$\int_0^\infty \frac{\sqrt x}{1+x^4} dx$ by residues 
Evaluate $\int_0^\infty \frac{\sqrt x}{1+x^4} dx$

I think I'm on the right path, but I'm not getting the right answer (which is $\frac{\pi}{4 \cos(\frac{\pi}{8})}$). Here is what I have done:
Define $f(z) = \frac{\sqrt z}{1+z^4}$ on the upper half circle $\alpha$. The singularities inside this half circle are $w_1 = e^{\frac{\pi i}{4}}$ and $w_2 = -e^{\frac{-\pi i}{4}} = - \bar{w_1}$.
Then the integral can be calculated as follows:
$$\oint_\alpha f(z) \,dz = 2\pi i (Res(f,w_1)+Res(f,w_2))$$
Calculating residues:
$Res(f,w_1) = \frac{\sqrt w_1}{4w_1^3} = \frac{e^{\frac{- \pi i}{8}}}{4i}$
$Res(f,w_2) = \frac{\sqrt w_2}{4w_2^3} = \frac{e^{\frac{ \pi i}{8}}}{4}$
From this, I can already see that this won't give me the right answer, since the final answer does not contain $i$ and my $cos$ is in the numerator ($e^{\frac{- \pi i}{8}} + e^{\frac{ \pi i}{8}} = 2cos(\frac{\pi}{8}))$
Some help would be appreciated! May be I made some mistakes calculating the residues? 
 A: Hint: The important thing is that when you choose a semicircle contour to evaluate a real integral, you must make sure that the integral along the semi-circular portion must tend to zero as you make the radius tend to infinity. This is necessary and sufficient that your answer is equal to the sum of residue. 
Showing that these parts go to zero as $R\to\infty$ is usually trivial, and when it is not trivial it almost always amounts to using Jordan's lemma.
A: Hint:
$\int^\infty_{-\infty} \frac{\sqrt{x}}{1+x^4} dx
= \int^\infty_{0} \frac{\sqrt{x}}{1+x^4} dx + \int^0_{-\infty} \frac{\sqrt{x}}{1+x^4} dx
$
Are you accounting for the last term?
A: $\newcommand{\Res}{\text{Res}}\newcommand{\Log}{\operatorname{Log}}$Let me start with the good news: your calculation of the residues are right!
You have taken the upper semi circle contour. I assume you use the principle value of the logarithm , i.e. $\Log(\cdot)$ , to define the complex square root function. After letting $R\to\infty$ everything goes alright (use the ML-lemma), so that you end up with:
\begin{align}
\int_{-\infty}^0 \frac{\sqrt[]{z}}{z^4+1}\,dz+\int_{0}^\infty \frac{\sqrt[]{t}}{t^4+1}\,dt = 2\pi i\sum_{i=1,2} \Res_{w=\omega_i}\frac{\sqrt[]{w}}{w^4+1}
\end{align}
I put there $z$ in the first integral to make sure you see that as a complex integral. Now notice that on the negative real line one has: $$\sqrt[]{z}=e^{\frac 1 2 \Log(z)} = e^{\frac 1 2 \left(\ln |z| - i\pi  \right)} = i \sqrt[]{|z|}$$
So we may write:
\begin{align}
\int_{-\infty}^0 \frac{\sqrt[]{z}}{z^4+1}\,dz = i\int_{-\infty}^0 \frac{\sqrt[]{|t|}}{t^4+1}\,dt = i\int_{0}^\infty \frac{\sqrt[]{t}}{t^4+1}\,dt
\end{align}
Hence by filling in the residues one gets:
\begin{align}
(1+i) \int_{0}^\infty \frac{\sqrt[]{t}}{t^4+1}\,dt = 2\pi i \left(\frac{\exp(-i\pi /8)}{i4} +\frac{\exp(i\pi/8)}{4} \right)
\end{align}
You say this does not look like the answer you must have? Well, notice that $\exp(-i\pi/8) = \exp(-i\pi/2) \exp(3i\pi/8)  = -i \exp(3i\pi/8)$ we get rid off the $i$ in the denominator and get:
\begin{align}
(1+i) \int_{0}^\infty \frac{\sqrt[]{t}}{t^4+1}\,dt &= 2\pi i \left(-\frac{\exp(i3\pi /8)}{4} +\frac{\exp(i\pi/8)}{4} \right) \\
&= 2\pi i \exp(2i\pi/8) \left(-\frac{\exp(i\pi /8)}{4} +\frac{\exp(-i\pi/8)}{4} \right)\\
&= \pi i \exp(i\pi/4) (-i \sin\left (\frac \pi 8\right) ) \\
&= \pi  \sin\left (\frac \pi 8\right) \frac{(1+i)}{\sqrt[]{2}}\\
\end{align}
Deviding both sides with $(1+i)$ yields:
\begin{align}
 \int_{0}^\infty \frac{\sqrt[]{t}}{t^4+1}\,dt = \frac{\pi \sin\left (\frac \pi 8\right)}{\sqrt[]{2}}
\end{align}
You still say this is not the same answer as it is written there? Well,
\begin{align}
\frac{\pi \sin\left (\frac \pi 8\right)}{\sqrt[]{2}} = \frac{\pi \sin\left (\frac \pi 8\right)\cos\left (\frac \pi 8\right)}{\sqrt[]{2} \cos\left (\frac \pi 8\right)} = \frac{\pi \sin\left (\frac \pi 4\right)}{2\sqrt[]{2}\cos\left (\frac \pi 8\right)} = \frac{\pi }{4\cos\left (\frac \pi 8\right)}
\end{align}
I don't think it is a problem you get a different looking representation of the same answer. Don't let that stop you from doing things.You have seen that I just did some simplifications that I am comfortable with and I still got an answer that did not look like the right answer.  I mean, who could know beforehand how one should simplify things to get to that particular form? I advice you to look up in WolframAlpha if you sometimes get something that looks different and it is always a nice exercise to get  from the one form to the other form. Have a nice day!
