# Find $\int_{ – \infty }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x$

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.

#### Solutions Collecting From Web of "Find $\int_{ – \infty }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x$"

Since the integrand is an even function, we can rewrite it as

\int_{-\infty}^\infty\dfrac{1}{1+x^4}\ dx=2\int_{0}^\infty\dfrac{1}{1+x^4}\ dx

In general, we have (click the formula below for the proof)

For the residue approach, you might refer to this link.

\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}

Consider contour, which is semicircle (origin at $0$ and radius $R$), more precisely:

$$\gamma=\{Re^{it}:t \in [0,\pi] \} \cup [-R,R]$$

According residue theorem:

$$\oint_{\gamma}\frac{1}{x^4+1}dz=2\pi i (\text{Res}_{b}\frac{1}{x^4+1}+\text{Res}_{a}\frac{1}{x^4+1})$$

Where $a=e^{\pi i/4}$ and $b=e^{2\pi i /4}$

But on the other hand:

$$\oint_{\gamma}\frac{1}{x^4+1}dz=\int_{-R}^{R}\frac{1}{x^4+1}dz+\int_{\gamma'(R)}\frac{1}{x^4+1}dz$$

where $\gamma'(R)=\{Re^{it}:t \in [0,\pi] \}$.

Lets try to calculate $\int_{\gamma'(R)}\frac{1}{x^4+1}dz$, for example using simple parametrisation:

$$\int_{\gamma'(R)}\frac{1}{x^4+1}dz=i\int_{0}^{\pi}Re^{it}\frac{1}{1+R^4e^{4it}}dt$$

It’s clear that:

$$\left|Re^{it}\frac{1}{1+R^4e^{4it}}\right| \to 0$$

when $R \to \infty$ for $t \in [0,\pi]$, so when $R \to \infty$:

$$0=\lim_{R \to \infty}\int_{\gamma'(R)}\frac{1}{x^4+1}dz=\lim_{R \to \infty}\oint_{\gamma}\frac{1}{x^4+1}dz-\int_{-R}^{R}\frac{1}{x^4+1}dz= \\ = \oint_{\gamma}\frac{1}{x^4+1}dz-\int_{-\infty}^{\infty}\frac{1}{x^4+1}dz$$

First, use the change of variable $x=\frac{1}{u} => dx = -\frac{1}{u^2}du$

$I = \int_{ – \infty }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x = 2*\int_{ 0 }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x = 2*\int_{ 0 }^{ + \infty } {\frac{u^2} {1 + {u^4}}} \;{\mathrm{d}}u$

=> $2*I = 2*\int_{ 0 }^{ + \infty } {\frac{1+x^2} {1 + {x^4}}} \;{\mathrm{d}}x => I= \int_{ 0 }^{ + \infty } {\frac{1+x^2} {1 + {x^4}}} \;{\mathrm{d}}x$

Now: Let: $t= x-\frac{1}{x}$ => $dt= \frac{1+x^2}{x^2}dx$ x->0, t->$-\infty$ ; x->$+\infty$ , t->$+\infty$

=> $I = \int_{ – \infty }^{ + \infty } {\frac{1+x^2} {1 + {x^4}}*\frac{x^2}{1+x^2}} \;{\mathrm{d}}t = \int_{ – \infty }^{ + \infty } {\frac{1} {x^{-2} + {x^2}}} \;{\mathrm{d}}t$

$x^2 + x^{-2} = (x-\frac{1}{x})^2 +2 = t^2 +2$

=> $I = \int_{ – \infty }^{ + \infty } {\frac{1} {2 + {t^2}}} \;{\mathrm{d}}t$

Let : t = $\sqrt{2}*v$

$I = \frac{1}{2}*\int_{ – \infty }^{ + \infty } {\frac{1} {1 + {(\frac{t}{\sqrt{2}})^2}}} \;{\mathrm{d}}t = \frac{1}{\sqrt{2}}*\int_{ – \infty }^{ + \infty } {\frac{1} {1 + {v^2}}} \;{\mathrm{d}}v = \frac{\pi}{\sqrt{2}}$

use that $1+x^4=- \left( x\sqrt {2}-{x}^{2}-1 \right) \left( x\sqrt {2}+{x}^{2}+1 \right)$ and make a partial fraction decomposition

If I did this problem, I would first prove that $\int_{-\infty}^{\infty}\frac{1}{1+x^4}dx$ converges, then we have

$$\int_{-\infty}^{\infty}\frac{1}{1+x^4}dx=\lim_{R\rightarrow\infty}\int_{-R}^{R}\frac{1}{1+x^4}dx=\lim_{R\rightarrow\infty}\left[\int_{C(R)^+}\frac{1}{1+z^4}dz-\int_{L(R)^+}\frac{1}{1+z^4}dz\right]$$

Where $L(R)$ is the upper part of the circle with radius $R$ and $C(R)$ is the closed contour made by $L(R)$ and the segment $[-R,R]$. Then I would evaluate
$\lim_{R\rightarrow\infty}\int_{C(R)^+}\frac{1}{1+z^4}dz$ by residual. The part $\lim_{R\rightarrow\infty}\int_{L(R)^+}\frac{1}{1+z^4}dz$ may be prove to be $0$ if we notice that $\left|\frac{1}{1+z^4}\right|\le\frac{2}{R^4}$ when $R$ approaches $\infty$.

One may be interested in the following solution:
\begin{eqnarray}
\int_{-\infty}^\infty\frac{1}{1+x^4}dx&=&2\int_{0}^\infty\frac{1}{1+x^4}dx\\
&=&-2\Im\int_{0}^\infty\frac{1}{x^2+i}dx\\
&=&-2\Im\frac{1}{\sqrt{i}}\arctan\frac{x}{\sqrt{i}}\bigg|_0^\infty\\
&=&-2\Im\frac{1}{\sqrt{i}}\frac{\pi}{2}\\
&=&-\pi\Im\frac{1}{\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}}\\
&=&\frac{\pi}{\sqrt{2}}.
\end{eqnarray}