Simplify result of $\int_0^{\infty} \frac{1}{1+x^n}dx$

Question

It is quite easy to show that (by using residue theorem) $$\int_0^{\infty} \frac{1}{1+x^n}dx = \frac{2 \pi i^{1+2/n}}{n(e^{2 \pi i / n} - 1)} $$ for $$n \ge 2$$

Is there any possibility to simplify $$\frac{2 \pi i^{1+2/n}}{n(e^{2 \pi i / n} - 1)}$$ or it is best result?

Thanks in advance!

Instead of $i^{2/n}$, it's probably better to write $e^{\pi i/n}$. Then you might notice that $$\frac{2i e^{\pi i/n}}{e^{2\pi i/n}-1} = \frac{2i}{(e^{2\pi i/n}-1)e^{-\pi i/n}} = \frac{2i}{e^{\pi i/n} - e^{-\pi i/n}} = \frac{1}{\sin \frac{\pi}{n}}.$$ — Daniel Fischer, May 04 '14 at 19:01
possible duplicate of Calculating the integral $\int_0^{\infty}{\frac{\ln x}{1+x^n}}$ using complex analysis — dustin, Mar 30 '15 at 19:53

score 7 · Answer 1 · answered May 04 '14 at 18:47

$\displaystyle\int_0^\infty\frac{dx}{1+x^n}=\frac\pi n\cdot\csc\frac\pi n$ . This can be easily shown by letting $t=\dfrac1{1+x^n}$ , then recognizing the expression of the beta function in the new integral, and then employing Euler's reflection formula for the $\Gamma$ function.

score 3 · Answer 2 · edited Apr 13 '17 at 12:20

3

Let $y = x^n$, then $\displaystyle dx = \frac{y^{(1-n)/n}}{n} dy $ and the integral changes into $$\frac 1 n \int_0^{\infty} \frac{y^{\frac 1 n - 1}}{1+y} dy = \frac{1}{n} \frac{\pi}{\sin \left( \frac \pi n \right)}$$ the integral is evaluated here.

edited Apr 13 '17 at 12:20

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answered May 04 '14 at 18:54

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Simplify result of $\int_0^{\infty} \frac{1}{1+x^n}dx$

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