Integral $\int_0^1\frac{\log(1-x)}{\sqrt{x-x^3}}dx$

Question

I have a trouble with this integral $$I=\int_0^1\frac{\log(1-x)}{\sqrt{x-x^3}}dx.$$ Could you suggest how to evaluate it?

Answer 1

Here is a proof of Cleo's answer.

Rewrite the integral as

$$ \begin{align*} I &= \int_0^1 \frac{\log(1-x)}{\sqrt{x}\sqrt{1-x^2}}dx \\ &= \int_0^1 \frac{\log(1-x^2)-\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx \\ &= \int_0^1 \frac{\log(1-x^2)}{\sqrt{x}\sqrt{1-x^2}}dx-\int_0^1 \frac{\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx \end{align*} $$ The first integral can be computed by calculating a derivative of the beta function. $$ \begin{align*} &\;\int_0^1 \frac{\log(1-x^2)}{\sqrt{x}\sqrt{1-x^2}}dx \\ &\stackrel{y=x^2}{=}\frac{1}{2}\int_0^1 \frac{\log(1-y)}{y^{\frac{3}{4}}\sqrt{1-y}}dy \\ &= \frac{1}{2}\frac{\partial}{\partial \alpha}\left\{B\left(\frac{1}{4},\alpha \right)\right\}_{\alpha=\frac{1}{2}}\\ &= \frac{\sqrt{\pi}\Gamma\left(\frac{1}{4}\right)}{2\Gamma\left(\frac{3}{4} \right)}\left\{ \psi_0\left(\frac{1}{2} \right)-\psi_0\left(\frac{3}{4} \right)\right\} \\ &= \frac{\Gamma\left(\frac{1}{4} \right)^2}{4\sqrt{2\pi}}\left(2\log 2-\pi \right) \end{align*} $$

The other integral can be evaluated by using equation $(3.22)$ of this paper.

$$ \begin{align*} &\;\int_0^1 \frac{\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx \\ &\stackrel{x=\sin^2 t}{=}2\int_0^{\frac{\pi}{2}}\frac{\log(1+\sin^2 t)}{\sqrt{1+\sin^2 t}}dt\\ &= \log(2) K(\sqrt{-1}) \\ &= \log(2)\frac{\Gamma\left(\frac{1}{4} \right)^2}{4\sqrt{2\pi}} \end{align*} $$ where $K(k)$ is the complete elliptic integral of the first kind. After combining everything, one gets

$$I=\frac{\Gamma\left(\frac{1}{4} \right)^2}{4\sqrt{2\pi}}\left(\log 2-\pi \right)\approx -3.20998$$

Answer 2

$$I=\frac{\Gamma\left(\frac14\right)^2}{4\,\sqrt{2\,\pi}}\Big(\ln2-\pi\Big).$$

Answer 3

You may not like this, but it is coincidentally similar:

$$1/4\int_{0}^{1}\frac{\log(x)}{x^{3/4}(1-x)^{1/2}}\mathrm dx-1/4\int_{0}^{1}\frac{\log(x)}{x^{1/2}(1-x)^{1/2}}\mathrm dx.$$

These can be solved with the Beta function without a whole lot of effort.

They evaluate to $$-\frac{\pi}{4}\beta(1/4,1/2)+\frac{\pi}{2}\log(2)$$

$$=\frac{-\pi^{5/2}\sqrt{2}}{4\Gamma^{2}(3/4)}+\frac{\pi}{2}\log(2)$$

$$\approx -3.029925329.....$$

Answer 4

Solution using the beta function and the gamma function

Start with a substitution, $x\rightarrow\frac{1-x}{1+x}$ $$I=\int_0^1\frac{\ln(1-x)}{\sqrt x\sqrt {1-x^2}}dx=\int_0^1\frac{\ln\left(\frac{2x}{1+x}\right)}{\sqrt x\sqrt{1-x^2}}dx=\frac{1}{2}\int_0^1\frac{\ln\left(\frac{2x(1-x)}{1+x}\right)}{\sqrt x\sqrt{1-x^2}}dx$$$$=\frac{1}{2}\int_0^1\frac{\ln\left(\frac{2x}{1+x}\right)}{\sqrt x\sqrt{1-x^2}}dx+\frac{1}{2}\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)}{\sqrt x\sqrt{1-x^2}}dx=(J+K)/2$$ Evaluate J and K separately. J can be evaluated using the beta function. $$J=\int_0^1\frac{\ln(2x)}{\sqrt x\sqrt{1-x^2}}dx=\frac{d}{ds}\int_0^1\frac{2^s x^{s-1/2}}{\sqrt{1-x^2}}dx\Bigg|_{s=0}=\frac{d}{ds}\left[2^{s-1}B\left(\frac{2s+1}{4},\frac{1}{2}\right)\right]\Bigg|_{s=0}=\frac{(2\ln2-\pi)\Gamma^2(1/4)}{4\sqrt{2\pi}}$$ For K, use the substitution $x\rightarrow\frac{1-\sqrt x}{1+\sqrt x}$ $$\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)}{\sqrt x\sqrt{1-x^2}}dx=\frac{1}{4}\int_0^1x^{-3/4}(1-x)^{-1/2}\ln (x)dx=\frac{1}4\frac{d}{ds}B\left(s+\frac{1}{4},\frac{1}{2}\right)\Bigg|_{s=0}$$$$=-\frac{\sqrt\pi \Gamma^2(1/4) }{4\sqrt2}$$ Substitute J and K into the equation for I $$I=\frac{(2\ln2-\pi)\Gamma^2(1/4)}{8\sqrt{2\pi}}-\frac{\sqrt\pi \Gamma^2(1/4) }{8\sqrt2}=\frac{(\ln2-\pi)\Gamma^2(1/4)}{4\sqrt{2\pi}}$$ $$I=\int_0^1\frac{\ln(1-x)}{\sqrt x\sqrt {1-x^2}}dx=\frac{(\ln2-\pi)\Gamma\left(\frac{1}{4}\right)^2}{4\sqrt{2\pi}}$$

Answer 5

There is a closed form expression for this integral and I am sure you will enjoy it $$\frac{3}{2} \pi \left(2 \text{Hypergeometric2F1}^{(0,0,1,0)}\left(\frac{3}{4},\frac{3}{4},1,-8\right)+\text {Hypergeometric2F1}^{(0,1,0,0)}\left(\frac{3}{4},\frac{3}{4},1,-8\right)+\text {Hypergeometric2F1}^{(1,0,0,0)}\left(\frac{3}{4},\frac{3}{4},1,-8\right)-\, _2F_1\left(\frac{3}{4},\frac{3}{4};1;-8\right) \log (4)\right)$$ This was found by a CAS (do not ask me how to arrive to this marvel !). The numerical value is $-3.209982474415127728669875$