Definite integral $\int_0^{\pi/4}\log\left(\tan{x}\right)\ dx$

Question

$$\int_0^{\pi/4}\log\left(\tan{x}\right)\ dx$$ My turn : $$I=\int_0^{\pi/4}\log\left(\sin{x}\right)-\log\left(\cos{x}\right)\ dx$$ $$I_1=\int_0^{\pi/4}\log\left(\sin{x}\right)\ dx$$ let $$u=2x$$ $$I_1=\frac{1}{2}\int_0^{\pi/2}\log\left(\sin{\frac{u}{2}}\right)\ du$$ But i could not evaluate the last integral ?

Answer 1

Under $\tan x\to x$, one has \begin{eqnarray} &&\int_0^{\pi/4}\ln(\tan x)dx\\ &=&\int_0^1\frac{\ln x}{1+x^2}dx\\ &=&\int_0^1\ln x\sum_{n=0}^\infty(-1)^nx^{2n}dx\\ &=&-\sum_{n=0}^\infty\int_0^1(-1)^nx^{2n}\ln xdx\\ &=&-\sum_{n=0}^\infty(-1)^n\frac{1}{(2n+1)^2}\\ &=&-C \end{eqnarray} where $C$ is the Catalan constant.

Answer 2

Another way is to use the Fourier expansion of $\log\sin x$ and $\log\cos x$ and integrate the expression term-wise. Since

$$\begin{align*}\log\cos x & =\sum\limits_{k\geq1}(-1)^{k-1}\frac {\cos 2kx}k-\log 2\\\log\sin x & =-\sum\limits_{k\geq1}\frac {\cos 2kx}k-\log 2\end{align*}$$

Therefore$$\begin{align*}I & =\int\limits_0^{\pi/4}dx\,\log\tan x\\ & =\int\limits_0^{\pi/4}dx\,\log\sin x-\int\limits_0^{\pi/4}dx\,\log\cos x\\ & =-\sum\limits_{k\geq1}\int\limits_0^{\pi/4}dx\,\frac {\cos 2kx}k+\sum\limits_{k\geq1}\int\limits_0^{\pi/4}dx\, (-1)^k\frac {\cos 2kx}k\end{align*}$$

Now integrate within the limits to get$$\begin{align*}I & =\frac 12\sum\limits_{k\geq1}\frac {(-1)^k\sin\left(\frac {\pi k}2\right)}{k^2}-\frac 12\sum\limits_{k\geq1}\frac {\sin\left(\frac {\pi k}2\right)}{k^2}\\ & =-\frac 12\sum\limits_{k\geq1}\frac {(-1)^k}{(2k-1)^2}-\frac 12\sum\limits_{k\geq1}\frac {(-1)^k}{(2k-1)^2}\\ & =-G\end{align*}$$Where $G$ is Catalan's Constant.

Answer 3

$$I(n)=\int_0^\frac{\pi}4 (\ln(\tan x))^n\,dx\underset{\tan x\to x}=\int_0^1 \frac{\ln^nx}{1+x^2}\,dx$$

$$\int_0^1 \frac{\ln^nx}{1+x^2}\,dx=\sum_{k=0}^\infty(-1)^k\color{blue}{\int_0^1 x^{2k}\ln^nx\,dx}$$

Put $-\ln(x)=t$ and apply Gamma function to derive below result, $$\int_0^1 x^{m}\ln^nx\,dx=\frac{(-1)^n n!}{(m + 1)^{(n + 1)}}\implies \color{blue}{\int_0^1 x^{2k}\ln^nx\,dx=\frac{(-1)^n n!}{(2k + 1)^{(n + 1)}}}$$

$$\sum_{k=0}^\infty(-1)^k\int_0^1 x^{2k}\ln^nx\,dx=(-1)^n\Gamma(n+1)\color{blue}{\sum_{k=0}^\infty(-1)^k\frac{ 1}{(2k + 1)^{(n + 1)}}}$$

Using the Dirichlet Beta function,

$$\displaystyle \beta (s)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)^{s}}\implies\color{blue}{\displaystyle \beta (s+1)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)^{s+1}}}}}$$

$$\therefore \int_0^\frac{\pi}4 (\ln(\tan x))^n\,dx = (-1)^n\Gamma(n+1)\beta(n+1)\underset{n=1}\implies \int_0^\frac{\pi}4 \ln(\tan x)\,dx = -\beta(2)=-G$$

• G : Catalan Constant, $G=\beta (2)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)^{2}}}$