Evaluate $\int_0^{\infty} \frac{dx}{(x+2)\sqrt{x-1}} $

Question

I would like to calculate the integral

$$\int_0^{\infty} \frac{dx}{(x+2)\sqrt{x-1}}$$

Integrating by parts doesn't seem to go anywhere useful, I can't find any trigonometric substitution that would work, and I cannot find this in any standard integral tables. Putting this into Mathematica gives a complex integral,

$$\int_0^{\infty} \frac{1}{x+2} \frac{1}{\sqrt{x-1}} = \frac{\pi - 2i \mathrm{arccsch(\sqrt{2})}}{\sqrt{3}}.$$

Am I missing something trivial?

Answer 1

Note: For real analysis this integral should be evaluated in $(0,\infty)$ as: $$I=\int_{1}^{\infty} \frac{dx}{(2+x)\sqrt{x-1}}$$ Klet $x-1=t^2$, then $$I=\int_{0}^{\infty}\frac {2t dt}{(3+t^2)t}= 2\int_{0}^{\infty} \frac{dt}{3+t^2}=\frac{2}{\sqrt{3}} \tan^{-1}(t/\sqrt{3})= \frac{\pi}{\sqrt{3}}$$

Note This integral is improper two ways (1) one limit is $\infty$, (2) at $x=1$ the integrand diverges. Note that the improper integral $$J=\int_{a}^{\infty} \frac{dx}{(x-a)^p}$$ is finite (converges) if $0<p<1.$ Here in this cae $p=1/2$.

Answer 2

I try the interval between 0 and 1. The approach here is to choose a branch cut for the negative square root and reduce than the integral to one containing only square root of positive numbers.

First of all I want to evaluate:

$I=\int_0^1 \frac{dx}{(x+2)\sqrt{x-1}}$

As discussed, we need to fix the choice of the branch for the square root. For positive numbers, we can choose the square root to be positive. For $a<0$:

$\sqrt{a}=\sqrt{ |a| }e^{i\pi/2}=i\sqrt{ |a| }$

The other branch is analogous. Now we can work on the integral. First we substitute $x'=x-1$ and renominating $x'$ with x:

$I= -i \int_{-1}^0 \frac{dx}{(x+3)\sqrt{-x}}$

now $x'=-x$ and renominating:

$I= -i\int_{0}^1 \frac{dx}{(3-x)\sqrt{x}}$

Almost done: $x=t^2$, $t'=\frac{t}{\sqrt{3}}$:

$I=-i\frac{2\sqrt{3}}{3} \int_{0}^{1/\sqrt{3}} \frac{dt}{1-t^2}$

The last integral can be evaluated by decomposition, leading to:

$I=-i\frac{1}{\sqrt{3}} ln\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)$

UPDATE We can also observe that :

$arccsch(y)=ln(\frac{1 \pm \sqrt{1+y^2}}{y})$

so that:

$arccsch(\sqrt{2})=ln(\frac{1 \pm \sqrt{3}}{\sqrt{2}})$

So since:

$ln\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)=ln\left(\frac{(\sqrt{3}+1)^2}{2}\right)=2ln\left(\frac{(\sqrt{3}+1)}{\sqrt{2}}\right)$

we see that the solution provided is equivalent to the one of Mathematica.