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

In summary, the conversation discusses methods for evaluating the integral $$\int_0^{\pi/2} (\log \sin x )^2 dx = \frac{1}{24} \left(\pi ^3+12 \pi \log^2(2)\right)$$ using gamma and digamma functions, as well as the Beta function. The conversation also mentions the possibility of a more elegant method for solving the integral.
  • #1
The Lord
4
0
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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  • #2
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?
 
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  • #3
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)
 
  • #4
Interesting problem...

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

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

\(\displaystyle \frac{d^2}{dp^2} B(p,q)\, \Bigg|_{p=q=1/2}\equiv\int_0^{\pi/2}\log^2(\sin x)\,dx\)
 
  • #5


To integrate the square of log-sine, we can use the substitution $u = \log(\sin x)$, which means $du = \frac{\cos x}{\sin x} dx$. This allows us to rewrite the integral as:

$$\int_0^{\pi/2} (\log \sin x )^2 dx = \int_{-\infty}^{0} u^2 \frac{e^u}{1-e^{2u}} du$$

Using partial fractions, we can rewrite the integrand as:

$$u^2 \frac{e^u}{1-e^{2u}} = \frac{1}{4}\left(\frac{u^2}{1-e^u} + \frac{u^2}{1+e^u}\right) + \frac{1}{2}\left(\frac{u}{1-e^u} - \frac{u}{1+e^u}\right) + \frac{1}{4}\left(\frac{1}{1-e^u} + \frac{1}{1+e^u}\right)$$

Integrating each term separately, we get:

$$\int_{-\infty}^{0} u^2 \frac{e^u}{1-e^{2u}} du = \frac{1}{4} \left(\frac{u^2}{1-e^u} - 2u + \log(1-e^u) + \frac{u^2}{1+e^u} + 2u + \log(1+e^u) + \log(1-e^u) + \log(1+e^u)\right)\bigg|_{-\infty}^{0}$$

Since $e^u$ approaches 0 as $u$ approaches $-\infty$, all the logarithmic terms evaluate to 0. This leaves us with:

$$\int_{-\infty}^{0} u^2 \frac{e^u}{1-e^{2u}} du = \frac{1}{4} \left(0 - 2(0) + 0 + \frac{0}{1+e^u} + 2(0) + 0 + 0 + 0\right) = 0$$

Therefore, we have shown that:

$$\int_0
 

FAQ: How Is the Integral of the Square of Log-Sine Calculated?

What is the "Integrate Square of Log-Sine" function?

The "Integrate Square of Log-Sine" function is a mathematical function that involves taking the integral of the square of a logarithmic sine function. It is commonly used in calculus and other areas of mathematics.

What is the formula for the "Integrate Square of Log-Sine" function?

The formula for the "Integrate Square of Log-Sine" function is ∫(ln(sin(x))^2)dx.

What are the applications of the "Integrate Square of Log-Sine" function?

The "Integrate Square of Log-Sine" function has various applications in physics, engineering, and other scientific fields. It can be used to solve problems involving motion, oscillations, and energy.

What is the process for integrating the "Integrate Square of Log-Sine" function?

To integrate the "Integrate Square of Log-Sine" function, you can use integration by parts or substitution methods. It is important to carefully follow the steps and apply appropriate techniques to solve the integral.

Are there any special properties or rules for the "Integrate Square of Log-Sine" function?

Yes, there are certain properties and rules that can make solving the "Integrate Square of Log-Sine" function easier. For example, the integral of the square of a logarithmic sine function is equal to the negative of the integral of the product of the logarithmic sine function and its derivative.

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