$\int _0 ^\infty \frac{1}{(x+1)(1-x) \sqrt{x}} dx$

Question

How do I have to compute the integral $\int _0 ^\infty \frac{1}{(x+1)(1-x) \sqrt{x}} dx$? I think maybe the residue theorem is useful, but there are poles at $0$ and $1$, so I don't see what do I have to do..

Answer 1

As noted, the integral $\int_0^\infty$ diverges because of the pole at $x=1$.

The indefinite integral is: $$ \int \!{\frac {1}{ \left( 1-x \right) \left( 1+x \right) \sqrt {x}}} \,{\rm d}x=-\frac{\ln \left( -1+\sqrt {x} \right)}{2} +\arctan \left( \sqrt {x} \right) +\frac{\ln \left( 1+\sqrt {x} \right) }{2}+ C $$ (of course possibly with different $C$ on opposite sides of $x=1$).

We may try for a "principal value" like this. $$ \lim_{a\searrow 0}\left(\int_0^{1-a}\frac{dx}{(1-x)(1+x)\sqrt{x}\;} +\int_{1+a}^{+\infty}\frac{dx}{(1-x)(1+x)\sqrt{x}\;}\right) = \frac{\pi}{2} $$

Answer 2

To elaborate on what @xpaul mentioned in the comments, You can use the substitution, $$x^{1/2} = u$$ Differentiating, $$2du = \frac{dx}{x^{1/2}}$$ Therefore the integral reduces to $$\int_{0}^{\infty}\frac{2du}{(1-u^{2})({1+u^{2})}}$$ Splitting the integral using partial fractions, $$-\int_{0}^{\infty}\frac{du}{u^{2}+1}+\int_{0}^{\infty}\frac{du}{u^{2}-1}$$ The first integral equals $\pi/2$ while the second one diverges