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. 
 A: Sorry but it could be solved much easily with beta and gamma functions. 
$$\int_0^{\infty} \frac {dx}{\sqrt[4] x (1+x^2)}=\frac 12 B\left(\frac 38,\frac 58\right)=\frac 12\Gamma(3/8)\Gamma(5/8)=\frac 12\pi\csc (3\pi/8)=\frac 12\pi\sec(\pi/8)$$
PS: Please don't downvote just because I used a different method. I just thought to introduce a little bit of advanced integration techniques other than used in complex analysis 
A: Yet another way: we may remove the branch point at the origin by setting $x=z^4$, leading to
$$ I = \int_{0}^{+\infty}\frac{dx}{\sqrt[4]{x}(1+x^2)}=\int_{0}^{+\infty}\frac{4z^2\,dz}{1+z^8}=\int_{0}^{1}\frac{4z^2\,dz}{1+z^8}+\int_{0}^{1}\frac{4z^4}{z^8+1}\,dz$$
then to
$$ I = 4\int_{0}^{1}\frac{(z^2+z^4)(1-z^8)}{1-z^{16}}\,dz=4\sum_{n\geq 0}\left[\tfrac{1}{16n+3}+\tfrac{1}{16n+5}-\tfrac{1}{16n+11}-\tfrac{1}{16n+13}\right].$$
Now the Dirichlet $L$-series appearing in the RHS can be computed by recalling that
$$ \sum_{n\geq 0}\left[\frac{1}{an+b}-\frac{1}{an+(a-b)}\right]=\frac{\pi}{a}\cot\left(\frac{\pi b}{a}\right) $$
holds by the reflection formula for the $\psi$ function / Herglotz trick. We get
$$ I = \frac{\pi}{4}\left[\cot\left(\frac{3\pi}{16}\right)+\cot\left(\frac{5\pi}{16}\right)\right]=\color{red}{\frac{\pi}{\sqrt{2+\sqrt{2}}}}. $$
A: 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 $\Gamma_4$ $\lim_{\epsilon\to 0} 2\pi
\epsilon / \epsilon^{1/4} / 1 = 0.$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{\infty}{\dd x \over
\root[{\large 4}]{x}\pars{1 + x^{2}}}}
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
{1 \over 2}\int_{0}^{\infty}{x^{-5/8} \over 1 + x}\,\dd x
\\[5mm] \stackrel{x + 1\ \mapsto\ x}{=}\,\,\,&\
{1 \over 2}\int_{1}^{\infty}{\pars{x - 1}^{-5/8} \over x}\,\dd x
\,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\,
{1 \over 2}\int_{1}^{0}{\pars{1/x - 1}^{-5/8} \over 1/x}\,
\pars{-\,{\dd x \over x^{2}}}
\\[5mm] = &\
{1 \over 2}\int_{0}^{1}x^{-3/8}\,\pars{1 - x}^{-5/8}\,\dd x =
{1 \over 2}\,{\Gamma\pars{5/8}\Gamma\pars{3/8} \over \Gamma\pars{1}} = {1 \over 2}\,{\pi \over \sin\pars{3\pi/8}}
\\[5mm] = &\
\bbx{{1 \over 2}\pi\sec\pars{\pi \over 8}} \approx 1.7002 \\ &
\end{align}
