Improper integration involving complex analytic arguments

I am trying to evaluate the following:

$\displaystyle \int_{0}^{\infty} \frac{1}{1+x^a}dx$, where $a>1$ and $a \in \mathbb{R}$

Any help will be much appreciated.

Solution 1:

Here is a complete answer. First, pick an integer $n > a$. Now, consider the change of variables $x^a = t^n$. Then $dx = \frac{n}{a} t^{n/a - 1}$, and so we have the equivalent integral $$ I = \frac{n}{a} \int_0^{\infty} \frac{ t^{n/a - 1} \, dt}{1 + t^n} $$ Now we use contour integration. Let $$ f(z) = \frac{n}{a} \frac{ z^{n/a - 1}}{1 + z^n}$$ defined on the branch cut with $z = e^{\theta i}$ and $-\epsilon < \theta < 2 \pi - \epsilon$ for some small $\epsilon > 0$. Now, consider the "pizza slice" contour $\Gamma$ given by the straight line segment connecting $z = 0$ to $z = R$, then a semicircular contour from $z = R$ to $z = Re^{2\pi i/n}$, then a line segment from $z = Re^{2 \pi i/n}$ to $z = 0$.

The first line segment integrates to $I_1$, the quantity that approaches the desired integral $I$ as $R \rightarrow \infty$. The other line segment, whose line integral is $I_2$, by using a change of variables, can be found to satisfy $$I_2 = - e^{2 \pi i/a} I_1$$ Now, taking $R \rightarrow \infty$, the contribution on the semicircular contour vanishes, and our integral becomes $$ \int_{\Gamma} f(z) \, dz = (1 - e^{2 \pi i /a}) I.$$ Now, we evaluate this directly by residue calculus. As $n > a$, then $f(z)$ is analytic on and inside our contour, except at the point $z_0 = e^{\pi i/n}$, which is a simple pole. Using L'Hopital's rule, we can evaluate the residue: $$ \mathrm{Res} = \lim_{z \rightarrow z_0} (z - z_0) f(z) = \frac{n}{a}\frac{e^{\pi i/a - \pi i/n} }{n e^{\pi i (n-1)/n}} = -\frac{e^{\pi i/a}}{a}$$ Therefore, by the residue theorem, we have $$ -2\pi i\frac{e^{\pi i/a}}{a} = (1 - e^{2 \pi i/a}) I$$ and dividing each side by $-2ie^{\pi i/a}$ yields $$ \frac{\pi}{a} = \sin(\pi/a) I \Longrightarrow I = \frac{\pi}{a} \csc(\pi/a).$$

Solution 2:

Use the change of variables $1+x^\alpha=\frac{1}{t}$ to cast the integral in terms of the beta function

$$ \frac{1}{\alpha}\int_{0}^{1}t^{-1/\alpha}(1-t)^{1/\alpha-1}= \frac{1}{\alpha}\Gamma\left(\frac{1}{\alpha}\right)\Gamma\left(1-\frac{1}{\alpha}\right) $$

Solution 3:

Here's another approach using contour integrals. There are some similarities with @ChristopherA.Wong's answer, though the transformation and contour are completely different.

Let $x = e^{2t/a}$. Then $$\begin{align*} I_1 &= \int_0^\infty \frac{dx}{1+x^a} \\ &= \frac{1}{a} \int_{-\infty}^\infty dt\, \frac{e^{(2/a-1)t}}{\cosh t} \\ &= \lim_{R\to\infty} \frac{1}{a} \int_{\gamma_1} dt\, \frac{e^{(2/a-1)t}}{\cosh t}. \end{align*}$$ The curve $\gamma_1$ is shown below. The integrand has poles at $(2n+1)i \pi/2$ for $n\in\mathbb{Z}$. Consider the integral along the contour, $I_\gamma = \sum_{k=1}^4 I_k.$ By Cauchy's residue theorem $$\begin{align*} I_\gamma &= 2\pi i \, \mathrm{Res}_{t=i\pi/2} \frac{1}{a} \frac{e^{(2/a-1)t}}{\cosh t} \\ &= \frac{2\pi i}{a} \frac{e^{(2/a-1)i\pi/2}}{\sinh i\pi/2} \\ &= -\frac{2\pi i}{a} e^{i\pi/a}. \end{align*}$$ The integrals $I_2$ and $I_4$ are exponentially suppressed in the limit $R\to\infty$ since $2/a-1 < 1$. The integral $I_3$ can be written in terms of $I_1$,
$$\begin{align*} I_3 &= \frac{1}{a}\int_{\gamma_3} dt\, \frac{e^{(2/a-1)t}}{\cosh t} \\ &= -e^{2\pi i/a} I_1. \end{align*}$$ Therefore, in the limit $I_\gamma = I_1 + I_3 = I_1(1-e^{2\pi i/a}) = -2\pi i e^{i\pi/a}/a$ and so $$I_1 = \frac{\pi \csc \pi/a}{a}.$$

Figure 1. The contour for $I_\gamma$.