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<math>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}</math> | <math>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}</math> | ||
| + | |||
| + | Now integrate both sides from 0 to 1 with respect to x. | ||
| + | |||
| + | <math>\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</math> | ||
| + | |||
| + | <math>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</math> | ||
Revision as of 10:54, 27 October 2008
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 $
