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I was playing around with the function $\dfrac{1}{1+x^2}$, and knowing that the integral over $(0,\infty)$ was $\dfrac{\pi}{2}$, I was hoping to see if there was some neat pattern to determining the value of the following generalized form: $$I=\int_0^\infty \frac{dx}{1+x^n}$$ for natural $n\ge2$. I ran some computations in Mathematica and got a consistent result of $\dfrac{\pi}{n}\csc\left(\dfrac{\pi}{n}\right)$, which also happens to be equivalent to $\dfrac{1}{n}\Gamma\left(\dfrac{1}{n}\right)\Gamma\left(1-\dfrac{1}{n}\right)=\mathrm{B}\left(\dfrac{1}{n},1-\dfrac{1}{n}\right)$.

Is there some neat trick I can use to get this result? I've considered integrating along a contour, but I'm somewhat rusty with complex methods. Also, the result was posted previously here.

I tried something like this instead: $$\frac{1}{1+x^n}=\begin{cases}\displaystyle \sum_{k=0}^\infty(-1)^kx^{nk}&\text{for }|x|<1\\\\ \displaystyle\sum_{k=0}^\infty (-1)^k x^{-n(k+1)}&\text{for }|x|>1\end{cases}$$ then integrating along the respective intervals $(0,1)$ and $(1,\infty)$.

Here's my attempt: $$\begin{align*}I &=\left\{\int_0^1+\int_1^\infty\right\}\frac{dx}{1+x^n}\\\\ &=\sum_{k=0}^\infty \left\{\int_0^1 (-1)^kx^{nk}\,dx+\int_1^\infty (-1)^kx^{-n(k+1)}\,dx\right\}\\\\ &=\sum_{k=0}^\infty \left\{(-1)^k\left[\frac{x^{nk+1}}{nk+1}\right]_{0}^{1}+(-1)^k\left[\frac{x^{-n(k+1)+1}}{-n(k+1)+1}\right]_1^\infty\right\}\\\\ &=\sum_{k=0}^\infty \left\{\frac{(-1)^k}{nk+1}-\frac{(-1)^{k}}{1-n(k+1)}\right\}\\\\ &=\left(1-\frac{1}{1-n}\right)+\left(-\frac{1}{n+1}+\frac{1}{1-2n}\right)\\ &\quad\quad+\left(\frac{1}{2n+1}-\frac{1}{1-3n}\right)+\left(-\frac{1}{3n+1}+\frac{1}{1-4n}\right)+\cdots\\\\ &=1-\frac{2}{1-n^2}+\frac{2}{1-4n^2}-\frac{2}{1-9n^2}+\cdots\\\\ &=\sum_{k=0}^\infty \frac{(-2)^k}{1-(kn)^2}\end{align*}$$ However, this series diverges by the ratio test, so I'm not sure if I made a mistake in my calculation or used some flawed reasoning.

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user170231
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    If you're just interested in showing it's equivalent to the beta function representation, that can be done almost immediately. But perhaps you already know this. As it happens, this gives the gamma function representation, and happens to exactly fit into a duplication formula to get the trig representation. – davidlowryduda Jan 11 '15 at 01:55
  • In the last line your denominator should be $(-1)^k2$. Write the sum as $1+\sum_{k=1}^\infty\frac{(-1)^k2}{1-k^2n^2}$. – Tim Raczkowski Jan 11 '15 at 01:56
  • @r9m Here was my reasoning: $\dfrac{1}{1+x^n}=\dfrac{1}{x^n}\cdot\dfrac{1}{1-\left(-x^{-n}\right)}=\sum\limits_{k=0}^\infty (-x^{-n})^k$, then $(-x^n)^k=(-1)^k(x^n)^k=(-1)^kx^{kn}$. Is this wrong? – user170231 Jan 11 '15 at 01:58
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    See here – Mhenni Benghorbal Jan 11 '15 at 02:00
  • @user170231: that is incorrect if $x > 1$. The geometric series expansion only holds for $|x| < 1$ – davidlowryduda Jan 11 '15 at 02:00
  • @user170231 If $|x|>1$, then $1/|x|<1$. So, $\frac{1}{1+x^n}=\frac{1/x^n}{1+1/x^n}$. So geometric series can be used. – Tim Raczkowski Jan 11 '15 at 02:03
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    @mixedmath I actually am only slightly familiar with the beta function and its associated identities, and was not aware of the connection until Mathematica listed it as a result. As for my mistake: I see where I went wrong, thanks everyone. – user170231 Jan 11 '15 at 02:04
  • I realize now I should have done some more research and expected this question to have already been asked before... – user170231 Jan 11 '15 at 02:08
  • This is covered in this question – robjohn Jan 11 '15 at 02:19

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