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I am trying to understand how to evaluate the following integral.

$$\int_{-\infty}^{\infty} \dfrac{1}{x^6+1}dx$$

We start by considering $f(z) = \dfrac{1}{z^6+1}dz.$ Then we find that $z_k = \exp \left (i \left (\frac{\pi}{6}+\frac{\pi k}{3}\right)\right )$, and simple poles occur when $k=0,1,2,3,4,5$.

Then

$$\text{Res}(f,z_k)= \dots =\dfrac{1}{6}\lim_{z \to z_k} z^{-5}= \dfrac{1}{6} \exp \left( -\frac{5 \pi i}{6} (1 +2k) \right)$$

I think that in the missing part, the following formula is used,

$$\text{Res}(f,c)=\frac{1}{(n-1)!}\lim_{z \to c} \frac{d^{n-1}}{dz^{n-1}} ((z-c)^n f(z)) $$

that $n=4$ and that a cancellation occurs (otherwise it would be really messy), but I'm not sure how to do it. How do I proceed?

Thanks for your help.

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This is rather peculiar: the denominator factors to $$ \prod_{k=0}^5 (z-z_k), $$ and the function has no real roots, so you take the three in the upper half-plane and find the residues at their poles. The three roots are simple, so the residue formula in its most basic form should be used, and you find something like $$ 2\pi i \left( \prod_{k=1}^5 \frac{1}{z_0-z_k} + \prod_{\substack{k=0 \\ k \neq 1}}^5 \frac{1}{z_1-z_k} + \prod_{\substack{k=0 \\ k \neq 2}}^5 \frac{1}{z_2-z_k} \right). $$ But as far as I know, there is no simpler way of doing this: you can make various simplifications, but you still end up having to push quite a lot of algebra through.


However, you don't need to do it that way: first, the integrand is symmetric, so the integral is twice the integral over $[0,\infty)$. Now substitute $x=y^{1/6}$, so $$ I = 2 \int_0^{\infty} \frac{1}{6} \frac{y^{-5/6}}{1+y} \, dy $$ There are a variety of ways of doing this integral, probably the easiest being to substitute $u=y/(1+y)$. Then $y= u/(1-u)= 1+\frac{1}{u-1}$, $dy = \frac{1}{(1-u)^2} \, du $, so $$ I = \frac{1}{3} \int_0^1 \left( \frac{u}{1-u} \right)^{-5/6} \frac{1-u}{1-u+u} \frac{1}{(1-u)^2} \, du = \frac{1}{3} \int_0^1 u^{-5/6} (1-u)^{-1/6} \, du = \frac{1}{3}\frac{\Gamma(1/6)\Gamma(5/6)}{\Gamma(1)} = \tfrac{1}{3}\pi\csc{\tfrac{1}{6}\pi}, $$ using the Beta function.

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  • $\begingroup$ Nice. I generalized your second proof to $n > 1$. $\endgroup$ – marty cohen May 12 '15 at 0:14
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I'll try to generalize Chappers's nice result and see if I get anywhere.

Let $J_n =\int_{0}^{\infty} \dfrac{1}{x^n+1}dx $, where $n > 1$. Note that this is half of the integral from $-\infty$ for even $n$, but is nicer to work with.

Let $x = y^{1/n}$, so $dx = \frac1{n}y^{1/n-1}dy$. Then $J_n =\int_{0}^{\infty}\frac1{n} \dfrac{y^{1/n-1}dy}{y+1} $.

As was done, let $u=\frac{y}{1+y}$ so $y = \frac{u}{1-u}$. Then $dy = \frac{1}{(1-u)^2} \, du$ so

$\begin{array}\\ J_n &=\frac1{n}\int_{0}^{1} \dfrac{\left(\frac{u}{1-u}\right)^{1/n-1}du}{(1-u)^2(1+u/(1-u))}\\ &=\frac1{n}\int_{0}^{1} \dfrac{u^{1/n-1}(1-u)^{1-1/n}du}{(1-u)(1-u+u)}\\ &=\frac1{n}\int_{0}^{1} u^{1/n-1}(1-u)^{-1/n}du\\ &=\frac1{n}B(1/n, -1/n+1) \quad\text{(the Beta function)}\\ &=\dfrac{\Gamma(1/n)\Gamma(-1/n+1)}{n\Gamma(1)}\\ &=\dfrac1{n}\Gamma(1/n)\Gamma(-1/n+1)\\ &=\dfrac1{n}\dfrac{\pi}{\sin(\pi/n)}\quad\text{(reflection formula)}\\ &=\dfrac{\pi\csc(\pi/n)}{n}\\ \end{array} $

and it works out fine.

