1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ Do you want to compute the integral? Or just to prove that it converges? $\endgroup$ – José Carlos Santos Apr 27 at 17:37
  • $\begingroup$ This integral would be represented with hypergeometric function and over (0,\infty) it has a closed form using gamma function you may check here $\endgroup$ – zeraoulia rafik Apr 27 at 17:39
  • $\begingroup$ @JoséCarlosSantos I only want to prove that it converges. $\endgroup$ – Manuel Vargas Apr 27 at 17:42
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ It's clear to me, but what can I do with the $\int_{0}^{1}\frac{1}{\sqrt{x^{5}+1}}dx$ part? $\endgroup$ – Manuel Vargas Apr 27 at 17:55
  • $\begingroup$ Nothing! It's a continuous function on $[0,1]$, right? Then it is integrable. $\endgroup$ – José Carlos Santos Apr 27 at 18:09
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.