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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!

Philipp
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Math Maniac
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4 Answers4

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

ling
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  • Excellent answer, thank you. (+1) – A-Level Student May 21 '21 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. – Philipp Jan 06 '22 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 08 '22 at 02:03
  • 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? – Philipp Jan 08 '22 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 09 '22 at 02:28
  • Ah ok, got it. Thanks :) Btw my previous comment was wrong, please ignore it. – Philipp Jan 09 '22 at 12:35
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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

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

Dispersion
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    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? – Math Maniac Sep 26 '18 at 17:22
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    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 $I'(0)$. – Dispersion Sep 26 '18 at 18:22
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    Thank you Zachary! – Math Maniac Sep 27 '18 at 00:22
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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}

Shimura Variety
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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

Y.D.X.
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