Evaluate ( if convergent) $\int\limits_0^\infty \frac{1}{\sqrt{x(1-x)}} dx $

Question

I would like to check whether the improper integral $$\int\limits_0^\infty \frac{1}{\sqrt{x(1-x)}} dx $$ is convergent or not. How can I check convergency? If convergent what is the integral value?

Proceed: If we substitute $x=\sin^2\theta$, hten we have

$$\int \frac1{\sqrt{x(1-x)}} \,dx = \sin^{-1} (2x-1) + C.$$

Added: for $x>1$, integrand is not defined. So if we take $0\leq x\leq 1$, then what we can say about its convergency?

Answer 1

The function $1/\sqrt{x(1-x)}$ is only defined over $(0,1)$, so the proposed integral makes no sense. If you want to compute $$ \int_0^1\frac{1}{\sqrt{x(1-x)}}\,dx $$ your substitution is correct: set $x=\sin^2\theta$, with $\theta\in(0,\pi/2)$, so $$ dx=2\sin\theta\cos\theta\,d\theta $$ and you have $$ \int\frac{1}{\sqrt{x(1-x)}}\,dx= \int\frac{2\sin\theta\cos\theta}{\sqrt{\sin^2\theta\cos^2\theta}}\,d\theta= \int 2\,d\theta=2\theta+c $$ Thus your improper integral becomes $$ \int_0^1\frac{1}{\sqrt{x(1-x)}}\,dx= \int_0^{\pi/2} 2\,d\theta=\Bigl[2\theta\Bigr]_0^{\pi/2}=\pi $$

Convergence follows also from the fact that $$ \int_0^1\frac{1}{\sqrt{x(1-x)}}\,dx= \int_0^{1/2}\frac{1}{\sqrt{x(1-x)}}\,dx + \int_{1/2}^1\frac{1}{\sqrt{x(1-x)}}\,dx = 2\int_0^{1/2}\frac{1}{\sqrt{x(1-x)}}\,dx $$ and this is convergent by comparison with $\int_0^{1/2}\frac{1}{\sqrt{x}}\,dx$

Answer 2

If $x>1$ , then $1-x<0$, henc $x(1-x)<0$ and $\sqrt{x(1-x)}$ is not defined !

Answer 3

$$ \int_{0}^{1}\!{\frac {1}{\sqrt {x \left( 1-x \right) }}}\,{\rm d}x= \pi $$

Perhaps try: Complete the square under the radical, trigonometric substitution.

Answer 4

Assuming you mean $\int_{0}^{\color{red}{1}}\frac{dx}{\sqrt{x(1-x)}}$ (the integral $\int_{1}^{+\infty}\frac{dx}{\sqrt{x(1-x)}}$ is blatantly divergent, no matter which determination of $\sqrt{\cdot}$ we pick) you may notice that by the substitution $x=u^2$ we have $$ \int_{0}^{1}\frac{dx}{\sqrt{x(1-x)}}=2\int_{0}^{1}\frac{du}{\sqrt{1-u^2}}=2\left[\arcsin u\right]_{0}^{1}=\pi$$ also because $\int_{0}^{1}\sqrt{1-u^2}\,du$ is the area of a quarter-circle, equal to $\frac{\pi}{4}$, and by integration by parts

$$ \frac{\pi}{4}=\int_{0}^{1}\sqrt{1-u^2}\,du=\left[u\sqrt{1-u^2}\right]_{0}^{1}+\int_{0}^{1}\frac{u^2}{\sqrt{1-u^2}}\,du=\int_{0}^{1}\frac{du}{\sqrt{1-u^2}}-\int_{0}^{1}\sqrt{1-u^2}\,du, $$ so that $$ \int_{0}^{1}\frac{du}{\sqrt{1-u^2}}=2\int_{0}^{1}\sqrt{1-u^2}\,du = \frac{\pi}{2}.$$