# Integral of $\frac{x^\alpha}{(1 + x^2)^2}$ from 0 to $\infty$ using contour integration for $-1 < \alpha <3$

I have tried to evaluate $\int_0^\infty \dfrac{x^\alpha}{(x^2 + 1)^2}\ dx$ for $-1 < \alpha < 3$ using the keyhole contour, but am not sure about my working, especially in defining the analytic branch of log $z$ when we consider the function $f(z) = \dfrac{z^\alpha}{(z^2 + 1)^2}$ where $z^\alpha = e^{\alpha\text{log}\ z}$. Below is my working:

Using the keyhole contour $K_{\varepsilon,M}$ and considering the complex function $\dfrac{z^\alpha}{(z^2 + 1)^2}$, we inspect the integral of the function over each part of the contour (namely $C_{\varepsilon},C_M, I_1$ and $I_2$) in the limit $\varepsilon \to 0,\ M \to \infty$. For $C_{\varepsilon}$, we have $|z^2 +1|$ to be the modulus of points lying on a circle of radius $\varepsilon^2$ centered at $z = 1$ and hence $|z^2 + 1| > 1/2$ for $\varepsilon$ small enough (e.g. $\varepsilon^2 < 1/2$). We thus have: \begin{align*} \int_{C_{\varepsilon}} \dfrac{z^\alpha}{(z^2 + 1)^2}\ dz \ll \pi\varepsilon\ \text{max}_{C_\varepsilon}\bigg|\dfrac{z^\alpha}{(z^2 + 1)^2}\bigg| \ll 4\pi\varepsilon^{\alpha + 1} \Rightarrow \lim_{\varepsilon \to 0}\int_{C_{\varepsilon}} \dfrac{z^\alpha}{(z^2 + 1)^2}\ dz = 0 \end{align*} For any $z \in C_M$, we have $|z^2 + 1| \geq |z^2| = M^2$ for $M$ big enough. We thus have: \begin{align*} \int_{C_M} \dfrac{z^\alpha}{(z^2 + 1)^2}\ dz \ll 2\pi M \ \text{max}_{C_M}\bigg|\dfrac{z^\alpha}{(z^2 + 1)^2}\bigg| \ll 2\pi M^{\alpha - 3} \Rightarrow \lim_{M \to \infty}\int_{C_M} \dfrac{z^\alpha}{(z^2 + 1)^2}\ dz = 0 \end{align*} For $I_1$, we have \begin{aligned} = \lim_{\varepsilon \to 0,\ M \to \infty}\int_{I_1}\dfrac{z^\alpha}{(z^2 + 1)^2}\ dz = \int_0^\infty \dfrac{x^\alpha}{(x^2 + 1)^2}\ dx \end{aligned}. As for $I_2$, we have $z^\alpha = e^{\alpha(\text{ln}x + 2\pi i)} = x^{\alpha}e^{2\pi i\alpha}$, and hence we have \begin{aligned} = \lim_{\varepsilon \to 0,\ M \to \infty}\int_{I_2}\dfrac{z^\alpha}{(z^2 + 1)^2}\ dz = -e^{2\pi i\alpha}\int_0^\infty \dfrac{x^\alpha}{(x^2 + 1)^2}\ dx \end{aligned}. Then, by the Residue Theorem, we have: \begin{align*} \int_0^\infty \dfrac{x^\alpha}{(x^2 + 1)^2}\ dx &= \frac{2\pi i}{1 - e^{2\pi i\alpha}}\sum_k \text{Res}\bigg(\dfrac{z^\alpha}{(z^2 + 1)^2}; z_k\bigg) \\ &= \frac{2\pi i}{1 - e^{2\pi i\alpha}}\bigg(\lim_{z \to i} \frac{d}{dz}\bigg(\dfrac{z^\alpha}{(z + i)^2}\bigg) + \lim_{z \to -i} \frac{d}{dz}\bigg(\dfrac{z^\alpha}{(z - i)^2}\bigg)\bigg) \\ &= \frac{2\pi i}{1 - e^{2\pi i\alpha}}\bigg(\frac{\alpha -1}{4}i^{\alpha -3} + \frac{\alpha -1}{4}(-i)^{\alpha -3}\bigg) \\ &= \frac{\pi i(\alpha -1)\text{cos}(\pi(\alpha - 3)/2)}{1 - e^{2\pi i\alpha}} \end{align*}