Note that I did almost nothing original here - I replaced $6$ with $n$ and followed Chappers's script.

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    $\begingroup$ I can do you one better: I proved an even more general case here: math.stackexchange.com/questions/1231316/… $\endgroup$ – Chappers May 12 '15 at 0:24
  • $\begingroup$ Very nice. Upvoted. Though in equation (1) you left off the "dx". $\endgroup$ – marty cohen May 12 '15 at 3:15
  • $\begingroup$ Goodness, how embarrassing... fixed. $\endgroup$ – Chappers May 12 '15 at 3:50
  • $\begingroup$ The demon proofreader strikes again. $\endgroup$ – marty cohen May 12 '15 at 3:51
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I would like to help with the algebra which needn't be difficult. Suppose we seek to evaluate $$\int_{-\infty}^{+\infty} \frac{1}{1+x^{2m}} dx$$ which does not have a pole on the real axis at $x=-1.$

I think the approach of using the residues of the poles in the upper half plane is not necessarily the best here, for an alternative consult e.g. this MSE link.

Note that the poles at $\rho_k = \exp(\frac{\pi i}{2m} + \frac{\pi i k}{m})$ are simple and hence the residues are given by

$$\left.\frac{1}{(x^{2m}+1)'}\right|_{x=\rho_k} = \left.\frac{1}{2mx^{2m-1}}\right|_{x=\rho_k} = \left.\frac{x}{2mx^{2m}}\right|_{x=\rho_k} = -\frac{1}{2m}\rho_k.$$

Therefore we get for the sum of the residues $$-\frac{1}{2m} \exp\left(\frac{\pi i}{2m}\right) \sum_{k=0}^{m-1} \exp\left(\frac{\pi i k}{m}\right) = -\frac{1}{2m} \exp\left(\frac{\pi i}{2m}\right) \frac{\exp(\pi i m/m)-1}{\exp(\pi i/m)-1} \\ = \frac{1}{m} \exp\left(\frac{\pi i}{2m}\right) \frac{1}{\exp(\pi i/m)-1} \\ = \frac{1}{m} \frac{1}{\exp(\pi i/2/m)-\exp(-\pi i/2/m)} \\ = \frac{1}{m\times 2i} \frac{1}{\sin(\pi/2/m)}.$$

This yields for the integral $$2\pi i \times \frac{1}{m\times 2i} \frac{1}{\sin(\pi/2/m)} = \frac{1}{m} \frac{\pi}{\sin(\pi/2/m)}.$$

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  • $\begingroup$ What if m is not an integer? My/Chapper's proof does not require this. $\endgroup$ – marty cohen May 12 '15 at 19:06
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You only need to consider the pole at $\exp(i\pi/6)$, because you can consider the contour from $0$ to $R$, from $R$ to $R\exp(i \pi/3)$ and then back to $0$ along the line $z = r\exp(i \pi/3)$ where $r$ goes from $R$ to $0$. In the limit of $R$ to infinity, the contour integral becomes:

$$I_c = \left[1-\exp\left(\frac{i\pi}{3}\right)\right] \int_0^{\infty}\frac{dx}{x^6+1} $$

The residue theorem yields:

$$I_c = 2\pi i\lim_{z\to\exp\left(i\pi/6\right)}\frac{z-\exp\left(\frac{i \pi}{6}\right)}{z^6+1} = \frac{i\pi}{3}\exp\left(-\frac{5i \pi}{6}\right)$$

Therefore

$$\int_{-\infty}^{\infty}\frac{dx}{x^6+1} =\frac{\pi}{3\sin\left(\frac{\pi}{6}\right)}=\frac{2\pi}{3} $$

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  • $\begingroup$ This approach is used at the MSE link that I included below. $\endgroup$ – Marko Riedel May 12 '15 at 18:53
  • $\begingroup$ @MarkoRiedel Yes, I see now. $\endgroup$ – Count Iblis May 12 '15 at 19:02

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