Compute integral $\int_{-\infty}^\infty \frac{x^4}{1+x^8} \, dx$

Question

Compute the integral.

$$\int_{-\infty}^\infty \frac{x^4}{1+x^8} \, dx$$

The answer at the back of the book is $$\frac{\pi}{4\sin(\frac{3\pi}{8})}$$

Answer 1

HINTS

Integrate around the usual semi-circular contour $$C := \{x \in \mathbb{R} : -R \le x \le R\} \cup \{R\mathrm{e}^{\mathrm{i}t} : 0 \le t \le \pi \}$$

Your function has simple poles, for $n=0,1,2,\ldots,7$, at

$$z=\cos\left(\frac{\pi}{8}+\frac{\pi n}{4}\right) + \mathrm{i} \sin \left(\frac{\pi}{8}+\frac{\pi n}{4}\right)$$

You'll need to find the residues of those in the upper-half plane, i.e. within the above contour.

Apply Cauchy's Residue Theorem, take the limit $R \to \infty$, and show that the integral along the circular part of the arc tends to zero as $R$ tends towards infinity.

The key fact is that $\displaystyle{\oint_C \mathrm{f}(z)~\mathrm{d}z= \int_{-R}^R \mathrm{f}(x)~\mathrm{d}x + \int_0^{\pi}}\mathrm{f}\left(R\mathrm{e}^{\mathrm{i}t}\right) \cdot \mathrm{i}R\mathrm{e}^{\mathrm{i}t}~\mathrm{d}t$.

Once $R > 1$ we are clear of all of the poles and so the integral around $C$ does not change; let it equal $L$. If we can show that the integral around the circular arc tends to zero as $R \to \infty$ we have $$L = \lim_{R \to \infty}\oint_C \mathrm{f}(z)~\mathrm{d}z= \int_{-\infty}^{\infty} \mathrm{f}(x)~\mathrm{d}x+ 0$$

Answer 2

As an alternative, integrate over a wedge contour in the first quadrant of radius $R$ and angle $\pi/4$. This contour encloses only one simple pole (at $z=e^{i \pi/8}$), so application of the residue theorem is simplified.

The contour integral is

$$\int_0^R dx \frac{x^4}{1+x^8} + i R \int_0^{\pi/4} d\theta \, e^{i \theta} \frac{R^4 e^{i 4 \theta}}{1+R^8 e^{i 8 \theta}} + e^{i \pi/4} \int_R^0 dt \frac{-t^4}{1+t^8}$$

I leave it to the reader to show that the second integral in fact vanishes as $R \to \infty$. Thus, by the residue theorem we have

$$\left (1+e^{i \pi/4} \right ) \int_0^{\infty} dx \frac{x^4}{1+x^8} = i 2 \pi \frac{e^{i 4 \pi/8}}{8 e^{i 7 \pi/8}}$$

Then,

$$\int_0^{\infty} dx \frac{x^4}{1+x^8} = 2 \pi \frac{e^{i \pi/8}}{8(1+e^{i \pi/4})} = \frac{\pi}{8\cos{(\pi/8)}} = \frac{\pi}{8\sin{(3\pi/8)}} $$

Answer 3

Due to parity, $\displaystyle\int_{-\infty}^\infty\frac{x^4}{1+x^8}~dx=2\int_0^\infty\frac{x^4}{1+x^8}~dx.~$ In general, all integrals of the form

$\displaystyle\int_0^\infty\frac{x^{n-1}}{(1+x^m)^p}~dx$ are solved by letting $t=\dfrac1{(1+x^m)^p},~$ then recognizing the expression

of the beta function in the new integral, and lastly applying Euler's reflection formula for

the $\Gamma$ function, finally arriving at $I=\displaystyle\frac1m\cdot B\bigg(p-\frac nm~,~\frac nm\bigg),~$ which for $p=1$ becomes

$\dfrac\pi m\cdot\csc\bigg(n\dfrac\pi m\bigg),~$ where $\csc x=\dfrac1{\sin x}~.~$ By replacing m and n, and taking into account

the fact that $\sin x=\sin(\pi-x)$, we ultimately get the desired result.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{-\infty}^{\infty}{x^{4} \over 1 + x^{8}} \,\dd x} = 2\int_{0}^{\infty}{x^{4} \over 1 + x^{8}} \,\dd x \\[5mm] \,\,\,\stackrel{x^{8}\ \mapsto\ x}{=}\,\,\, & {1 \over 4}\int_{0}^{\infty}{x^{\color{red}{5/8} - 1} \over 1 + x} \,\dd x \end{align} Note that $\ds{{1 \over 1 + x} = \sum_{k = 0}^{\infty} \color{red}{\Gamma\pars{1 + k}}{\pars{-x}^{k} \over k!}}$. Then, \begin{align} &\!\!\!\!\!\bbox[5px,#ffd]{\int_{-\infty}^{\infty}{x^{4} \over 1 + x^{8}} \,\dd x} = {1 \over 4}\, \Gamma\pars{\color{red}{5 \over 8}} \Gamma\pars{1 -\color{red}{5 \over 8}}\,,\ \pars{\substack{\ds{Ramanujan's}\\[0.5mm] \ds{Master}\\[0.65mm] \ds{Theorem}}} \\[5mm] = &\ \bbx{\pi \over 4\sin\pars{3\pi/8}} \approx 0.8501 \\ & \end{align}

Answer 5

Note

\begin{align} \int_{-\infty}^\infty \frac{x^4}{1+x^8} dx & =-\frac1{2i}\int_{-\infty}^\infty \left(\frac{1}{1+i x^4} - \frac{1}{1-i x^4} \right)dx \\ & = -\text{Im} \int_{-\infty}^\infty \frac{1}{1+i x^4}dx \overset{t=i^{1/4}x} = -\text{Im} \left( e^{-i\frac\pi8}\int_{-\infty}^\infty \frac{1}{1+t^4}dt\right)\\ &= -\text{Im} \left( e^{-i\frac\pi8}\cdot \frac\pi{\sqrt2} \right)= \frac\pi{\sqrt2}\cdot\sin\frac\pi8= \frac\pi2\sqrt{1-\frac1{\sqrt2}} \end{align}