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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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    $\begingroup$ Would you settle for expanding $1/(1-x)$ in a power series, and working out $\int_0^1x^k\sin(\pi x)\,dx$? $\endgroup$ – Gerry Myerson Jun 5 '13 at 13:46
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    $\begingroup$ $$\left(\sum_{k=0}^\infty a_k \right) \left(\sum_{k=0}^\infty b_k \right) = \sum_{n=0}^\infty \sum_{k=0}^n a_k b_{n-k}$$ $\endgroup$ – gt6989b Jun 5 '13 at 13:46
  • $\begingroup$ @GerryMyerson +1, an interesting idea :) $\endgroup$ – gt6989b Jun 5 '13 at 13:47
  • $\begingroup$ @gt6989b How did you obtain your result? $\endgroup$ – CAF Jun 5 '13 at 13:49
  • $\begingroup$ @GerryMyerson That is the first thing I tried but kept having to do multiple integ. by parts - I'll try again though. Should that be $$\sum_{k=0}^{\infty} \int_0^1 x^k \sin(\pi x) dx?$$ $\endgroup$ – CAF Jun 5 '13 at 13:52
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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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  • $\begingroup$ That's nice - but how do you prove it algebraically? $\endgroup$ – CAF Jun 5 '13 at 14:19
  • $\begingroup$ @CAF I don't understand. Each $a_ib_j$ term is counted in both sides exactly once, and there are no other terms present. $\endgroup$ – gt6989b Jun 5 '13 at 15:16
  • $\begingroup$ Sorry, I meant how do you prove it without resorting to the above geometric reasoning with the equidistant straight lines. $\endgroup$ – CAF Jun 5 '13 at 15:32
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    $\begingroup$ @CAF Algebraically, you count the terms in both the left-hand side and the right-hand side, noting that the LHS has $a_i b_j$ exactly once for each distinct $i,j$, and no other terms. Similarly, the RHS will include exactly one $a_i b_j$ term for each unique pair $(i,j)$. This is easy to see. If $i+j = N$, it's not possible to find such a term in any element of the outer sum that is not $N$. And in that summation, it will get used exactly once for each $(i,j)$. $\endgroup$ – gt6989b Jun 5 '13 at 18:42

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