Evaluating $\int_0^{\infty}\frac1{x^9+1}~\mathrm{d}x$

Question

How can we evaluate this integral? $$\int_0^{\infty}\frac1{x^9+1}~\mathrm{d}x$$

Answer 1

Let me extend my comment and obtain the answer for a more general integral $$I(\gamma)=\int_0^{\infty}\frac{dx}{1+x^{\gamma}},\qquad \gamma>1.$$ The change of variables $$y=\frac{1}{1+x^{\gamma}},\qquad x=\left(\frac{1-y}{y}\right)^{1/\gamma},\qquad dx=-\frac{1}{\gamma y^2}\left(\frac{1-y}{y}\right)^{1/\gamma-1}dy$$ transforms it into \begin{align} I(\gamma)&=\frac{1}{\gamma}\int_0^1 y^{-1/\gamma}\left(1-y\right)^{1/\gamma-1}dy=\\ &=\frac1\gamma B\left(1-\frac1\gamma,\frac1\gamma\right)=\\ &=\frac1\gamma\Gamma\left(1-\frac1\gamma\right)\Gamma\left(\frac1\gamma\right)=\\ &=\frac{\pi}{\gamma \sin\frac{\pi}{\gamma}}, \end{align} where at the first step we use the beta function, at the second its expression in terms of gamma functions, and at the third Euler's reflection formula.

Answer 2

How can we evaluate this integral ?

By letting $t=\dfrac1{x^9+1}$ , and recognizing the expression of the beta function in the new integral,

then applying Euler's reflection formula for the $\Gamma$ function to that expression in order to finally

arrive at $\displaystyle\int_0^\infty\frac{x^{n-1}}{x^m+1}dx=\frac\pi m\cdot\csc\bigg(n\cdot\frac\pi m\bigg)$, which for $n=1$ and $m=9$ becomes $\dfrac\pi{9\cdot\sin\dfrac\pi9}$

Answer 3

Using contour integration, in this answer, it is shown that $$ \frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right)=\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x $$ Plugging in $n=0$ and $m=9$ yields $$ \int_0^\infty\frac{\mathrm{d}x}{1+x^9}=\frac\pi9\csc\left(\frac\pi9\right) $$

Answer 4

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} &\color{#00f}{\large\int_{0}^{\infty}{\dd x \over x^{9} + 1}} =\int_{0}^{\infty}\pars{\int_{0}^{\infty}\expo{-\pars{x^{9} + 1}\xi}\,\dd\xi}\,\dd x \\[3mm]&=\int_{0}^{\infty}\expo{-\xi} \pars{\overbrace{\int_{0}^{\infty}\expo{-\xi x^{9}}\,\dd x} ^{\ds{\mbox{Set}\ t \equiv \xi x^{9}\ \imp\ x = \xi^{-1/9}t^{1/9}}}}\,\dd\xi \\[3mm]&=\int_{0}^{\infty}\expo{-\xi}\pars{\int_{0}^{\infty}\expo{-t}\xi^{-1/9}\, {1 \over 9}\,t^{-8/9}\,\dd t}\,\dd\xi \\[3mm]&={1 \over 9}\pars{\int_{0}^{\infty}\xi^{-1/9}\expo{-\xi}\,\dd\xi} \pars{\int_{0}^{\infty}t^{-8/9}\expo{-t}\,\dd t} ={1 \over 9}\,\Gamma\pars{8 \over 9}\Gamma\pars{1 \over 9} \end{align} where $\ds{\Gamma\pars{z}}$ is the Gamma Function.

By using the Euler Reflection Formula ${\bf\mbox{6.1.17}}$: \begin{align} &\color{#00f}{\large\int_{0}^{\infty}{\dd x \over x^{9} + 1}} =\color{#00f}{\large{1 \over 9}\,\pi\,\csc\pars{\pi \over 9}} \end{align}

Answer 5

Denote $a_k= \frac{(2k-1)\pi}9$ and integrate via partial fraction \begin{align} & \int_{0}^{\infty} \frac{1}{1+x^{9}} dx =\int_{0}^{1} \frac{1+x^{7}}{1+x^{9}}dx \\ =&\int_0^1 \sum_{k=1}^{4} \frac{\frac49\sin^2 a_k}{x^2-2x \cos a_k+1 }\ dx =-\sum_{k=1}^{4} \frac{4\pi k}{81} \sin a_k\\ =& \ \frac d{dt}\bigg(\frac29\sum_{k=1}^{4} \cos a_k t\bigg)_{t=1} =\frac d{dt}\bigg(\frac{\sin 2\pi t}{18\sin\frac{\pi t}9} \bigg)_{t=1} =\frac{\pi}{9} \csc \frac{\pi}{9} \end{align}

Answer 6

Quickest answer, without refering to an answer. Lets take the counter clockwise contour on $\Bbb C$ with rays $\theta=\frac{2\pi}{9}$ and $\theta=0$. Now, $w=e^{\frac{\pi i}{9}}$ is the only pole of $f(z)=\frac{1}{z^9+1}$ of order $1$ inside the contour. By residue throrem, $$(-w^2+1)I=2\pi i\frac{1}{9w^8}.$$ Hence, $I=\int_0^\infty f(x)dx=\frac\pi 9\csc\frac\pi 9.$