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

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!

• Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. Sep 26, 2018 at 9:46
• $\log\sin x$ has no elementary antiderivative, DUIS will not help.
– user65203
Sep 26, 2018 at 9:58
• And by the way, this formula is not used as an integration technique.
– user65203
Sep 26, 2018 at 10:04
• Ok thanks Yves Daoust... Can you tell me what should i learn about integration after duis? Sep 26, 2018 at 10:05

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

• Excellent answer, thank you. (+1) May 21, 2021 at 16:57
• Note that $\int\limits_0^{\frac{\pi}{2}}\ln (\sin x)dx$ is an improper integral. Without proving or at least assuming its existence it makes no sense to apply integration by substitution. Jan 6, 2022 at 20:22
• @Philipp Yes, you are right. We should first prove the existence of the integral, and it is clear by noting that $x^\varepsilon\ln \sin x\to 0$ as $x\to 0$.
– ling
Jan 8, 2022 at 2:03
• Just to make sure that I understood your comment right, if $0<x\leq1$, then $\ln(\sin(x))\leq x^{\epsilon}\ln(\sin(x))<0$, where $\epsilon\in\mathbb{R}$. Therefore integral $\int\limits_0^{1}x^{\epsilon}\ln(\sin(x)) dx$ exists and of course $\int\limits_0^{\frac{\pi}{2}}x^{\epsilon}\ln(\sin(x)) dx$. So by comparison test $\int\limits_0^{\frac{\pi}{2}}\ln (\sin x)dx$ exists. Is this what you had in mind? Jan 8, 2022 at 10:11
• @Philipp Not exactly. What I mean is that for any fixed $0<\varepsilon<1$, there is $|\ln\sin x|\leq\frac{C}{x^\varepsilon}$ for $x$ sufficiently close to $0$, then the integral exists by comparison test.
– ling
Jan 9, 2022 at 2:28

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

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

• I use hit and trial method for defining the function on which we have to apply DUIS. Is hit and trial the only way or do we use some method for defining that function? Sep 26, 2018 at 17:22
• There is no general way to find an integrand on which to apply differentiation under the integral sign, but there are some tricks you can use--and with experience, you will have more tools at your disposal and be able to perhaps combine them in effective ways. I will also add that this can be a highly creative and imaginative process, and I find it incredibly fun and rewarding. An example of a technique I used in this answer: when you have an integral of the form $\int \log f(x) dx$, try to compute the integral $I(a)=\int f(x)^a dx$, differentiate it with respect to $a$, finding $I'(0)$. Sep 26, 2018 at 18:22
• Thank you Zachary! Sep 27, 2018 at 0:22

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}

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,

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