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Evaluate $$I=\int_{0}^{\infty} \frac{dx}{1+x^{2017}}$$

I have splitted the integral as:

$$I=\int_{0}^{1} \frac{dx}{1+x^{2017}}+\int_{1}^{\infty} \frac{dx}{1+x^{2017}}=I_1+I_2$$

where $$I_2=\int_{1}^{\infty} \frac{dx}{1+x^{2017}}$$ For $I_2$ i used substitution $x=\frac{1}{v}$ then we get

$$I_2=\int_{1}^{0}\frac{-v^{2015}\:dv}{1+v^{2017}}=\int_{0}^{1}\frac{x^{2015}\:dx}{1+x^{2017}}$$ hence

$$I=I_1+I_2=\int_{0}^{1}\frac{(1+x^{2015})\:dx}{1+x^{2017}}$$

Any clue here?

Umesh shankar
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    There must be an answer in the site for $$\int_{0}^{\infty} \frac{1}{1+x^a} \text{d} x$$ – Ghartal Jun 01 '17 at 17:37
  • @Ghartal Related question. It seems like the answer is $$\int_{0}^{\infty} \frac{1}{1+x^a} \text{d} x=\frac{\pi}{a}\csc\left(\frac \pi {a}\right)$$ – John Doe Jun 01 '17 at 17:39
  • Aha! There you go! – Ghartal Jun 01 '17 at 17:41
  • I'm voting to close this question as off-topic because this is just a particular case of a well-known identity, proved by Lucian here: https://math.stackexchange.com/questions/781186/simplify-result-of-int-0-infty-frac11xndx?noredirect=1&lq=1 – Jack D'Aurizio Jun 01 '17 at 17:43
  • after the hint above we get $$\frac{\pi \csc \left(\frac{\pi }{2017}\right)}{2017}$$ – Dr. Sonnhard Graubner Jun 01 '17 at 17:47
  • @JohnDoe: I was not certain about that, it is not a duplicate strictly speaking, just a particular case of a broader question, so I used the custom template. – Jack D'Aurizio Jun 01 '17 at 17:48
  • @JackD'Aurizio Oh, ok I see. I have chosen duplicate, since the description of that choice said "this has an answer already", which I deemed to be the case here. But I understand your choice too. – John Doe Jun 01 '17 at 17:51

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