$\int_0^\infty \frac{1}{1+x^4}dx$ using the Residue Theorem I'm trying to evaluate the integral
$$\int_0^\infty \frac{1}{1+x^4}dx $$
using the Residue Theorem.

My approach:
Let's consider
$$\oint_\Gamma f$$
with $f(z)=\frac{1}{1+z^4}$ and $\Gamma = \Gamma_1 + \Gamma_2$, where:

*

*$\Gamma_1:[-R,R]\rightarrow \mathbb{C}$, with $\Gamma_1(t)=t$

*$\Gamma_2:[0, \pi] \rightarrow \mathbb{C}$, with $\Gamma_2(t)=Re^{it}$
So basically $\Gamma$ is the semicircle centers in the origin with imaginary part greater or equal to zero.
First we need to find the isolated singularities $\alpha_i$ of the function $f$. This singularities are the solution of the equation $1 + z^4 = 0$:

Let's call:

*

*$\alpha_1 = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$

*$\alpha_2 = -\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$

*$\alpha_3 = -\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$

*$\alpha_4 = \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$
So now we have:
$$\oint_\Gamma f = 2 \pi i \sum_i \text{Res}(f,\alpha_i) \text{Ind}_\Gamma(\alpha_i)$$
All singularities are poles of order one, with:
$$\text{Res}(f,\alpha_i)=\frac{1}{1+4\alpha_i^3}$$
So we end up with:
$$\oint_\Gamma f = 2 \pi i \sum_i \frac{\text{Ind}_\Gamma(\alpha_i)}{1+4\alpha_i^3} $$
Because of the shape of our curve $\Gamma$ we have that $\text{Ind}_\Gamma(\alpha_3)=\text{Ind}_\Gamma(\alpha_4)=0$
So we end up with:
$$\oint_\Gamma f = 2 \pi i \underbrace{\left( \frac{1}{1+4\alpha_1^3} + \frac{1}{1+4\alpha_2^3} \right)}_{:=\xi} $$
Now we can work on the left side of this expression:
$$\int_{\Gamma_1} f + \int_{\Gamma_2} f = 2 \pi i \xi$$
We have that:
$$\int_{\Gamma_1}f = \int_{-R}^R \frac{1}{1 + t^4} dt$$
And we also know that
$$\begin{align}
\int_{\Gamma_2}f &\leq \int_0^\pi \left|\frac{Rie^{it}}{1 + R^4e^{4it}} \right| dt
\\
\\ &= \int_0^\pi \frac{R}{\left|1 + R^4e^{4it}\right|}  dt
\\
\\ &= \int_0^\pi \frac{1}{\left| \frac{1}{R} + R^3e^{4it}\right|}  dt
\end{align}$$
If we let $R \to \infty$ we have that $\int_{\Gamma_1} f = \int_{-\infty}^\infty \frac{dt}{1 + t^4}$ and $\int_{\Gamma_2} f = 0$ and because $\int_{-\infty}^\infty \frac{dt}{1 + t^4} = 2 \int_{0}^\infty \frac{dt}{1 + t^4}$, we end up with:
$$\int_{0}^\infty \frac{dt}{1 + t^4} = \pi i \xi$$

The thing is that $\pi i \xi$ is a complex number, so what did I do wrong?
 A: The poles lying inside the contour are $\alpha_1=i\pi/4, \alpha_2=i3\pi/4$.
$Res[f(z),\alpha_1]=\frac{1}{4z^3}\rvert_{\alpha_1}=\frac{1}{4}e^{-3i\pi/4}$
$Res[f(z),\alpha_2]=\frac{1}{4z^3}\rvert_{\alpha_2}=\frac{1}{4}e^{-9i\pi/4}$
$ \text{After using the estimation principle to squeeze the integral on upper part to zero}$
$\therefore I=2\pi i\times \left [\frac{1}{4}e^{-3i\pi/4}+\frac{1}{4}e^{-9i\pi/4}\right ]$
$=\pi i/2\times \left (\frac{-1}{\sqrt 2} -i\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2}-i\frac{1}{\sqrt 2}\right )$
$=\frac{\pi }{\sqrt 2}$
Now you can use the symmetry of integrand function to change the limits to $(0,\infty)$ so that the value of required integral is $\frac{\pi}{2\sqrt 2}$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[15px,#ffd]{\int_{0}^{\infty}{\dd x \over 1 + x^{4}}}}$

I'll consider $\ds{\int_{\mathcal{P}}{\dd z \over 1 + z^{4}}}$ where
$\ds{\mathcal{P}}$ is a contour in the first quadrant. Namely,
$$
\mathcal{P} \equiv
\pars{0,R}\cup R\expo{\ic\pars{0,\pi/2}}\cup\pars{R\expo{\ic\pi/2},0}
\,,\qquad R > 1
$$
There is just one single pole inside $\ds{\mathcal{P}}$: $\ds{\expo{\ic\pi/4}}$.

Then,
\begin{align}
\int_{\mathcal{P}}{\dd z \over 1 + z^{4}} & =
\left. 2\pi\ic\,{1 \over 4z^{3}}\right\vert_{\ z\ =\ \exp\pars{\ic\pi/4}} =
\left. {1 \over 2}\,\pi\ic\,{z \over z^{4}}
\right\vert_{\ z\ =\ \exp\pars{\ic\pi/4}} =
-\,{1 \over 2}\,\pi\ic\expo{\ic\pi/4}
\\[5mm] & =
\int_{0}^{R}{\dd x \over 1 + x^{4}}\ +\
\overbrace{\int_{0}^{\pi/2}{R\expo{\ic\theta}\ic\,\dd\theta \over
1 + R^{4}\expo{4\ic\theta}}}
^{\ds{\stackrel{\mrm{as}\ R\ \to\ \infty}{\LARGE \to}\ \color{red}{\large 0}}}\ +\
\int_{R}^{0}{\expo{\ic\pi/2}\dd r \over 1 + \pars{r\expo{\ic\pi/2}}^{4}}
\\[5mm] & \stackrel{R\ \to\ \infty}{\to}\,\,\,
\pars{1 - \ic}\int_{0}^{\infty}{\dd x \over 1 + x^{4}}
\\[5mm] \implies &
\int_{0}^{\infty}{\dd x \over 1 + x^{4}} =
-\,{1 \over 2}\,{\pi\ic\expo{\ic\pi/4} \over 1 - \ic} =
\bbox[15px,#ffd,border:1px solid navy]{{\root{2} \over 4}\,\pi}\
\approx\ 1.1107
\end{align}

Note that
$\ds{0 < \verts{\int_{0}^{\pi/2}{R\expo{\ic\theta}\ic\,\dd\theta \over
1 + R^{4}\expo{4\ic\theta}}}_{\ R\ > 1} <
{1 \over 2}\,\pi\,{R \over R^{4} - 1}}$
