Integration of $\ln\sin x$ from 0 to$ \frac{\pi}{2}$by DUIS

Question

How can we evaluate the integration $$\int_0^{\frac{\pi}{2}}\ln \sin x\,dx$$ by using DUIS(Differentiating under the integral sign)?

This question popped in my head when I was reading an article about DUIS as $\ln |\sin x|$ is the integral of $\cot x$.

Although I am in 12th Standard, I am keen to learn any new and interesting concepts and techniques so please tell me if there are any related to the question. Thanks!

Answer 1

You can start by defining $$I(a)=\int_0^{\pi/2} \sin^a x\,dx$$ Your desired integral is then just $I'(0)$. Now, in order to evaluate $I(a)$ in closed form, we will have to use the Beta function and its connection to the Gamma function:

\begin{align} I(a)&=\int_0^{\pi/2} \sin^a x\,dx \\ &=\frac{1}{2}B\left(a/2+1/2,1/2\right) \\ &=\frac{\Gamma\left(a/2+1/2\right)\Gamma\left(1/2\right)}{2\Gamma(a/2+1)} \\ &= \frac{\sqrt{\pi}}{2}\frac{\Gamma\left(a/2+1/2\right)}{\Gamma(a/2+1)} \end{align} Differentiating $I(a)$ and letting $a\to 0$ then yields

\begin{align} I'(a)\Big|_{a=0}&=\frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma\left(a/2+1/2\right)\left(\psi^{(0)}\left(a/2+1/2\right) - \psi^{(0)}(a/2+1)\right)}{\Gamma(a/2)}\Biggr|_{a=0} \\ &=-\frac{\sqrt{\pi}}{2}\cdot \sqrt{\pi}\log 2 \\ &=-\frac{\pi}{2}\log 2 \end{align}

And we can conclude that

$$\int_0^{\pi/2} \log\sin x\,dx = -\frac{\pi}{2} \log 2$$

Answer 2

There is another way to evaluate the integral.

Firstly, we have $$\int_0^{\frac{\pi}{2}}\ln \sin x\ dx\overset{t=\frac{\pi}{2}-x}{=}\int_0^{\frac{\pi}{2}}\ln \cos t\ dt,$$ and $$\int_{\frac{\pi}{2}}^{\pi}\ln \sin x\ dx\overset{t=x-\frac{\pi}{2}}{=}\int_0^{\frac{\pi}{2}}\ln \cos t\ dt.$$ Then \begin{align} 2\int_0^{\frac{\pi}{2}}\ln \sin x\,dx &=\int_0^{\frac{\pi}{2}}\ln \sin x\,dx+\int_0^{\frac{\pi}{2}}\ln \cos x\,dx \\ &=\int_0^{\frac{\pi}{2}}\ln \sin 2x\,dx-\frac{\pi}{2}\ln 2 \\ &=\frac{1}{2}\int_0^{\pi}\ln \sin x\ dx-\frac{\pi}{2}\ln 2\\ &=\frac{1}{2}\left(\int_0^{\frac{\pi}{2}}\ln \sin x\ dx+\int_{\frac{\pi}{2}}^{\pi}\ln \sin x\ dx\right)-\frac{\pi}{2}\ln 2\\ &=\int_0^{\frac{\pi}{2}}\ln \sin x\ dx-\frac{\pi}{2}\ln 2. \end{align}

That is $$\int_0^{\frac{\pi}{2}}\ln \sin x\ dx=-\frac{\pi}{2}\ln 2.$$

Answer 3

We can also derive by power series / Fourier series in $\mathbb{C}$.

$$ \ln(1+z) = z - \frac{z^2}{2} + \frac{z^3}{3} \pm \cdots, \quad |z| \leq 1 \land z \neq -1. $$

Substitute $z = e^{2i \theta}$ and take the real part, we get

$$ \ln 2 + \ln\cos\theta = \cos(2\theta) - \frac{\cos(4\theta)}{2} + \frac{\cos(6\theta)}{3} \pm\cdots. $$

(Hint: $\ln z = \ln |z| + i \arg z$, and $|1+e^{2i\theta}| = 2 \cos \theta$.)

Now integrate both sides from $0$ to $\frac\pi2$:

• RHS: They all vanish because $\cos(2n \theta)$ is anti-symmetric about $\theta = \frac\pi4$.
• LHS:
  • $\ln 2$ integrates to $\frac\pi2 \ln 2$.
  • $\int_0^{\pi/2} \ln\cos \theta\ \mathrm{d}\theta = \int_0^{\pi/2} \ln\sin x\ \mathrm{d}x$ by $x = \frac\pi2 - \theta$.

Therefore,

$$ \int_0^{\pi/2} \ln\sin x\ \mathrm{d}x = -\frac\pi2 \ln 2. $$

Further reading

• Tristan Needham, Visual Complex Analysis.

  §2.3 of the book discusses power series, and §2.3.7 is Fourier series.

• 泰勒猫爱丽丝, Why Taylor series is just Fourier series?（为什么泰勒级数与傅里叶级数是一回事）

  Original source.

  (The video is not in English, but math formulae are sufficient for understanding.)

Answer 4

First we begin with an equivalence: \begin{align}\int_{0}^{\pi\over 2}\ln(\sin x)\ dx=x\ln (\sin x)\ dx \biggr|_{0}^{\pi\over2}-\int_{0}^{\pi\over2}x\cot x\ dx\implies \int_{0}^{\pi\over 2}\ln(\sin x)\ dx=-\int_{0}^{\pi\over2}x\cot x\ dx \end{align} We integrate the second integral by Leibnitz rule \begin{align}\int_{0}^{\pi\over2}x\cot x\ dx=\int_{0}^{\pi\over2}\frac{\arctan{\tan x}}{\tan x}\ dx\end{align} \begin{align}I'(a)=\int_{0}^{\pi\over2}\frac{\partial}{\partial a}\frac{\arctan{(a\tan x)}}{\tan x}\ dx=\int_{0}^{\pi\over2}\frac{1}{a^2\tan^2x+1}\ dx\end{align} With the substitution $\tan x = \xi$ (and so $dx=\frac{d\xi}{\xi^2+1}$): \begin{align}\frac{a^2}{a^2-1}\int_{0}^{\infty}\frac{d\xi}{a^2\xi^2+1}-\frac{1}{a^2-1}\int_{0}^{\infty}\frac{d\xi}{\xi^2+1}=\frac{a}{a^2-1}\int_{0}^{\infty}\frac{d\xi}{\xi^2+1}-\frac{1}{a^2-1}\int_{0}^{\infty}\frac{d\xi}{\xi^2+1}\end{align} As the two integrals in the above expression would both equal $\pi\over2$, we would get: \begin{align}I'(a)=\frac{\pi}{2(a+1)}\implies I(1)=\frac{\pi}{2}\ln (2)\end{align} So,\begin{align}\int_{0}^{\pi\over 2}\ln(\sin x)\ dx=-\frac{\pi}{2}\ln (2)\end{align}