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Why $\lim\limits_{R \to +\infty} \int_{[0,Re^{i2\pi/n}]}\frac{1}{1+z^n}dz$ = $e^{i2\pi/n}$ $\int_0^{+\infty} \frac{1}{1+x^n}dx$

(The integration is on the ray: ${[0,Re^{i2\pi/n}]}$)

As it is shown in the following topic: Show that $\int_0^ \infty \frac{1}{1+x^n} dx= \frac{ \pi /n}{\sin(\pi /n)}$ , where $n$ is a positive integer. in number (3).

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jojox
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    Just parametrise the ray by $z = e^{2\pi i /n}\cdot x$. Note the relation between $dz$ and $dx$ and between $z^n$ and $x^n$. – Daniel Fischer Dec 18 '15 at 12:14
  • But by parametrising the ray by $z = e^{i(2 \pi)/n} . x$, x goes from 0 to 1, i need it to go from 0 to R in order to apply the limit and obtain the integrale of $1/(1+x^n)$ from 0 to +$\infty$.

    And when i try $z = e^{i(2 \pi)/n} . x * \frac1 R$, the second term of the equality doesn't appear easily..

     – jojox Dec 18 '15 at 12:29
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    No, to reach the point $Re^{2\pi i/n}$, we need $x = R$ in that parametrisation. – Daniel Fischer Dec 18 '15 at 12:32
  • Ahh you're right! you used :

    h(t) = $ \frac {b-t} {b-a} z_{A} + \frac {b-t} {b-a} z_{B} $ to parametrize a segment [AB]...

    Thank you !

     – jojox Dec 18 '15 at 12:42

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