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I am trying to calculate $\displaystyle\int_{0}^{\infty}\frac{x^{n}}{1 + x^{2n}} {\rm d}x$, $n$ being a positive integer greater than $2$:

  • I already got the case where $n$ is $even$ and I am unsure what contour I should use when $n$ is $odd$.
  • I tried calculating numerically a few cases and it seems that the formula I got for $n$ $even$ also applies to $n$ $odd$, so maybe there is some reasoning where I do not have to redo the integral again $?$.
Felix Marin
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user21417
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  • Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc. –  Oct 08 '21 at 03:13
  • What contour did you use when $n$ is even? – Sarvesh Ravichandran Iyer Oct 08 '21 at 03:21
  • I used upper half circle for n even. But for $n$ odd the integrand would be an odd function. – user21417 Oct 08 '21 at 03:22
  • Without doing any work myself, have you tried, say, a quarter circle with the imaginary axis? – A. Thomas Yerger Oct 08 '21 at 03:38
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    I get it, and it makes sense, but for expressions with denominator $\frac 1{1+x^k}$ for some $k$, I'm left thinking about the wedge contour , which is basically a contour that would go around exactly one pole of the denominator. See e.g. this answer for details. – Sarvesh Ravichandran Iyer Oct 08 '21 at 03:39
  • In fact, this is probably a duplicate, see here for some details. The contour details from the answer are missing, but you can see the links and try to reconstruct the answer. – Sarvesh Ravichandran Iyer Oct 08 '21 at 03:42
  • Thank you. I have not seen this contour before. It seems to work for even n as well and the residue is far easier to compute. – user21417 Oct 08 '21 at 04:28

1 Answers1

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Too long for a comment

Let's consider a general case: $$I(m,k)=\int_{0}^{\infty}\frac{x^{m}}{1 + x^k}dx; \,\,m\in (-1;k-2)$$ Then $$\int_{0}^{\infty}\frac{x^{n}}{1 + x^{2n}}dx=I(n,2n)$$

I would recommend to make substitution in the integral first: $$I(m,k)= (t=x^k)\,\,\frac{1}{k}\int_0^\infty\frac{t^{\frac{m+1}{k}-1}}{1+t}dt$$ Then we can easily evaluate the integral via Beta function, or using complex integration: $$I(m,k)= (x=\frac{1}{1+t})\,\,\frac{1}{k}\int_0^1 x^{-\frac{m+1}{k}}(1-x)^{\frac{m+1}{k}-1}dx$$ $$=\frac{1}{k}B\Big(1-\frac{m+1}{k};\frac{m+1}{k}\Big)=\frac{\Gamma(1-\frac{m+1}{k})\Gamma(\frac{m+1}{k})}{k}=\frac{\pi}{k}\,\frac{1}{\sin\frac{\pi(m+1)}{k}}$$ $$I(n;2n)=\frac{\pi}{2n}\frac{1}{\sin\frac{\pi(n+1)}{2n}}=\frac{\pi}{2n}\frac{1}{\cos\frac{\pi}{2n}}$$

If we want to use the complex integration to evaluate $I(m,k)$, we can integrate along the keyhole contour with the cut $[0;\infty]$:

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Let's denote $a=\frac{m+1}{k}-1\,\, (a<0)$, then $$\frac{1}{k}\oint\frac{z^a}{1+z}dz=I(m,k)-e^{2\pi ia}I(m,k)=2\pi iRes_{z=e^{\pi i}}\frac{1}{k}\frac{z^a}{1+z}$$ $$I(m,k)=-\frac{\pi}{k}\frac{1}{\sin \pi a}=\frac{\pi}{k}\,\frac{1}{\sin\frac{\pi(m+1)}{k}}$$

Svyatoslav
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