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I'm trying to compute the following integral:

\begin{align} \int_{0}^{\infty} \frac{dx}{1 + x^n}, \quad n \geq 2. \end{align}

Consider the function $f(z) = \frac{1}{1 + z^n}$. Consider the following contour, $\Gamma_n$:

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We then have,

\begin{align} \int_{\Gamma_n} = \int_{0}^{R} \frac{dz}{1 + z^n} + \int_{\Gamma_n^{1}} \frac{dz}{1 + z^n} + \int_{\Gamma_n^{2}} \frac{dz}{1 + z^n}, \end{align}

where $\Gamma_n^{1}$ denotes an arc of the circle, $R e^{i \theta}, \theta \in [0, 2\pi] $. I have shown that the integral along this arc goes to zero as $R$ goes to $\infty$. I'm having trouble with the integral along the arc $\Gamma_n^{2}$. Here's what I have done:

Parametrize $\Gamma_n^{2}$ as,

\begin{align} z(t) = \Big (1 - \frac{t}{R} \Big ) R e^{\frac{2 \pi i}{n}}, \quad t \in [0, R]. \end{align}

We then have,

\begin{align} dz(t) & = - e^{\frac{2 \pi i}{n}} dt, \\ 1 + z(t)^n & = 1 + R^n \Big (1 - \frac{t}{R} \Big )^n = 1 + R^n \sum_{k = 0}^{n} (-1)^k \Big ( \frac{t}{R} \Big )^k \\ & = 1 + R^n - t R^{n-1} + t^2 R^{n-2} - \cdots + (-1)^n t^n. \end{align}

Hence, we have that,

\begin{align} \int_{\Gamma_n^{2}} \frac{dz}{1 + z^n} = - e^{\frac{2 \pi i}{n}} \int_{0}^{\infty} \frac{dt}{1 + R^n - t R^{n-1} + t^2 R^{n-2} - \cdots + (-1)^n t^n}. \end{align}

I'm not sure how to proceed from here. The answer given states that this integral converges to,

\begin{align} - e^{\frac{2 \pi i}{n}} \int_{0}^{\infty} \frac{dx}{1 + x^n}. \end{align}

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    $\begingroup$ Is $\Gamma_n^2$ the straight line from $0$ to $R\exp(2\pi i/n)$? Then that's a jolly peculiar way of parametrising it. How about $t\exp(2\pi i/n)$? $\endgroup$ – Lord Shark the Unknown Jul 17 '18 at 13:09
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    $\begingroup$ Maybe it will be easier to use the parametrization $z(t) = Re^{2\pi i/n} t$, with $t\in [0,1]$, and then invert the boundaries of integration. $\endgroup$ – giobrach Jul 17 '18 at 13:10
  • $\begingroup$ @LordSharktheUnknown I'll try that. $\endgroup$ – user82261 Jul 17 '18 at 13:11
  • $\begingroup$ @giobrach Initially, I used this parameterization, but I wasn't sure how I'll I get the limits from $0$ to $\infty$ in this case. I thought both the limits and the integrand must depend on $R$. $\endgroup$ – user82261 Jul 17 '18 at 13:12
  • $\begingroup$ @user82261 Actually, when you’re trying to prove that an integral on one of the contour components goes to zero, things usually work out much nicer when the limits are fixed. $\endgroup$ – giobrach Jul 17 '18 at 13:18
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Note that $\Gamma_n^1$ is an arc of a circle, not a semicircle (in general).

Besides,\begin{align}\int_{\Gamma_n^2}\frac{\mathrm dz}{1+z^n}&=\int_0^R\frac{-e^{\frac{2\pi i}n}}{1+\left(1-\frac tR\right)^nR^n}\,\mathrm dt\\&=-e^{\frac{2\pi i}n}\int_0^R\frac{\mathrm dt}{1+(R-t)^n}\\&=-e^{\frac{2\pi i}n}\int_0^R\frac{\mathrm dt}{1+t^n}\\&=-e^{\frac{2\pi i}n}\int_0^R\frac{\mathrm dz}{1+z^n}.\end{align}Can you take it from here?

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  • $\begingroup$ Oh, I overlooked that a change of variable would do the trick here. Yes, I can. Thanks. $\endgroup$ – user82261 Jul 17 '18 at 13:22

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