MHB How Is the Integral of the Square of Log-Sine Calculated?

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The integral of the square of the logarithm of the sine function, given by $$\int_0^{\pi/2} (\log \sin x)^2 dx$$, is evaluated using differentiation of the sine integral in terms of the gamma function. The result is shown to be $$\frac{1}{24} \left(\pi^3 + 12 \pi \log^2(2)\right)$$. The discussion highlights the tedious nature of differentiating gamma and digamma functions to arrive at the solution. An alternative method involving the Beta function is suggested, which simplifies the process by differentiating the Beta function twice with respect to its parameters. The conversation emphasizes the complexity of the integral and the potential for more elegant solutions.
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Prove that

$$\int_0^{\pi/2} (\log \sin x )^2 dx = \frac{1}{24} \left(\pi ^3+12 \pi \log^2(2)\right)$$
 
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Hi! :D
I think I know how to solve this integral but my method is a little tedious. Anyway I will post my solution.

Let $\displaystyle I(n) = \int_0^{\pi/2}\sin ^n (x)dx$

$I(n)$ can be evaluated in terms of gamma function.

$$ I(n)= \int_0^{\pi/2}\sin ^n (x)dx = \frac{\sqrt{\pi} \Gamma \left( \frac{1+n}{2}\right)}{2\Gamma \left( 1+\frac{n}{2}\right)}$$

Differentiating with respect to n

$$ I'(n) = \int_0^{\pi/2}\sin^n (x) \log(\sin x) dx = \\ \frac{\sqrt{\pi} \Gamma \left( \frac{1+n}{2}\right)}{4\Gamma \left( 1+\frac{n}{2}\right)} \left( \psi \left( \frac{1+n}{2}\right) -\psi\left( 1+\frac{n}{2}\right)\right) \tag{1}$$

Let n=0

$$ I'(0) = \int_0^{\pi/2}\log(\sin x)dx = \frac{\sqrt{\pi} \sqrt{\pi}}{4} \left( \gamma -\gamma -2\ln(2)\right) = \frac{-\pi}{2}\ln(2)$$

$\gamma$ is the Euler-Mascheroni Constant. For the Square of Log-Sine we must differentiate (1) again.

$$I''(n)=\int_0^{\pi/2}\sin^n(x) (\log(\sin x))^2 dx = \frac{\sqrt{\pi } \Gamma\left(\frac{1+n}{2}\right) \psi\left(1+\frac{n}{2}\right)^2}{8 \Gamma\left(1+\frac{n}{2}\right)}- \\ \frac{\sqrt{\pi } \Gamma\left(\frac{1+n}{2}\right) \psi\left(1+\frac{n}{2}\right) \psi\left(\frac{1+n}{2}\right)}{4 \Gamma\left(1+\frac{n}{2}\right)}+ \frac{\sqrt{\pi } \Gamma\left(\frac{1+n}{2}\right) \psi\left(\frac{1+n}{2}\right)^2}{8 \Gamma\left(1+\frac{n}{2}\right)}- \\ \frac{\sqrt{\pi } \Gamma\left(\frac{1+n}{2}\right) \psi_1 \left(1+\frac{n}{2}\right)}{8 \Gamma \left(1+\frac{n}{2}\right)} + \frac{\sqrt{\pi } \Gamma\left(\frac{1+n}{2}\right)\psi_1\left(\frac{1+n}{2}\right)}{8 \Gamma\left(1+\frac{n}{2}\right)} \tag{2}$$

$\psi_1(z)$ is the PolyGamma Function. Put n=0.

$$I''(0) =\int_0^{\pi/2}\left( \log \sin x\right)^2 dx = \frac{1}{24}\left( \pi^3 +2\pi \log^2(2)\right)$$

Here, I have used

$\psi_1 \left( \frac{1}{2}\right)=\frac{\pi^2}{2}$

$\psi_1 \left( 1\right) = \frac{\pi^2}{6}$

$\psi \left( \frac{1}{2}\right) = -\gamma -2\ln(2)$

$\psi(1) = -\gamma$

and $\Gamma \left( \frac{1}{2}\right)=\sqrt{\pi}$

As I mentioned before, differentiating all the gamma and digamma can be tedious. Maybe Mr.Lord has a more elegant method in mind?
 
Last edited:
sbhatnagar said:
As I mentioned before, differentiating all the gamma and digamma can be tedious. Maybe Mr.Lord has a more elegant method in mind?

Great! This was my approach as well. (Yes)
 
Interesting problem...

The thing to do here is to consider the Beta function:

$$B(p,q)=2\int_0^{\pi/2}\sin^{2p-1}x\cos^{2q-1}\,dx\equiv \frac{2\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$$Now differentiate $$\frac{2\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$$ twice w.r.t $$p$$, which will give an expression in terms of the Beta function, Digamma, and Trigamma functions. Then set $$q=p=1/2$$, since:

$$\frac{d^2}{dp^2} B(p,q)\, \Bigg|_{p=q=1/2}\equiv\int_0^{\pi/2}\log^2(\sin x)\,dx$$
 
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