# Establish $\int_0^{\infty} \frac{x^a}{x^2 + b^2}dx = \frac{\pi b^{a-1}}{2 \cos(\pi a /2)}$ when $-1 < a < 1$

My attempt at a solution: (this is homework, btw)

Let $f(z) = \frac{z^a}{z^2 + b^2}dz$ then the singularities of $f$ occur at $\pm ib$. $$Res(f; ib) = \frac{z^a}{z + ib} \biggr |_{ib} = \frac{(ib)^a}{2ib}$$ $$Res(f; -ib) = \frac{z^a}{z - ib} \biggr |_{-ib} = \frac{(-ib)^a}{-2ib}$$ We sum the residues and multiply by $2 \pi i$ to give the value of a contour integral containing the two poles. $$2 \pi i \cdot \biggl ( \frac{(ib)^a}{2ib} - \frac{(-ib)^a}{2ib} \biggr ) = \pi b^{a-1} (i^a - (-i)^a) = \pi b^{a-1} (e^{\pi i a/2} - e^{- \pi i a/2}) = \pi b^{a-1} 2 i \sin (\pi i a/2)$$ We use a circle contour centered at the origin with the origin cut out (since $a$ can be negative)

Also we make the branch cut for $x^a$ along the positive real axis.

$\gamma = [r, R]$

$\mu_R = Re^{i \theta}$, where $0 < \theta < 2\pi$

$\kappa = [-R, -r]$

$\mu_r = re^{i \theta}$, where $0 < \theta < 2\pi$

$$\pi b^{a-1} 2 i \sin (\pi i a/2) = \int_r^R \frac{x^a}{x^2 + b^2}dx + \int_0^{2\pi} \frac{(Re^{i \theta})^a}{(Re^{i \theta})^2 + b^2}i R e^{i \theta} d \theta + \int_{R}^{r} -\frac{x^a}{x^2 + b^2}dx + \int_{2\pi}^0 \frac{(re^{i \theta})^a}{(re^{i \theta})^2 + b^2}i r e^{i \theta} d \theta$$ As $r \rightarrow 0$ and $R \rightarrow \infty$ we have that the second and fourth integrals go to zero. Thus, $$\pi b^{a-1} 2 i \sin (\pi i a/2) = 2 \int_0^{\infty} \frac{x^a}{x^2 + b^2}dx \implies \int_0^{\infty} \frac{x^a}{x^2 + b^2}dx = \frac{\pi b^{a-1} i \sin (\pi i a/2)}{2}$$

So where did I go wrong?

• For the integral below the real axis, you don't get $x^a$, you get $e^{2\pi ia}x^a$, so overall $(1-e^{2\pi ia})\int_0^\infty \frac{x^a}{x^2+b^2}\,dx$. Also, you wrote $e^{i\pi a/2} - e^{-i\pi a/2} = \sin (\pi i a/2)$, but it actually is $2i\sin (\pi a/2)$. – Daniel Fischer Dec 3 '13 at 23:54
• This looks like a change of variables away from this question or this question. – robjohn Dec 4 '13 at 0:03
• Thanks. I also used $i^a = e^{a\pi i/2}$ and $(-i)^a = e^{-a \pi i/2}$ but this crosses my branch cut. If I instead use $(-i)^a = e^{3a \pi i/2}$ then things work out better, i.e., I can use the double angle sine identity, as in my corrected answer I posted. Thanks again. – user105994 Dec 4 '13 at 1:17

Setting $m=2$ in this answer, it is shown that $$\frac{\pi}{2}\csc\left(\pi\frac{a+1}{2}\right)=\int_0^\infty\frac{x^a}{1+x^2}\,\mathrm{d}x$$ Thus, \begin{align} \int_0^\infty\frac{x^a}{b^2+x^2}\,\mathrm{d}x &=\frac{\pi}{2}\csc\left(\pi\frac{a+1}{2}\right)b^{1-a}\\ &=\frac{\pi\,b^{1-a}}{2\cos\left(\frac{\pi a}{2}\right)} \end{align}

My corrected answer (thanks to Daniel Fischer and robjohn):

We use a circle contour centered at the origin with the origin cut out (since $a$ can be negative).

(Also we make the branch cut for $x^a$ along the positive real axis.)

$\gamma = [r, R]$

$\mu_R = Re^{i \theta}$, where $0 < \theta < 2\pi$

$\kappa = [-R, -r]$

$\mu_r = re^{i \theta}$, where $0 < \theta < 2\pi$ $$\int_{\Gamma} f = \int_r^R \frac{x^a}{x^2 + b^2}dx + \int_0^{2\pi} \frac{(Re^{i \theta})^a}{(Re^{i \theta})^2 + b^2}i R e^{i \theta} d \theta + \int_{R}^{r} \frac{e^{2 \pi i a}x^a}{x^2 + b^2}dx + \int_{2\pi}^0 \frac{(re^{i \theta})^a}{(re^{i \theta})^2 + b^2}i r e^{i \theta} d \theta$$ As $r \rightarrow 0$ and $R \rightarrow \infty$ we have that the second and fourth integrals go to zero. Thus,

$$\int_{\Gamma} f = (1 - e^{2 \pi i a}) \int_0^{\infty} \frac{x^a}{x^2 + b^2}dx$$ Now we calculate the residues of $f$ occur at $\pm ib$. $$Res(f; ib) = \frac{z^a}{z + ib} \biggr |_{ib} = \frac{(ib)^a}{2ib} \qquad Res(f; -ib) = \frac{z^a}{z - ib} \biggr |_{-ib} = \frac{(-ib)^a}{-2ib}$$ We sum the residues and multiply by $2 \pi i$ to give the value of a contour integral containing the two poles. $$2 \pi i \cdot \biggl ( \frac{(ib)^a}{2ib} - \frac{(-ib)^a}{2ib} \biggr ) = \pi b^{a-1} (i^a - (-i)^a) = \pi b^{a-1} (e^{\pi i a/2} - e^{3 \pi i a/2})$$ Then \begin{align*} \int_0^{\infty} \frac{x^a}{x^2 + b^2}dx &= \frac{ \pi b^{a-1} (e^{\pi i a/2} - e^{3 \pi i a/2}) }{1 - e^{2 \pi i a}}\\ &= \frac{ \pi b^{a-1} (e^{-\pi i a/2} - e^{ \pi i a/2}) }{e^{- \pi i a} - e^{ \pi i a}}\\ &= \frac{ \pi b^{a-1} \sin (\pi a/2)}{\sin( \pi a)}\\ &= \frac{ \pi b^{a-1} \sin (\pi a/2)}{\sin( \pi a)}\\ &= \frac{ \pi b^{a-1} \sin (\pi a/2)}{2 \sin( \pi a/2) \cos( \pi a/2)}\\ &= \frac{ \pi b^{a-1}}{2\cos( \pi a/2)}\\ \end{align*}