Finding $\displaystyle \int^{\frac{\pi}{4}}_{0}\ln(1+\cos x)\mathrm dx$
What I tried
Put $\displaystyle I =\int^{\frac{\pi}{4}}_{0}\ln(1+\cos x)\mathrm dx$
\begin{align*}
I&=\int^{\frac{\pi}{4}}_{0}\ln(1+\cos x)\mathrm dx\\
&=\int^{\frac{\pi}{4}}_{0}\ln\bigg(2\cos^2\frac{x}{2}\bigg)\mathrm dx\\
&=\frac{\pi}{4}\ln(2)+4\int^{\frac{\pi}{8}}_{0}\ln(\cos x)\mathrm dx\\
&=\frac{\pi}{4}\ln(2)+4\left[\left[x\ln(\cos x)\right]^{\frac{\pi}{8}}_{0}+\int^{\frac{\pi}{8}}_{0}\ln(\tan x)\mathrm dx\right]\\
&=\frac{\pi}{8}\ln(2)+\frac{\pi}{2}\ln\left(\cos \frac{\pi}{8}\right)+4\int^{\frac{\pi}{8}}_{0}\ln(\tan x)\mathrm dx
\end{align*}
How do I solve it? Help me, please.
Best Answer
\begin{align}J=\int_0^{\frac{\pi}{8}}\ln\left(\tan x\right)\,dx\end{align}
Perform the change of variable $y=\left(\sqrt{2}+1\right)\tan x$,
\begin{align}J&=(\sqrt{2}-1)\int_0^{1}\frac{\ln\big((\sqrt{2}-1)x\big) }{1+(\sqrt{2}-1)^2x^2}\,dx\\ &=(\sqrt{2}-1)\int_0^{1}\frac{\ln\big(\sqrt{2}-1\big) }{1+(\sqrt{2}-1)^2x^2}\,dx+(\sqrt{2}-1)\int_0^{1}\frac{\ln x }{1+(\sqrt{2}-1)^2x^2}\,dx\\ &=\frac{\pi}{8}\ln\big(\sqrt{2}-1\big)+(\sqrt{2}-1)\int_0^{1}\frac{\ln x }{1+(\sqrt{2}-1)^2x^2}\,dx\\ &=\frac{\pi}{8}\ln\big(\sqrt{2}-1\big)+\frac{1}{2}\big(\sqrt{2}-1\big)\int_0^1\frac{\ln x}{1-i\big(1-\sqrt{2}\big)x}\,dx+\frac{1}{2}\big(\sqrt{2}-1\big)\int_0^1\frac{\ln x}{1-i\big(\sqrt{2}-1\big)x}\,dx\\ &=\boxed{\frac{\pi}{8}\ln\big(\sqrt{2}-1\big)+\frac{1}{2}i\text{Li}_2\left(i\big(\sqrt{2}-1\big)\right)-\frac{1}{2}i\text{Li}_2\left(i\big(1-\sqrt{2}\big)\right)} \end{align}
NB:
For $0<|z|<1$ complex,
\begin{align}\int_0^1 \frac{\ln x}{1-zx}\,dx&=\int_0^1 \left(\sum_{n=0}^{\infty}(xz)^n\right)\ln x\,dx\\ &=\sum_{n=0}^{\infty} z^n\left(\int_0^1 x^n\ln x\,dx\right)\\ &=-\sum_{n=0}^{\infty} \frac{z^n}{(n+1)^2}\\ &=-\frac{1}{z}\sum_{n=0}^{\infty} \frac{z^{n+1}}{(n+1)^2}\\ &=-\frac{1}{z}\text{Li}_2(z) \end{align}
PS:
if you prefer,
\begin{align}\boxed{J=\frac{\pi}{8}\ln\big(\sqrt{2}-1\big)-\Im{\left(\text{Li}_2\left(i\big(\sqrt{2}-1\big)\right)\right)}}\end{align}