TITLE: Find $\int_{ - \infty }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x$
QUESTION [7 upvotes]: How can we find the integral:
$$\int_{ - \infty }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x$$
I tried to find and got it to be $\cfrac{\pi}{\sqrt2}$. Am I correct? Please help me with an appropriate method. I tried to use sum of residue.

REPLY [4 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large\int_{-\infty}^{\infty}{\dd x \over x^{4} + 1}}
=2\int_{0}^{\infty}{\dd x \over x^{2} + 1/x^{2}}\,{1 \over x^{2}}\,\dd x\ =\
\underbrace{\overbrace{%
2\int_{0}^{\infty}{1 \over \pars{x - 1/x}^{2} + 2}\,{1 \over x^{2}}\,\dd x}
^{\ds{\color{#c00000}{x\ \mapsto\ {1 \over x}}}}}_{\ds{\color{#00f}{{\cal I}_{0}}}}
\\[5mm]&=2\int_{\infty}^{0}{1 \over \pars{1/x - x}^{2} + 2}\,x^{2}\,
\pars{-\,{\dd x \over x^{2}}}
=\underbrace{2\int_{0}^{\infty}{1 \over \pars{x - 1/x}^{2} + 2}\,\dd x}
_{\ds{\color{#00f}{{\cal I}_{1}}}}
\end{align}

As $\ds{\color{#00f}{{\cal I}_{0}}\ \mbox{and}\ \color{#00f}{{\cal I}_{1}}}$ are equal to the original integral we'll have:

\begin{align}&\color{#66f}{\large\int_{-\infty}^{\infty}{\dd x \over x^{4} + 1}}
={\color{#00f}{{\cal I}_{0}} + \color{#00f}{{\cal I}_{1}} \over 2}
=\int_{0}^{\infty}{1 \over \pars{x - 1/x}^{2} + 2}\,\pars{1 + {1 \over x^{2}}}
\,\dd x
\\[5mm]&=\overbrace{\int_{x\ =\ 0}^{x\ \to\ \infty}{1 \over \pars{x - 1/x}^{2} + 2}
\,\dd\pars{x - {1 \over x}}}
^{\ds{\color{#c00000}{t \equiv x - {1 \over x}}}}
=\int_{-\infty}^{\infty}{\dd t \over t^{2} + 2}
=\root{2}\ \overbrace{\int_{0}^{\infty}{\dd t \over t^{2} + 1}}
^{\ds{\color{#c00000}{\pi \over 2}}}
=\color{#66f}{\large{\root{2} \over 2}\,\pi}
\end{align}