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This question comes from the MIT Integration Bee 2022 (Semi-Final). It can be simplified as $$\int^\infty_{-\infty}\frac{1}{(x-\cot(x))^2+1}dx$$ or $$\int^\infty_{-\infty}\frac{\sin^2x}{x^2\sin^2x-x\sin2x+1}dx$$ The official website gives the answer is $\pi$. I tried to use Wolfram Alpha to evaluate this integral times $1/\pi$, but it does not give the value "1". So how to tackle this problem? Thank you.

HeyFan
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    Glasser's master theorem ( https://art1lib.org/book/48978411/d4be66 ) $\displaystyle \int_{-\infty}^{\infty}F(u)dx=\int_{-\infty}^{\infty}F(x)dx\tag{}$ where $u=x-\sum_{j=1}^{n-1}\frac{a_j}{x-C_j}$ ; $ a_j $- positive and $ C_j $ - real constants. The theorem is applicable even for the infinite series. $\displaystyle \pi \cot x=\frac{1}{z} +\sum_{n=1}^\infty\Big(\frac{1}{z-\pi n}+\frac{1}{z-\pi n}\Big)\tag{}$ The requirements of the theorem are met, therefore $\displaystyle \int_{-\infty}^\infty\frac{dx}{1+(x-\cot x)^2}=\int_{-\infty}^\infty\frac{dx}{1+x^2}=\pi\tag*{}$ – Svyatoslav Aug 07 '22 at 15:20
  • @Svyatoslav If anyone talks you into converting that into an answer, make sure you don't mix $x$ & $z$ in one equation, as in your formula for the cotangent. – J.G. Aug 07 '22 at 15:29
  • @J.G. - thank you! My apology for the typo. Yes, it has to be $\pi \cot z$ – Svyatoslav Aug 07 '22 at 15:51
  • Thanks for your suggestions. – HeyFan Aug 07 '22 at 16:56
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    This may help:https://math.stackexchange.com/questions/1015462/a-strange-integral-int-infty-infty-dx-over-1-leftx-tan-x-rig/1015739#1015739 – Olivier Oloa Aug 07 '22 at 17:32
  • @OlivierOloa. Long time no speak ! Are you OK ? Cheers :-) – Claude Leibovici Aug 08 '22 at 06:24
  • @ClaudeLeibovici Thank you very much! Je ne suis malheureusement plus très présent sur le site... J'espère que vous allez bien ! Cheers! – Olivier Oloa Aug 08 '22 at 22:20

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