Evaluating the integral $\int_0^{\infty}\frac{dx}{\sqrt[4]{x}(1+x^2)}$ using Residue Theorem I need to evaluate the integral $$\int_0^{\infty}\frac{dx}{\sqrt[4]{x}(1+x^2)}$$
I've been given the hint to use the keyhole contour. So I would first choose the principal branch of $\sqrt[4]{\cdot}$, then I have the "keyhole" around $0$, giving me
$$2\pi i\left(\text{Res}_if+\text{Res}_{-i}f\right)= \\ \int_{\gamma_{R,\epsilon}}\frac{dz}{\sqrt[4]{z}(1+z^2)}=\int_{\gamma_R}f(z)dz-\int_{\gamma_\epsilon}f(z)dz+\int^R_{\epsilon} f(z)dz - \int^R_{\epsilon} f(-z)dz$$
($f$ is the integrand) Then taking $R \rightarrow \infty$ and $\epsilon \rightarrow 0$, should give me my result. But I happened to check out the integral in Wolfram Alpha and it gives the integral as $\frac{1}{2}\pi\sec\left(\frac{\pi}{8}\right)$, which is not what I get. Can I get some help? I'm sure I've gone wrong somewhere, and I'm pretty new at using these arguments, so help or insights will be nice. 

Keeping it simple we introduce
$$f(z) = \exp(-(1/4)\mathrm{Log}(z)) \frac{1}{1+z^2}$$
with the  branch cut of  the logarithm on  the positive real  axis and
argument from  $0$ to  $2\pi.$ The  slot of the  keyhole rests  on the
positive real axis and the contour is traversed counter-clockwise. Let
the  segment above  the  real  axis be  $\Gamma_1,$  the large  circle
$\Gamma_2$, the  segment below the  positive real axis  $\Gamma_3$ and
the small circle around the origin $\Gamma_4.$ 
We get for $\Gamma_1$ in the limit
$$J = \int_0^\infty \frac{1}{\sqrt[4]{x}} \frac{1}{1+x^2} \; dx,$$
i.e.  the target integral.  The contribution from the circlular
components vanishes in the limit. We get below the cut on $\Gamma_3$
in the limit
$$\exp(-(1/4)2\pi i)
\int_\infty^0 \frac{1}{\sqrt[4]{x}} \frac{1}{1+x^2} \; dx
\\= - \exp(-(1/2)\pi i)
\int_0^\infty \frac{1}{\sqrt[4]{x}} \frac{1}{1+x^2} \; dx
= iJ.$$
We have for the first residue at the  pole $z=i$
$$\left.\exp((-1/4)\mathrm{Log}(z))\frac{1}{z+i}\right|_{z=i}
= \exp((-1/4)\pi i/2 )\frac{1}{2i}$$
and for the second one
$$\left.\exp((-1/4)\mathrm{Log}(z))\frac{1}{z-i}\right|_{z=-i}
= -\exp((-1/4)3\pi i/2)\frac{1}{2i}.$$
Collecting everything we have
$$(1+i) J = 2\pi i \frac{1}{2i}
(\exp(-\pi i/8) - \exp(-3\pi i/8)).$$
This is
$$J = \pi (\exp(-\pi i/8) - \exp(-3\pi i/8))
\frac{1}{\sqrt{2}} \exp(-\pi i /4)
\\ = \frac{\sqrt{2}}{2} \pi (\exp(-3\pi i/8) - \exp(-5\pi i/8))
\\ = \frac{\sqrt{2}}{2} \pi \exp(-4\pi i/8)
(\exp(\pi i/8) - \exp(-\pi i/8))
\\ = \frac{\sqrt{2}}{2} \pi \exp(-\pi i/2)
\times 2i \sin(\pi/8).$$
The end result is
$$\bbox[5px,border:2px solid #00A000]{
\sqrt{2} \times \pi \times \sin(\pi/8).}$$
Remark. As per the contribution from the circles vanishing, we get
for the large circle $\Gamma_2$ $\lim_{R\to\infty} 2\pi R / R^{1/4} /
R^2 = 0$  and for the small one $\Gamma_4$ $\lim_{\epsilon\to 0} 2\pi
\epsilon / \epsilon^{1/4} / 1 = 0.$
