Compute $\int_0^{\infty} \frac{1}{\sqrt{x^3 + x}}$

Question

Could you tell me how to compute $$\int_0^{\infty} \frac{1}{\sqrt{x^3 + x}}dx$$ I have really no idea how to do this and I've tried for a couple of hours.

Answer 1

Well, the first thing you can do is this:

$$\int_0^{\infty} \frac{1}{\sqrt{x} \sqrt{1+x^2}}dx=2\int_0^{\infty} \frac{1}{\sqrt{1+t^4}}dt$$

Here we used a substitution $t=\sqrt{x}$.

Now this integral is not simple, but it has a closed form solution:

$$\int_0^{\infty} \frac{1}{\sqrt{1+t^4}}dt=\frac{4}{\sqrt{\pi}} \left( \Gamma \left( \frac{5}{4} \right) \right)^2 $$

The answer looks like Beta function to me, since:

$$B(x,y)=\frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$$

Here it seems that $x=y=\frac{5}{4}$ (or, using the main property of Gamma function, $x=y=\frac{1}{4}$)

Now the question is how to transform the integral into a Beta function form. Here is the appropriate integral form of the Beta function:

$$B (x,y)=\int_0^{\infty} \frac{u^{x-1}}{(1+u)^{x+y}}du,~~~~~x,y>0$$