Consider the integral: ∫01sin⁡(πx)1−xdx I want to do this via power series and obtain an exact solution. In power series, I have ∫01(∑{n=0}∞(−1)n(πx)2n+1(2n+1)!⋅∑{n=0}∞)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.

Question

Consider the integral:  ∫01sin⁡(πx)1−xdx  I want to do this via power series and obtain an exact solution.  In power series, I have  ∫01(∑{n=0}∞(−1)n(πx)2n+1(2n+1)!⋅∑{n=0}∞)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 · Accepted Answer

Let&#039;s take a more abstract case, trying to multiply  ∑{k=0}∞an  and  ∑{k=0}∞bn. Note that In the resulting sum, we will have aibj for all possibilities of  i,j ∈N.One way to make it compact is to sum across diagonals. Think about an integer lattice in the first quadrant of R2. 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(∑k=0∞an)(∑k=0∞bn)=∑{n=0}∞∑j,k,along  x+y=nakbj=∑{n=0}∞∑{k=0}∞akbn−k

Onlaceing · Accepted Answer

I am trying to solve and it does not work, if you can, then please help

Consider the integral: \int_0^1\frac{sin(\pi x)}{1-x}dx I want to do this via power

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