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The integral

$$ \int_0^{\infty}\frac{1}{1 + x^{k}} \,{\rm d}x $$ is

  • $\frac{\pi}{2}$ for $k = 2$,
  • $\frac{2\pi}{3\sqrt{3}}$ for $k=3$,
  • $\frac{\pi}{2\sqrt{2}}$ for $k=4$,
  • $\frac{\pi}{3}$ for $k=6$.

Is there a general rule?

Rodrigo de Azevedo
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Nico Schlömer
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    Also https://math.stackexchange.com/q/4370370 – Gary Sep 09 '22 at 09:59
  • You can express the denominator as a product of linear factors, split into partial fractions and combine complex conjugate parts if you want to try doing this yourself. – Mark Bennet Sep 09 '22 at 09:59

1 Answers1

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Use beta integral with Gamma functions as$$2\int_{0}^{\pi/2} \sin^m t \cos^n t dt=\frac{\Gamma(\frac{m+1}{2})\Gamma(\frac{n+1}{2})}{\Gamma(\frac{m+n+2}{2})}.$$ Use $x=\tan^{2/k} t \implies dx=\frac{2}{k} \tan^{3/k-1} \sec^2t dt$. Then $$I_k=\int_{0}^{\infty} \frac{dx}{1+x^k}=\frac{2}{k}\int_{0}^{\pi/2} \sin ^{2/k-1} t \cos^{1-2/k} t dt=\frac{1}{k}\Gamma(1/k)\Gamma(1-1/k)=\frac{\pi}{k}\csc(\pi/k).$$ We have used $\Gamma(z) \Gamma(1-z)=\pi \csc(\pi/z).$

Gary
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Z Ahmed
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    Please don not answer questions that are duplicates. Post your answer (if it is different from existing ones) to the original question instead. – Gary Sep 09 '22 at 10:26