi x)}{1-x} dx$$ I want to do this via power series and obtain an exact solution.
In power series, I have $$\int_0^1 \left( \sum_{n=0}^{\infty} (-1)^n \frac{(\pi x)^{2n+1}}{(2n+1)!} \cdot \sum_{n=0}^{\infty} x^n \right)\,\,dx$$ 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.
Many thanks
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$\begingroup$Let's take a more abstract case, trying to multiply $\sum_{k=0}^\infty a_n$ and $\sum_{k=0}^\infty b_n$. 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. $(n,0),(n-1,1),\ldots,(k,n-k)\ldots,(0,n)$. So we can renumber the summation based on these diagonals, getting
$$ \left(\sum_{k=0}^\infty a_n\right) \left(\sum_{k=0}^\infty b_n \right) = \sum_{n=0}^\infty \sum_{j,k\text{ along } x+y=n} a_k b_j = \sum_{n=0}^\infty \sum_{k=0}^n a_k b_{n-k}. $$
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