Integral of $\frac{x}{\sqrt{1+x^5}}$

Question

I am trying to calculate the following integral: $\displaystyle\int_0^\infty \frac{x}{\sqrt{1+x^5}}\, dx$

But I can't seem to find a primitive for that function. I was trying to find a good substitution, but was unable to. Also, attempting to use parts becomes a dead end. What can I do?

there exists no antiderivative in the known elementary functions — Dr. Sonnhard Graubner, Nov 13 '14 at 17:41
@Dr.SonnhardGraubner : If you can prove that either from scratch or by showing how to reduce it to some well known case, then you could post that as an answer. ${}\qquad{}$ — Michael Hardy, Nov 13 '14 at 17:43
Mathematica gives $(2 \Gamma[2/5] \Gamma[11/10])/\sqrt{\pi}$. — Cheerful Parsnip, Nov 13 '14 at 17:44
Interesting... The primitive contains a hypergeometric function — Adnan, Nov 13 '14 at 17:45
I understand. What if I wanted to determine whether that integral converges or diverges? Could I do that without having to calculate the antiderivative? — John, Nov 13 '14 at 17:47
Maple 18 says this here $1/2,{x}^{2}{\mbox{$_2$F$_1$}(2/5,1/2;,7/5;,-{x}^{5})}$ — Dr. Sonnhard Graubner, Nov 13 '14 at 17:49
@John: yes, use that it is bounded above by $\frac{x}{\sqrt{x^5}}$ on $[1,\infty)$. — Cheerful Parsnip, Nov 13 '14 at 17:51
For the interested people, the primitive is: $\frac 12x^2\cdot H\bigg(\frac 25, \frac 12, \frac 75, -x^5\bigg)$, where the $H(a,b,c,d)$ is the hypergeometrical funcion. — Adnan, Nov 13 '14 at 17:54
So, if $x\geq1$, $\frac{x}{\sqrt{1+x^5}} \leq \frac{x}{\sqrt{x^5}} = \sqrt{\frac{1}{x^3}}$ which converges. So by the comparison test, the original integral also converges. Thanks a lot to everyone! — John, Nov 13 '14 at 17:58

score 3 · Answer 1 · answered Nov 13 '14 at 18:16

$$\int_{0}^{+\infty}\frac{x}{\sqrt{1+x^5}}\,dx = \frac{2}{5}\int_{0}^{+\infty}x^{-1/5}(1+x^2)^{-1/2}\,dx = \frac{2}{5}\int_{0}^{\pi/2}\sin\theta^{-1/5}\cos\theta^{-4/5}\,d\theta,$$

$$\int_{0}^{+\infty}\frac{x}{\sqrt{1+x^5}}\,dx=\frac{2}{5}\int_{0}^{1}t^{-1/5}(1-t^2)^{-9/10}\,dt=\frac{1}{5}\int_{0}^{1}u^{-3/5}(1-u)^{-9/10}\,du,$$

$$\int_{0}^{+\infty}\frac{x}{\sqrt{1+x^5}}\,dx=\frac{1}{5}\frac{\Gamma(2/5)\,\Gamma(1/10)}{\Gamma(1/2)}=\frac{\Gamma(2/5)\,\Gamma(1/10)}{5\sqrt{\pi}}.$$

score 2 · Answer 2 · answered Nov 13 '14 at 18:30

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large\ \overbrace{\int_{0}^{\infty}{x \over \root{1 + x^{5}}} \,\dd x}^{\ds{\color{#c00000}{x^{5}\ \mapsto x}}}} =\int_{0}^{\infty}{x^{1/5} \over \root{1 + x}}\,{1 \over 5}\,x^{-4/5}\,\dd x ={1 \over 5}\int_{0}^{\infty}{x^{-3/5} \over \root{1 + x}}\,\dd x \\[5mm]&={1 \over 5}\ \overbrace{% \int_{1}^{\infty}\pars{x - 1}^{-3/5}x^{-1/2}\,\dd x} ^{\ds{\color{#c00000}{x\ \mapsto\ {1 \over x}}}} ={1 \over 5}\int_{1}^{0}\pars{{1 \over x} - 1}^{-3/5}x^{1/2}\, \pars{-\,{\dd x \over x^{2}}} \\[5mm]&={1 \over 5}\int_{0}^{1}\pars{1 - x}^{-3/5}x^{-9/10}\,\dd x ={1 \over 5}\,{\Gamma\pars{2/5}\Gamma\pars{1/10} \over \Gamma\pars{1/2}} =\color{#66f}{\large{\Gamma\pars{2/5}\Gamma\pars{1/10} \over 5\root{\pi}}} \approx {\tt 2.3812} \end{align}

@John You're welcome. Thanks. – Felix Marin Nov 13 '14 at 18:58 — Felix Marin, Nov 13 '14 at 18:58

score 2 · Answer 3 · answered Nov 13 '14 at 18:31

You can make the substituion $x^{5}=u\Rightarrow dx=\frac{du}{5u^{4/5}}$ and obtain $$\frac{1}{5}\int_{0}^{\infty}\frac{u^{-3/5}}{\sqrt{1+u}}du=\frac{1}{5}\int_{0}^{\infty}\frac{u^{(2/5)-1}}{\left(1+u\right)^{(2/5)+(1/10)}}du=\frac{1}{5}\textrm{B}\left(\frac{2}{5},\frac{1}{10}\right)$$where B is the Beta function.

Integral of $\frac{x}{\sqrt{1+x^5}}$

3 Answers3