Integrating $\int_{-\infty}^\infty \frac{1}{1 + x^4}dx$ with the residue theorem

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $$\bbox[15px,#ffe,border:2px dotted navy]{\ds{% \mbox{There is an interesting 'real integration' which I want to recall here}}} $$ \begin{align} \mc{J} & \equiv \int_{-\infty}^{\infty}{\dd x \over 1 + x^{4}} = 2\int_{0}^{\infty}{1 \over 1/x^{2} + x^{2}}\,{1 \over x^{2}}\,\dd x \\[5mm] & = \overbrace{2\int_{0}^{\infty}{1 \over \pars{x - 1/x}^{2} + 2}\, {1 \over x^{2}}\,\dd x}^{\ds{\mc{J}}}\label{1}\tag{1} \\[5mm] & \stackrel{x\ \mapsto\ 1/x}{=}\,\,\, \underbrace{2\int_{0}^{\infty}{\dd x \over \pars{1/x - x}^{2} + 2}} _{\ds{\mc{J}}} \label{2}\tag{2} \end{align} With \eqref{1} and \eqref{2} RHS: \begin{align} \mc{J} & = {\mc{J} + \,\mc{J} \over 2} = \int_{0}^{\infty}{1 \over \pars{x - 1/x}^{2} + 2}\, \pars{1 + {1 \over x^{2}}}\,\dd x \,\,\,\stackrel{\pars{x - 1/x}\ \mapsto\ x}{=}\,\,\, \int_{-\infty}^{\infty}{\dd x \over x^{2} + 2} \\[5mm] & \,\,\,\stackrel{x/\root{2}\ \mapsto\ x}{=}\,\,\, \root{2}\int_{0}^{\infty}{\dd x \over x^{2} + 1} =\ \bbox[15px,#ffe,border:2px dotted navy]{\ds{{\root{2} \over 2}\,\pi}} \end{align}

I thought that it might be instructive to present an alternative and efficient approach. To do so, we first note that from the even symmetry that

$$\int_{-\infty}^\infty \frac{1}{1+x^4}\,dx=2\int_{0}^\infty \frac{1}{1+x^4}\,dx \tag 1$$

We proceed to evaluate the integral on the right-hand side of $(1)$.

Next, we move to the complex plane and choose as the integration contour, the quarter circle in the upper-half plane with radius $R$. Then, we can write

$$\begin{align} \oint_C \frac{1}{1+z^4}\,dz&=\int_0^R \frac{1}{1+x^4}\,dx+\int_0^\pi \frac{iRe^{i\phi}}{1+R^4e^{i4\phi}}\,d\phi+\int_R^0 \frac{i}{1+(iy)^4}\,dy\\\\ &=(1-i)\int_0^R\frac {1}{1+x^4}\,dx+\int_0^\pi \frac{iRe^{i\phi}}{1+R^4e^{i4\phi}}\,d\phi \tag 2 \end{align}$$

As $R\to \infty$, the second integral on the right-hand side of $(2)$ approaches $0$ while the first approaches the $1/2$ integral of interest. Hence, we have

$$\bbox[5px,border:2px solid #C0A000]{\lim_{R\to \infty}\oint_C \frac{1}{1+z^4}\,dz=\frac{1-i}2 \int_{-\infty}^\infty \frac{1}{1+x^4}\,dx} \tag 3$$

Now, since $C$ encloses only the pole at $z=e^{i\pi/4}$, the Residue Theorem guarantees that

$$\begin{align}\oint_C \frac{1}{1+z^4}\,dz&=2\pi i \text{Res}\left(\frac{1}{1+z^4}, z=e^{i\pi/4}\right)\\\\ &=\frac{2\pi i}{4e^{i3\pi/4}}\\\\ &=\frac{\pi e^{-i\pi/4}}{2}\\\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{\pi(1-i)}{2\sqrt 2}} \tag 4 \end{align}$$

Finally, equating $(3)$ and $(4)$, we find that

$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{1}{1+x^4}\,dx=\frac{\pi}{\sqrt{2}}}$$

as expected.