How to compute the integral $\int^{\pi/2}_0\ln(1+\tan\theta)d\theta$?

Question

How to compute the integral $\int^{\pi/2}_0\ln(1+\tan\theta)d\theta$. If we let $t=\tan\theta$, then the integral becomes to

$$\int^{\pi/2}_0\ln(1+\tan\theta)d\theta=\int_0^\infty\frac{\ln(1+t)}{1+t^2}dt$$. Can we calculate this integral explicitly?

Answer 1

$$\begin{align}\int_0^{\pi/2} d\theta \, \log{(1+\tan{\theta})} &=\int_0^{\pi/2} d\theta \, \log{(\sin{\theta}+\cos{\theta})} - \int_0^{\pi/2} d\theta \, \log{(\cos{\theta})} \\ &= \int_0^{\pi/2} d\theta \, \log{\left [\sqrt{2}\cos{\left (\theta-\frac{\pi}{4} \right )}\right ]} - \int_0^{\pi/2} d\theta \, \log{(\cos{\theta})} \\ &= \frac{\pi}{4} \log{2} + \int_0^{\pi/2} d\theta \, \log{\left [\cos{\left (\theta-\frac{\pi}{4} \right )}\right ]} - \int_0^{\pi/2} d\theta \, \log{(\cos{\theta})}\\ &= \frac{\pi}{4} \log{2} + \int_{-\pi/4}^{\pi/4} d\theta \, \log{\left (\cos{\theta}\right )} - \int_0^{\pi/2} d\theta \, \log{(\cos{\theta})}\\ &= \frac{\pi}{4} \log{2} + \int_{0}^{\pi/4} d\theta \, \log{\left (\cos{\theta}\right )} - \int_{\pi/4}^{\pi/2} d\theta \, \log{(\cos{\theta})} \\ &= \frac{\pi}{4} \log{2} + \int_{0}^{\pi/4} d\theta \, \log{\left (\cos{\theta}\right )} - \int_{0}^{\pi/4} d\theta \, \log{\left (\sin{\theta}\right )}\end{align} $$

Now use the Fourier series representations:

$$-\log(\sin(\theta))=\sum_{k=1}^\infty\frac{\cos(2k \theta)}{k}+\log(2)$$

and

$$-\log(\cos(\theta))=\sum_{k=1}^\infty(-1)^k\frac{\cos(2k \theta)}{k}+\log(2)$$

Substituting, exchanging the respective sums and integrals, we get

$$\begin{align}\int_0^{\pi/2} d\theta \, \log{(1+\tan{\theta})} &= \frac{\pi}{4} \log{2} + \sum_{k=1}^{\infty} \frac1{2 k^2} \left [1-(-1)^k \right ] \sin{\frac{\pi}{2} k} \\ &= \frac{\pi}{4} \log{2} + \sum_{k=1}^{\infty} \frac{(-1)^k}{(2 k+1)^2} \\ &= \frac{\pi}{4} \log{2} + G\end{align} $$

where $G$ is Catalan's constant.

Answer 2

$$$$ Let us start to calculate it. \begin{eqnarray} &&\int_0^{\pi/2}\ln(1+\tan \theta)d\theta \\ &=&\int_0^{\pi/2}\ln(\frac{\sin \theta+\cos \theta}{\cos \theta})d\theta\\ &=&\int_0^{\pi/2}[ \ln(\sqrt{2}\cos (\theta-\pi/4))-\ln (\cos \theta) d\theta ]\\ &=& \frac{\pi}{4}\ln2+\int_0^{\pi/2}\ln(\cos \theta-\pi/4)d\theta-\int_0^{\pi/2}\ln (\cos \theta) d\theta \\ &=&\frac{\pi}{4}\ln2+\int_{-\pi/4}^{\pi/4}\ln\cos (\theta)d\theta-\int_0^{\pi/2}\ln (\cos \theta) d\theta \\ &=&\frac{\pi}{4}\ln2+2\int_{0}^{\pi/4}\ln\cos (\theta)d\theta-\int_0^{\pi/2}\ln (\cos \theta) d\theta \\ &=&\frac{\pi}{4}\ln2+G \end{eqnarray}

We know the following integrals:

$$$$ \begin{eqnarray} \int_{0}^{\pi/4}\ln\cos (\theta)d\theta&=&-\frac{\pi}{4}\ln2+\frac{G}{2}\\ \int_{0}^{\pi/2}\ln\cos (\theta)d\theta &=& -\frac{\pi}{2}\ln2\\ \end{eqnarray} where $G$ is Catalan constant.

Answer 3

Let $$ I(a)=\int_0^{\pi/2}\ln(1+a\tan x)dx. $$ Then $I(0)=0$ and \begin{eqnarray} I'(a)&=&\int_0^{\pi/2}\frac{\tan x}{1+a\tan x}dt\\ &=&\int_0^{\pi/2}\frac{\sin x}{\cos x+a\sin x}dx. \end{eqnarray} Let $$ A=\int_0^{\pi/2}\frac{\sin x}{\cos x+a\sin x}dx, B=\int_0^{\pi/2}\frac{\cos x}{\cos x+a\sin x}dx. $$ Then it is easy to check $$ -A+aB=\ln a, B+aA=\frac{\pi}{2}, $$ from which we have $$ I'(a)=A=\frac{a\pi-2\ln a}{2(1+a^2)}. $$ Noting we have $$ I(1)=\int_0^1\frac{a\pi-2\ln a}{2(1+a^2)}da=\frac{\pi}{4}\ln2+G. $$