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Show that: $$\int_0^{\pi/4}\ln ^3\sin x\text{d}x=\frac34\text{Im}\left(\text{Polylog}\left(4,i\right)\right)-3\text{Im}\left(\text{Polylog}\left(4,\frac12+\frac{i}{2}\right)\right)-\frac{23}{128}\pi^3\ln 2+\frac32\text{Im}\left(\text{Polylog}\left(3,\frac12+\frac{i}{2}\right)\right)\ln 2-\frac38G\ln^22-\frac5{16}\pi\ln^32-\frac38\pi\zeta\left(3\right)$$

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Where did this arise? –  Potato Feb 19 '13 at 0:12

2 Answers 2

A related problem. Here is a different form for the solution in terms of the hypergeometric function
$$\int_0^{\frac{\pi}{4}}\ln ^3\sin x\text{d}x = -\frac{\pi\,\ln^3(2)}{32}-\frac{3}{8}\sqrt {2} \left(\ln^2( 2 ) + 4\,\ln \left( 2 \right) +8 \right)\times$$

$${_7F_6\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\bar{\alpha}, \alpha\, ;\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\bar{\beta} , \beta;\,\frac{1}{2}\right )}\sim -6.037581109, $$

where $i=\sqrt{-1},$

$$ \alpha = \frac{3\ln(2)+2+2i }{2\ln(2)}, \quad \beta = \frac{\ln(2)+2+2i }{2\ln(2)}, $$

and $ \bar{\alpha},\bar{\beta} $ are the complex conjugates of $\alpha,\beta$.

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I've noticed you post challenging integrals. Sorry this is rather belated, but I just found it while nosing around the site.

These log-sin integrals are tedious, especially with pi/4.

There is an "identity" that can be used to evaluate them, but it is laborious.

Note that $\displaystyle \ln(1-e^{i\theta})=\ln(2\sin(x/2))-\frac{i}{2}(\pi -x)$

So, it can be squared, cubed, etc. and solved for the integral in question to derive the result.

We can use the formula:

$\displaystyle \int_{0}^{\frac{\pi}{4}}\left(\ln(2\sin(x/2))-\frac{i}{2}(\pi -x)\right)^{3}$

$\displaystyle=-i\sum_{k=0}^{3}(-1)^{k+1}\frac{6}{(3-k)!}\cdot \ln^{3-k}(1-e^{\pi i/4})Li_{k+1}(1-e^{\pi i/4})$.

Where $Li$ represents the polylog.

But, it is rather nasty.

The $\frac{\pi}{3}$ case is among the easiest and $\frac{\pi}{4}$ is notoriously difficult.

I see part of your solution involves the imaginary part of $Li_{4}(i)$

This is equal to $\displaystyle \frac{\psi^{(3)}(1/4)-\psi^{(3)}(3/4)}{1536}$, which is the third derivative of digamma. Some of the polylogs have closed forms especially for 2 and 3. Higher order polylogs tend not to have closed forms.

Take for instance $\displaystyle Li_{3}(e^{\frac{\pi i}{3}})=\frac{\zeta(3)} {3}+\frac{5{\pi}^{3}i}{162}$

Check this out:

formula that can be used to integrate powers of log-sin

that formula can be handy for integrating various log-sin integrals if the upper limit is pi/2.

But, if you would be satisfied with

$\displaystyle \int_{0}^{\frac{\pi}{4}}\ln^{3}(\sin(2t))dt$, then it can be done by using

$\displaystyle \int_{0}^{1}\frac{\ln^{3}(x)}{\sqrt{1-x^{2}}}dx$

and making the sub $x=\sin(2t)$

and by differentiating

$\displaystyle \int_{0}^{1}\frac{t^{a-1}}{\sqrt{1-t^{2}}}dt=\frac{1}{2}\beta\left(\frac{1}{2},\frac{a}{2}\right)$

with respect to 'a' three times and let a=1.

As seen here along with other very nice methods:

Integrate square of the log-sine integral: $\int_0^{\frac{\pi}{2}}\ln^{2}(\sin(x))dx$

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