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I'm going to make use of the series $\displaystyle \sum_{n=0}^\infty \frac 1{n+1}$.

and that $\displaystyle \int_0^\infty \frac{ \sin^2 x} x \, dx = \sum_{n=0}^\infty\int_{n\pi}^{(n+1)\pi} \frac{ \sin^2 x} x \,dx$

If I use variable substitution $t=x-n\pi$ it gives

$$\tag 1 \sum_{n=0}^\infty\int_0^\pi \frac{\sin^2 t }{n\pi+t}dt $$

gives

$$\tag 2\frac 1 \pi\sum_{n=0}^\infty\int_0^\pi \frac{\sin^2 t }{n+1} \, dt$$

$$\tag 3 \frac 1 \pi \int_{n=0}^\pi \sin^2 t \;dt\cdot \sum_{n=0}^\infty \frac 1 {1+n}$$

I don't really know how to explain this or what i have done. If someone knows how to solve this.

  • 0
    You have a $n+1$ when you should have a $n+\dfrac t\pi$ in $(2)$2012-11-04
  • 0
    You might be also interested in this:http://math.stackexchange.com/q/67198/94642012-11-04

1 Answers 1