Revision as of 10:54, 27 October 2008 by Jhunsber (Talk)

We will start this from the beginning with the series:

$ 1+r+r^2+r^3+...+r^N=\frac{1}{1-r}-\frac{r^{N+1}}{1-r} $

From here we substitute $ r=-x^2 $ to get

$ 1-x^2+x^4-x^6+...+(-1)^Nx^{2N}=\frac{1}{1+x^2}-\frac{(-1)^{N+1}x^{2(N+1)}}{1+x^2} $

Now integrate both sides from 0 to 1 with respect to x.

$ \int_0^11-x^2+x^4-x^6+...+(-1)^Nx^{2N}dx=\int_0^1\frac{1}{1+x^2}dx-\int_0^1\frac{(-1)^{N+1}x^{2(N+1)}}{1+x^2}dx $

$ 1-1/3+1/5-1/7+...+\frac{(-1)^N}{2N+1}=\frac{\pi}{4}+\int_0^1\frac{(-1)^{N}x^{2(N+1)}}{1+x^2}dx $

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