# Help with $\int_{0}^{\infty}\frac{1}{\sqrt{x^{5}+1}}dx$

$$\int_{0}^{\infty}\frac{1}{\sqrt{x^{5}+1}}dx$$

So far,

$$\int\frac{1}{\sqrt{x^{5}+1}}dx=\int\frac{1}{\sqrt{x^{4}\left(x+\frac{4}{x^{4}}\right)}}dx=\int\frac{1}{x^{2}\sqrt{x+\frac{4}{x^{4}}}}dx$$

but from there I do not know how to proceed, I have tried several changes of variable but I have not obtained anything clear.

I only can work with real methods. Any suggestion? Thanks!

Edit: It's sufficiente to prove that it converges

• Do you want to compute the integral? Or just to prove that it converges? – José Carlos Santos Apr 27 at 17:37
• This integral would be represented with hypergeometric function and over (0,\infty) it has a closed form using gamma function you may check here – zeraoulia rafik Apr 27 at 17:39
• @JoséCarlosSantos I only want to prove that it converges. – Manuel Vargas Apr 27 at 17:42

Use the fact that$$\lim_{x\to\infty}\frac{\frac1{\sqrt{x^5+1}}}{\frac1{x^{5/2}}}=1$$and that $$\int_1^\infty\frac1{x^{5/2}}\,\mathrm dx$$ converges.

• It's clear to me, but what can I do with the $\int_{0}^{1}\frac{1}{\sqrt{x^{5}+1}}dx$ part? – Manuel Vargas Apr 27 at 17:55
• Nothing! It's a continuous function on $[0,1]$, right? Then it is integrable. – José Carlos Santos Apr 27 at 18:09

Hint: put, $$x^5=t$$ and use $$\int_{0}^\infty \frac{x^{m-1}}{(1+x)^{m+n}}~dx=\beta(m,n)$$

The convergence is straightforward since $$\frac{1}{\sqrt{x^5+1}}$$ is positive and bounded by $$\min\left(1,\frac{1}{x^{5/2}}\right)$$, so the integral is positive and less than $$\frac{5}{3}$$. Of course we can state something fairly more accurate. By Euler's Beta function our integral equals

$$\int_{0}^{+\infty}\frac{z^{-4/5}\,dz}{\sqrt{z+1}}=\frac{1}{5}\int_{1}^{+\infty}\frac{(z-1)^{-4/5}}{\sqrt{z}}\,dz = \frac{1}{5}\int_{0}^{1}z^{-7/10}(1-z)^{-4/5} dz=\color{red}{\frac{\Gamma(3/10)\Gamma(2/10)}{5\sqrt{\pi}}}$$ or, by exploiting the Weierstrass product for the $$\Gamma$$ function, $$\frac{5}{3}\prod_{n\geq 1}\left(1+\frac{3}{10m}\right)^{-1}\left(1+\frac{2}{10n}\right)^{-1}\left(1+\frac{1}{2n}\right)=\color{red}{\frac{5}{3}\prod_{n\geq 1}\left(1+\frac{3}{50n^2+25n}\right)^{-1}}.$$ By creative telescoping it turns out that the value of the integral is $$\leq\frac{25}{16}=\left(\frac{5}{4}\right)^2$$, with a very small absolute error.

I would start by breaking the integral into two pieces \begin{align*} \int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x^{5}+1}} = \int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{x^{5}+1}} + \int_{1}^{\infty}\frac{\mathrm{d}x}{\sqrt{x^{5}+1}} \end{align*} Thus, according to the change of variables $$u = 1/x$$, one has that \begin{align*} \int_{1}^{\infty}\frac{\mathrm{d}x}{\sqrt{x^{5}+1}} = \int_{0}^{1}\frac{\sqrt{u}}{\sqrt{u^{5}+1}}\mathrm{d}u \end{align*}

Therefore \begin{align*} \int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x^{5}+1}} = \int_{0}^{1}\frac{\sqrt{x}+ 1}{\sqrt{x^{5}+1}}\mathrm{d}x \end{align*}

Since the integrand is continuous on $$[0,1]$$, the integral converges.

Hopefully this helps.