Prove that : $\int_0^{1}\frac{1-\sqrt{1-x^{4}}}{x^{2}\sqrt{1-x^4}}dx=1-\frac{\sqrt{2}\pi^{\frac{3}{2}}}{\Gamma(\frac{1}{4})^2}$

Question

Prove that :

$$\int_0^{1}\frac{1-\sqrt{1-x^{4}}}{x^{2}\sqrt{1-x^4}}dx=1-\frac{\sqrt{2}\pi^{\frac{3}{2}}}{\Gamma(\frac{1}{4})^2}.$$

I know how to use the definition of the Beta and Gamma functions so my problem here is when I divide the integral I find: $$\int_0^{1}\frac{1}{x^2}dx=\infty!!$$

(divergent)

I tried using $y=\frac{1}{x}$ but didn't get the answer.

I also tried $y=x^{4}$ and got the same problem (a divergent integral).

Answer 1

I begin by making the following observation: $$\frac{d}{dx} \left (\frac{1 - \sqrt{1 - x^4}}{x} \right )= \frac{x^4 + 1 - \sqrt{1 - x^4}}{x^2 \sqrt{1 - x^4}}.$$

So, for your integral, one can write \begin{align} \int_0^1 \frac{1 - \sqrt{1 - x^4}}{x^2 \sqrt{1 - x^4}} \, dx &= \int_0^1 \frac{(x^4 + 1 - \sqrt{1 - x^4}) - x^4}{x^2 \sqrt{1 - x^4}} \, dx\\ &= \int_0^1 \left (\frac{1 - \sqrt{1 - x^4}}{x} \right )' \, dx -\underbrace{\int_0^1 \frac{x^2}{\sqrt{1 - x^4}} \, dx}_{x \, \mapsto \, \sqrt[4]{x}}\\ &= 1 - \frac{1}{4} \int_0^1 x^{-\frac{1}{4}} (1 - x)^{-\frac{1}{2}} \, dx\\[1ex] &= 1 - \frac{1}{4} \operatorname{B} \left (\frac{3}{4}, \frac{1}{2} \right )\\[1ex] &= 1 - \frac{1}{4} \frac{\Gamma (\frac{3}{4}) \Gamma (\frac{1}{2})}{\Gamma (\frac{5}{4})}\\ &= 1 - \sqrt{\pi} \frac{\Gamma (\frac{3}{4})}{\Gamma (\frac{1}{4})}\\ &= 1 - \frac{\pi \sqrt{2 \pi}}{\Gamma^2 (\frac{1}{4})}, \end{align} as required. Here $\Gamma (\frac{3}{4}) \Gamma (\frac{1}{4}) = \pi \sqrt{2}$, has been used.

Answer 2

Substitute $u = x^4$ to get

$$ \int_{0}^{1} \frac{1-\sqrt{1-x^4}}{x^2\sqrt{1-x^4}} \, \mathrm{d}x = \frac{1}{4} \int_{0}^{1} \frac{1-(1-u)^{1/2}}{u^{5/4}(1-u)^{1/2}} \, \mathrm{d}u. $$

Now we regularize the integral by replacing $u^{5/4}$ by $u^s$ for $s < 1$. Then

\begin{align*} \frac{1}{4} \int_{0}^{1} \frac{1-(1-u)^{1/2}}{u^{s}(1-u)^{1/2}} \, \mathrm{d}u &= \frac{1}{4}\left(B\left(1-s,\tfrac{1}{2}\right)-\frac{1}{1-s}\right) \\ &= \frac{1}{4}\left(\frac{\Gamma(1-s)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{3}{2}-s\right)}-\frac{1}{1-s}\right). \end{align*}

Note that the left-hand side defines an analytic function for $\operatorname{Re}(s) < 2$, while the right-hand side is a meromorphic function on all of $\mathbb{C}$ with the removable singularity at $s = 1$. So both sides give rise to analytic functions on $\operatorname{Re}(s) < 2$, and so, they are equal on this region by the principle of analytic continuation. Therefore we can simply plug $s = 5/4$ to get

\begin{align*} \frac{1}{4} \int_{0}^{1} \frac{1-(1-u)^{1/2}}{u^{5/4}(1-u)^{1/2}} \, \mathrm{d}u &= 1 + \frac{\Gamma\left(-\frac{1}{4}\right)\Gamma\left(\frac{1}{2}\right)}{4\Gamma\left(\frac{1}{4}\right)} \\ &= 1 - \frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{1}{4}\right)}, \end{align*}

which simplifies to the desired expression by Euler's reflection formula $\Gamma(z)\Gamma(1-z) = \pi\csc(\pi z)$.