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Possible Duplicate:
Please help me to evaluate $\int\frac{dx}{1+x^{2n}}$.

Is there any trick to evaluate

$$\int_{-\infty}^\infty \frac{{\rm d} x}{x^{2n}+1}?$$

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marked as duplicate by Marvis, amWhy, Belgi, Zev Chonoles Jan 10 '13 at 23:09

This question was marked as an exact duplicate of an existing question.

See here. – Mhenni Benghorbal Jan 11 '13 at 4:28
up vote 0 down vote accepted

This is just a try ...

Since the function is even, we have $2 \int_{0}^\infty \frac{1}{x^{2n}+1}dx$

$$ \int_0^1 \frac{1}{x^{2n}+1} dx + \int_1^\infty\frac{1}{x^{2n}+1} dx \\ = \int_0^1 \sum_{k=0}^\infty (-1)^k (x)^{2kn} + \int_1^\infty \frac{1}{x^{2n}} \sum_{k=0}^\infty (-1)^kx^{-2nk} \\ = \sum_{k=0}^\infty (-1)^k \left[\frac{(x)^{2kn+1}}{2kn+1}\right]_0^1 + \sum_{k=0}^{\infty}(-1)^k \left[ \frac{x^{-2nk-2n+1}}{-2nk-2n+1}\right]_1^\infty \\ $$

So, we have $ $

$$=\sum_{k=0}^\infty(-1)^k \frac{1}{2kn+1} + \sum_{k=0}^\infty (-1)^k \frac{1}{2nk+2n-1} \\ =\sum_{k=0}^\infty (-1)^k \left( \frac{1}{2kn+1} + \frac{1}{2n(k+1) - 1}\right)$$

$$ \int_{-\infty}^\infty \frac{{\rm d} x}{x^{2n}+1}? = 2 \sum_{k=0}^\infty (-1)^k \left( \frac{1}{2kn+1} + \frac{1}{2n(k+1) - 1}\right)$$

Or on complex analysis this is just special case of this problem.

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Try shifting some indices. Then you get $$ 2 \sum_{k=0}^\infty (-1)^k \left( \frac{1}{2kn+1} + \frac{1}{2n(k+1) - 1}\right)$$ $$ 2 \sum_{k=0}^\infty (-1)^k \frac{1}{2kn+1} - 2\sum_{k=0}^\infty (-1)^{k+1} \frac{1}{2n(k+1) - 1}$$ $$ 2 \sum_{k=0}^\infty (-1)^k \frac{1}{2kn+1} - 2\sum_{k=1}^\infty (-1)^{k} \frac{1}{2nk - 1}$$ $$ 2 \sum_{k=0}^\infty (-1)^k \frac{1}{2kn+1} + 2\sum_{k=1}^\infty (-1)^{-k} \frac{1}{2n(-k) + 1}$$ $$2 \sum_{k\in \Bbb Z}^\infty (-1)^k \frac{1}{2kn+1}$$ It seems sums of this form inevitably lead to trigonometric functions. – Pedro Tamaroff Jan 10 '13 at 23:26
@PeterTamaroff thanks for the hint!! – Santosh Linkha Jan 10 '13 at 23:27
@PeterTamaroff how do you know it's a trigonometric function? could you give me any links? – Santosh Linkha Jan 10 '13 at 23:32
See the duplicate of this. It is a "partial fraction decomposition" sum over the "singularities" of the cosecant of $\pi/2n$. – Pedro Tamaroff Jan 10 '13 at 23:35
@PeterTamaroff awesome!! thanks ... – Santosh Linkha Jan 10 '13 at 23:40

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