Intereting Posts

Fourier Transform of $\exp{(A\sin(x))}$
Why is the sum of the rolls of two dices a Binomial Distribution? What is defined as a success in this experiment?
find the value of limit
Proving that a convex function is Lipschitz
Prove that if fewer than $n$ students in class are initially infected, the whole class will never be completely infected.
Equilateral triangle inscribed in a triangle
An integral to prove that $\log(2n+1) \ge H_n$
Applications of ultrafilters
Integer induction without infinity
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circles, power of point, cross ratios
Calculate Third Point of Triangle
Exotic maps $S_5\to S_6$
Limit of a recursive sequence $s_n = (1-\frac{1}{4n^2})s_{n-1}$
Proof that the hypergeometric distribution with large $N$ approaches the binomial distribution.

after lots of discussion and help from @Chris’sis I have tried integrating it in varied ways but it involves something that is new to me:

$$ I=\int_0^{\pi/2}\log^2(\sin^2x)\sin^2x{\rm d}x$$ It could be written as:

$$I=\int_0^{\pi/2}4\log^2(\sin x)\sin^2x{\rm d}x$$

I know:

$$\int_0^{\pi/2}\log(\sin x){\rm d}x=-\frac{\pi}2\ln 2$$

And couldn’t possibly work out this:

$$\int_0^{\pi/2}\log^2(\sin x){\rm d}x$$

I know there’ a question on last one on MSE, but the evaluation techniques I’m unfamiliar to.

For more info see here.

- Integral of $\sin (x^3)dx$
- Using contour integration, or other means, is there a way to find a general form for $\int_{0}^{\infty}\frac{\sin^{n}(x)}{x^{n}} \, dx$?
- $ \int_1^2\int_1^2 \int_1^2 \int_1^2 \frac{x_1+x_2+x_3-x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4 $
- Evaluation of the integral $\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$
- Evaluate $\int^1_0 \log^2(1-x) \log^2(x) \, dx$
- Prove the integral of $f$ is positive if $f ≥ 0$, $f$ continuous at $x_0$ and $f(x_0)>0$
- Interesting problem of finding surface area of part of a sphere.
- Value of an integral related to Stirling's formula
- Find the density of the sum of two uniform random variables
- Integral ${\large\int}_0^1\frac{\ln^2\ln\left(\frac1x\right)}{1+x+x^2}dx$

**HINT**: As I mentioned in chat, one of the natural ways is to use beta function in trigonometric form that immediately leads to the desired solution

$$\int_0^{\pi/2} \sin^a(x)\cos^b(x) \ dx=\frac{1}{2}B \left(\frac{1}{2}(a+1),\frac{1}{2}(b+1)\right)$$

Thus,

$$I=4\lim_{b\to 0} \lim_{a\to2}\frac{\partial^2}{\partial a^2}\left(\frac{1}{2}B \left(\frac{1}{2}(a+1),\frac{1}{2}(b+1)\right)\right)$$

$$=\pi\log^2(2)-\pi\log(2)+\frac{\pi^3}{12}-\frac\pi2.$$

Using

$$

\log(\sin(x))=-\log(2)-\sum_{k=1}^\infty\frac{\cos(2kx)}{k}

$$

and

$$

\sin^2(x)=\frac{1-\cos(2x)}2

$$

and

$$

\cos(2kx)\cos(2x)=\frac{\cos((2k+2)x)+\cos((2k-2)x)}2

$$

we get, remembering the orthogonality of $\cos(2jx)$ and $\cos(2kx)$,

$$

\begin{align}

&\int_0^{\pi/2}\log^2(\sin^2(x))\sin^2(x)\,\mathrm{d}x\\

&=4\int_0^{\pi/2}\log^2(\sin(x))\sin^2(x)\,\mathrm{d}x\\

&=4\log(2)^2\int_0^{\pi/2}\sin^2(x)\,\mathrm{d}x\\

&+4\log(2)\sum_{k=1}^\infty\int_0^{\pi/2}\frac{2\cos(2kx)-\cos((2k+2)x)-\cos((2k-2)x)}{2k}\mathrm{d}x\\

&+\sum_{k=1}^\infty\sum_{j=1}^\infty\int_0^{\pi/2}\frac{\cos(2jx)}j\frac{2\cos(2kx)-\cos((2k+2)x)-\cos((2k-2)x)}{k}\,\mathrm{d}x\\

&=\pi\log^2(2)-\pi\log(2)+\frac{\pi^3}{12}-\frac\pi2

\end{align}

$$

For historical context, here is Ramanujan’s solution.

\begin{align}

\int_0^{\pi/2}\log^m(\sin(x))dx & = \int_0^1\frac{\log^m(x)}{\sqrt{1-x^2}}dx\tag{1}\\

&= (-1)^m\int_0^{\infty}t^m\sum_{n=0}^{\infty}\frac{(1/2)_ne^{-(2n+1)t}}{n!}dt\tag{2}\\

&= (-1)^mm!\sum_{n=0}^{\infty}\frac{(1/2)_n}{n!(2n+1)^{m+1}}\tag{3}

\end{align}

In $(1)$, let $x=e^{-t}$. In $(2)$, $(1/2)_n$ is the following sequence $(1/2)_0 = 1$ and $(1/2)_n = (1/2)(1/2 + 1)\cdots(1/2 + n – 1)$. In $(3)$, we have

$$

\int_0^{\infty}t^me^{-(2n+1)t}dt=\frac{m!}{(2m+1)^{m+1}}

$$

Now taking $m = 2$, we arrive at

$$

\int_0^{\pi/2}\log^2(\sin(x))dx = 2\sum_{n=0}^{\infty}\frac{(1/2)_n}{n!(2n+1)^{3}}

$$

Complex approach different from robjohn’s link in the comments to the OP. From the basic trig identity $\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$, we can write $-2i\sin(x) = e^{ix}(e^{-ix}-e^{ix}) = 1 – e^{2ix}$. If we consider the contour I mention here, there will be no poles in the contour. By Cauchy’s integral formula, the integral over the contours will be zero.

\begin{align}

\log(1-e^{2ix}) &= \log[e^{ix}(e^{-ix}-e^{ix})]\\

&= ix + \log(2) – \frac{\pi}{2}i+\log(\sin(x))

\end{align}

Then we can write the integral as (where I dropped the imaginary parts since we are solving a real integral)

\begin{align}

\int_0^{\pi/2}\Bigl[ix + \log(2) – \frac{\pi}{2}i+\log(\sin(x))\Bigr]^2

&=\int_0^{\pi/2}\Bigl[x^2-\pi x+\frac{\pi^2}{4}-\log^2(2)-2\log(2)\log(\sin(x))-\log^2(\sin(x))\Bigr]dx\\

\int_0^{\pi/2}\log^2(\sin(x))dx &=\int_0^{\pi/2}\Bigl[x^2-\pi x+\frac{\pi^2}{4}-\log^2(2)-2\log(2)\log(\sin(x))\Bigr]dx

\end{align}

As mentioned in the OP or in the link above, we already know $\int_0^{\pi/2}\log(\sin(x))dx=-\frac{\pi}{2}\log(2)$. Therefore, the RHS becomes

$$

\frac{\pi^3}{24}+\frac{\pi}{2}\log^2(2)

$$

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