where the keyhole contour is given as !Keyhole Contour 1

Any help in correcting my working or giving hints or posting your solution is appreciated.

Complex Analysis Approach

Consider the function

$$f(z) = \frac{z^{\alpha}}{(z^2+1)^2}$$

Note that $z^{\alpha} = e^{\alpha \log(z)}$ and choose the branch cut on the imaginary axis where $$\log(z) = \log|z|+ i\theta , \,\,\,\theta \in (-3\pi/2,3\pi/2)$$

Consider the following contour $$\int^{-r}_{-R}f(z) \,dz +\int^{R}_{r}f(z) \,dz+ \int_{C_R}f(z)\,dz+\int_{c_r}f(z)\,dz =2\pi i \mathrm{Res}(f,i)$$

As you showed the integrals on the semi-circles go to zero as $r \to 0, R \to \infty$ hence

$$\int^{0}_{-\infty}\frac{e^{\alpha(\log|x|+i\pi)}}{(x^2+1)^2}\,dx+\int_{0}^{\infty}\frac{e^{\alpha(\log|x|)}}{(x^2+1)^2}\,dx = 2\pi i \mathrm{Res}(f,i)$$

$$(1+e^{\alpha \pi i})\int_{0}^{\infty}\frac{x^{\alpha}}{(x^2+1)^2}\,dx = 2\pi i \mathrm{Res}(f,i)$$

Note that

$$\mathrm{Res}(f,i) = \lim_{z \to i}\frac{d}{dz} \frac{e^{\alpha \log(z)}}{(z+i)^2} = \lim_{z \to i} \frac{\alpha z^{\alpha-1}(z+i)-2z^{\alpha}}{(z+i)^3} = \frac{2e^{i\alpha \pi/2}(\alpha-1)}{-8 i }$$

$$\int_{0}^{\infty}\frac{x^{\alpha}}{(x^2+1)^2}\,dx=\frac{\pi e^{i\alpha \pi/2}(1-\alpha)}{2(1+e^{\alpha \pi i}) } = \frac{\pi (1-\alpha)}{4\cos\left( \frac{\pi}{2}\alpha\right)}$$

Real Analysis Approach

$$\int_{0}^{\infty}\frac{x^{\alpha}}{(x^2+1)^2}\,dx$$

Let $x^2 = t$

\begin{align}\frac{1}{2}\int_{0}^{\infty}\frac{t^{(\alpha-1)/2}}{(t+1)^2}\,dt &= \frac{1}{2}\Gamma \left(\frac{\alpha +1}{2} \right)\Gamma \left(2-\frac{\alpha +1}{2}\right) \tag{1}\\ & = \frac{(1-\alpha)}{4}\Gamma \left(\frac{\alpha +1}{2} \right)\Gamma \left(1-\frac{\alpha +1}{2}\right) \tag{2}\\ & = \frac{\pi (1-\alpha)}{4 \sin\left((1-\alpha)\pi/2 \right)} \tag{3}\\ & = \frac{\pi (1-\alpha)}{4\cos\left( \frac{\pi}{2}\alpha\right) }\tag{4}\end{align}

(1). $$\int^\infty_0 \frac{t^{x-1}}{(1+t)^{x+y}}\,dt = B(x,y)$$ (2). $$\Gamma(x+1) = x \Gamma(x)$$ (3). $$\Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin(\pi x)}$$

• @JackD'Aurizio, thanks Jack. May 22, 2017 at 12:22