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

Question

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.

Answer 1

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

Answer 2

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 $\infty$.

Answer 3

Since the integrand is an even function, we can rewrite it as \begin{equation} \int_{-\infty}^\infty\dfrac{1}{1+x^4}\ dx=2\int_{0}^\infty\dfrac{1}{1+x^4}\ dx \end{equation} In general, we have (click the formula below for the proof) \begin{equation} \int_0^\infty\dfrac{1}{1+x^n}\ dx=\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)},\qquad\mbox{for}\qquad n>0 \end{equation} For the residue approach, you might refer to this link.

Answer 4

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 $R \to \infty$:

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

Answer 5

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 = $\sqrt{2}*v$

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

Answer 6

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}

Answer 7

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{#c00000}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}