Problem's solving the integral $\int_0^\infty \frac{1} {x^{1/3}(x+a)} dx$ where a is a constant

Question

I found this integral

$$ \int_{0}^{\infty} \frac{1} {x^{1/3}(x+a)} \, \mathrm{d}x $$

as part of an example of differentiation under the integral sign in the book Advanced Calculus Explored, by Hamza E. Asamraee. There is only the hint that you can use the substitution:

$$y= x^{1/3}$$

so in the end:

$$ \mathrm{d}x = 3y^{2}\mathrm{d}y$$

And after performing the substitution the integral becomes:

$$ \int_{0}^{\infty} \frac{3y^2} {y(y^3+a)} \, \mathrm{d}y = \int_{0}^{\infty} \frac{3y} {(y^3+a)} \, \mathrm{d}y $$

After this you can factorize the denominator in the form

$$y^3+a = (y+a^{1/3})(y^2-a^{1/3}y+a^{2/3})$$

and use partial fractions so that the integral becomes:

$$ -\frac{1}{a^{1/3}} \int_{0}^{\infty} \frac{\mathrm{d}y}{y+a^{1/3}} + \int_{0}^{\infty} \frac{y+a}{y^2-a^{1/3}y+a^{2/3}} \, \mathrm{d}y $$

And the problem is that these two integrals diverge but in the book or even in wolfram alpha the result is:

$$ \frac{2\pi}{a^{1/3}\sqrt{3}} $$

But I don't know if I made a mistake in the substitution or in which part, so I request for help :c

Answer 1

Hint:

WLOG $a=1$ (you can rescale the variable).

$$\int\left(\frac{y+1}{y^2-y+1}-\frac{1}{y+1}\right)dy=\int\left(\frac{y-\frac12}{y^2-y+1}+\frac{\frac32}{y^2-y+1}-\frac{1}{y+1}\right)dy \\=\ln\frac{\sqrt{y^2-y+1}}{y+1}+\int \frac{\frac32}{y^2-y+1}dy.$$

The first term will cancel out, and the remaining integral is solved by an arc tangent.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ This is a simple application of Ramanujan's Master Theorem: \begin{align} &\bbox[5px,#ffd]{\left.\int_{0}^{\infty} {\dd x \over x^{1/3}\pars{x + a}}\,\right\vert_{\,a\ >\ 0}} = {1 \over a^{1/3}}\int_{0}^{\infty} {x^{\color{red}{2/3} - 1} \over x^{1/3}\pars{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]{\left.\int_{0}^{\infty} {\dd x \over x^{1/3}\pars{x + a}}\,\right\vert_{\,a\ >\ 0}} = {1 \over a^{1/3}}\,\Gamma\pars{\color{red}{2 \over 3}} \Gamma\pars{1 - \color{red}{2 \over 3}} \\[5mm] = &\ {1 \over a^{1/3}}\,{\pi \over \sin\pars{\pi/3}} = {1 \over a^{1/3}}\,{\pi \over \root{3}/2} = \bbx{{2\root{3} \over 3}\,\pi\,a^{-1/3}} \\ & \end{align}