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

Question

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?

Answer 1

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, $$ \begin{align} \int_0^\infty\frac{x^a}{b^2+x^2}\,\mathrm{d}x &=\frac{\pi}{2}\csc\left(\pi\frac{a+1}{2}\right)b^{1-a}\\ &=\frac{\pi\,b^{1-a}}{2\cos\left(\frac{\pi a}{2}\right)} \end{align} $$

Answer 2

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*}