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?
$\newcommand{\Res}{\text{Res}}\newcommand{\Log}{\operatorname{Log}}$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!