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$$\int_0^\infty \frac{1}{1+x^{17}}\cdot \frac{1}{1+x^2}\ dx$$

I have checked the solution which used omega (a cube root of unity). But I required a simple answer to my question.

Ѕᴀᴀᴅ
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suraj kumar behera
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    Set $x=\tan y$ and use https://math.stackexchange.com/questions/2518305/mit-integration-bee-2017-problem-int-0-pi-2-frac-1-1-tan2017-x-dx?noredirect=1&lq=1 – lab bhattacharjee Feb 02 '18 at 04:14
  • I suspect that all of the poles $e^{\frac{\pi(2n+1)}{17}i}$ are going to cancel out, leaving just the poles $i,-i$ to evaluate. – Doug M Feb 02 '18 at 04:36

1 Answers1

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$$I=\int^{\infty}_{0}\frac{1}{(1+x^{17})}\cdot \frac{1}{1+x^2}dt\cdots \cdots (+)$$

Put $\displaystyle x=\frac{1}{t}$ and $d =-\frac{1}{t^2}dt$

$$I=\int^{\infty}_{0}\frac{t^{17}}{(1+t^{17})}\cdot \frac{1}{1+t^2}dt$$

$$I=\int^{\infty}_{0}\frac{x^{17}}{(1+x^{17})}\cdot \frac{1}{1+x^2}dt\cdots \cdots (++)$$

So $$2I=\int^{\infty}_{0}\frac{1}{1+x^2}dx$$

DXT
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