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?
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wolram was unable to find a solution in terms of functions, I doubt this is solvable. – Guy Mar 07 '14 at 05:37
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http://www.wolframalpha.com/input/?i=integrate+((log(1-x)%2F(x-x%5E3)%5E(1%2F2))+from+0+to+1 – Guy Mar 07 '14 at 05:37
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Are you really sure it's not by any chance $x^2$ in the denominator ? – Lucian Mar 07 '14 at 05:51
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@Lucian Nope, it's $x^3$. – Marty Colos Mar 07 '14 at 16:51
4 Answers
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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
$$I=\frac{\Gamma\left(\frac{1}{4} \right)^2}{4\sqrt{2\pi}}\left(\log 2-\pi \right)\approx -3.20998$$
Shobhit Bhatnagar
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You may not like this, but it is coincidentally similar:
$$1/4\int_{0}^{1}\frac{log(x)}{x^{3/4}(1-x)^{1/2}}dx-1/4\int_{0}^{1}\frac{log(x)}{x^{1/2}(1-x)^{1/2}}dx$$
These can be solved with the Beta function without a whole lot of effort.
They evaluate to $$-\frac{\pi}{4}\beta(1/4,1/2)+\frac{\pi}{2}log(2)$$
$$=\frac{-\pi^{5/2}\sqrt{2}}{4\Gamma^{2}(3/4)}+\frac{\pi}{2}log(2)$$
$$\approx -3.029925329.....$$
Cody
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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 $-3.209982474415127728669875$
Claude Leibovici
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It's easy to arrive at this expression, either by expanding the natural logarithm into its Taylor series, and then switching the order of summation and integration, or by noticing that $$J(n)=\displaystyle\int_0^1\frac{(1-x)^n}{\sqrt{x(1+x)}}dx\iff I=\int_0^1\frac{\log(1-x)}{\sqrt{x-x^3}}dx=J'\bigg(-\frac12\bigg).$$ – Lucian Mar 07 '14 at 07:05
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@ClaudeLeibovici Thanks. But I hoped to find a closed form that does not contain unevaluated derivatives, sums, integrals etc. – Marty Colos Mar 07 '14 at 16:54
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@MartyColos.You are welcome. I did not expect anything ! Just horrified by what a CAS is able to produce. Cheers. – Claude Leibovici Mar 07 '14 at 17:14