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

Question

How can we evaluate the integration $$ \int_{0}^{\pi/2}\ln\left(\sin\left(x\right)\right) \,{\rm d}x $$ by using DUIS ( differentiation under the integral sign ) ?.

This question popped into my head when I read an article about DUIS as $\ln\left(\left\vert\sin\left(x\right)\right\vert\right)$ is the integral of $\cot\left(x\right)$.

Although I am in $12$th Standard, I am keen to learn new and exciting concepts and techniques, so please tell me if you have any questions related to this. Thanks !.

Answer 1

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 2

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 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}

Answer 5

The evaluation of the integral (using the DUIS) has already been provided. Here, I present a different approach. It uses a beautiful product of sine functions, i.e., $\displaystyle\prod_{k=1}^{n-1}\sin\left(\frac{k \pi}{n}\right) = \frac{n}{2^{n-1}}$.

Note that $f(x)=\ln(\sin(x))$ is symmetric about $x=\frac\pi2$ since $f\left(2.\frac\pi2-x\right)=f(\pi-x)=f(x)$. So, $\displaystyle\int_0^\pi\ln(\sin(x))\ \mathrm dx=2\int_0^\frac\pi2\ln(\sin(x))\ \mathrm dx=2\int_{\frac\pi2}^\pi\ln(\sin(x))\ \mathrm dx$.

Now, the given integral can be evaluated as: $$\displaystyle\begin{align*}\int_0^\frac\pi2\ln(\sin(x))\ \mathrm dx &=\frac12\int_0^\pi\ln(\sin(x))\ \mathrm dx\\&=\frac12\displaystyle\lim_{n\rightarrow\infty}\bigg[\bigg(\frac{\pi-0}{n}\bigg)\sum_{k=1}^{n-1}\ln\bigg(\sin\bigg(0+\frac{(\pi-0)k}{n}\bigg)\bigg)\bigg]\\ &=\displaystyle\frac\pi2\lim_{n\rightarrow\infty}\bigg[\frac{1}{n}\ln\bigg(\prod_{k=1}^{n-1}\sin\bigg(\frac{k\pi}{n}\bigg)\bigg)\bigg]\\ &=\displaystyle\frac\pi2\lim_{n\rightarrow\infty}\bigg[\frac{1}{n}\ln\bigg(\frac{n}{2^{n-1}}\bigg)\bigg]\\ &=\displaystyle\frac\pi2\lim_{n\rightarrow\infty}\bigg[\frac{\ln n}{n}-\bigg(1-\frac{1}{n}\bigg)\ln2\bigg]\\ \color{red}{\int_0^\frac\pi2\ln(\sin(x))\ \mathrm dx}&\color{red}{=\displaystyle-\frac\pi2\ln2}\end{align*}$$

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Alternatively, $$\begin{aligned}\mathcal I=\int_0^\frac\pi2\ln(\sin(x))\ \mathrm dx&=\frac12\left(\int_0^\frac\pi2\ln(\sin(x))\ \mathrm dx+\int_0^\frac\pi2\ln\left(\sin\left(\frac\pi2-x\right)\right)\ \mathrm dx\right)\\&=\frac12\left(\int_0^\frac\pi2\ln(\sin(x))\ \mathrm dx+\int_0^\frac\pi2\ln(\cos(x))\ \mathrm dx\right)\\&=\frac12\int_0^\frac\pi2\ln(\sin(x)\cos(x))\ \mathrm dx\\&=\frac12\int_0^\frac\pi2\ln(\sin(2x))\ \mathrm dx-\frac12\int_0^\frac\pi2\ln(2)\ \mathrm dx\\&=\frac{1}{2\cdot2}\int_0^\pi\ln(\sin(u))\ \mathrm du-\frac\pi4\ln2\qquad\text{substitute } u=2x\\\mathcal I&=\frac{\mathcal I}2-\frac\pi4\ln2\\\frac{\mathcal I}2&=-\frac\pi4\ln2\\\color{red}{\int_0^\frac\pi2\ln(\sin(x))\ \mathrm dx=\mathcal I}&\color{red}{=-\frac\pi2\ln2}& \end{aligned}$$

Hence, the required integral is $$\color{blue}{\boxed{\boxed{\int_0^\frac\pi2\ln(\sin(x))\ \mathrm dx=-\dfrac\pi2\ln2}}}$$

Hope this helps!