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This problem will help me to solve the very hard problem

$$\int_0^{\infty} \frac {x\sin^{2n+1}x}{x^2+1}dx$$

mrtaurho
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Ahmad Albaw
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    Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :) – mrtaurho Oct 11 '18 at 10:21
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    Downvoter should be care about good questions. s/he should be expert in integration. Moderators must remind downvoter about him/her votes. – Nosrati Oct 11 '18 at 12:02
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    @Nosrati It's not clear what you are saying. However, one can see that this question is a problem statement and no more, which are generally discouraged. See How to ask a good question for more info. – Simply Beautiful Art Oct 11 '18 at 13:08
  • @SimplyBeautifulArt Some questions are too hard such that questioner may not has any idea about them. – Nosrati Oct 11 '18 at 13:22
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    @Nosrati If you read the link, you'll find there's a lot more you can add to the question other than what they've tried. But leaving a question so blank is in very poor taste... – Simply Beautiful Art Oct 11 '18 at 13:24

1 Answers1

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Claim: For any $n\in\Bbb N$, $$J_n:=\int_0^\infty \frac{x\sin^{2n+1}x}{1+x^2}\,\mathrm dx=\frac {(-1)^n\pi}{(2e)^{2n+1}}\sum_{k=0}^n\binom{2n+1}{k}(-e^2)^k.$$

One may use the power-reduction formula

For any $n\in \Bbb N$ and $x\in\Bbb R$, $$\sin^{2n+1}x=\frac 1{2^{2n}}\sum_{k=0}^n(-1)^{n+k}\binom{2n+1}{k}\sin(2n-2k+1)x.\tag{*}$$

Thus, $$J_n=\frac {(-1)^n}{2^{2n}}\sum_{k=0}^n(-1)^{k}\binom{2n+1}{k}I_{2n-2k+1},$$ where $$I_m:=\int_0^\infty \frac{x\sin mx}{1+x^2}\,\mathrm dx=\frac\pi2e^{-m},\tag{**}$$ here is the proof of $(**)$. Therefore the claim follows.

We can further "simplify" the expression using the hypergeometric function by considering \begin{align*} \sum_{k=0}^n\binom{2n+1}{k}x^k=&\sum_{k=0}^{2n+1}\binom{2n+1}{k}x^k-\sum_{k=n+1}^{2n+1}\binom{2n+1}{k}x^k\\ =&(1+x)^{2n+1}-x^{n+1}\sum_{k=0}^{n}\binom{2n+1}{n+k+1}x^k\\ %=&(1+x)^{2n+1}-\binom{2n+1}{n+1}x^{n+1}\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{(n+2)_k}(-x)^k\\ =&(1+x)^{2n+1}-\binom{2n+1}{n+1}x^{n+1} \,_2F_1(-n,1;n+2;-x). \end{align*} We arrive the result

$$J_n=\frac {(-1)^n\pi}{(2e)^{2n+1}}\left((1-e^2)^{2n+1}-\binom{2n+1}{n+1}(-e^2)^{n+1} \,_2F_1(-n,1;n+2;e^2)\right).$$

Tianlalu
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