When I met the integral $\int_{0}^{\frac{\pi}{2}} \ln (1+\sin x-\cos x) d x$, I evaluated it by squaring $ 1+\sin x -\cos x. $ $ \begin{aligned}\int_{0}^{\frac{\pi}{2}} \ln (1+\sin x-\cos x) d x&=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \ln (1+\sin x-\cos x)^{2} d x \\&=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \ln [2(1+\sin x-\cos x-\sin x \cos x)] d x \\&=\frac{1}{2}\left[\int_{0}^{\frac{\pi}{2}} \ln 2 d x+ \int_{0}^{\frac{\pi}{2}} \ln [(1+\sin x)(1-\cos x)] d x\right]\end{aligned} \tag*{} $ Now we deal with the last integral $ \displaystyle \begin{aligned}\int_{0}^{\frac{\pi}{2}} \ln [(1+\sin x)(1-\cos x)] d x&=\int_{0}^{\frac{\pi}{2}} \ln (1+\sin x) d x+\int_{0}^{\frac{\pi}{2}} \ln (1-\cos x) d x \\&=\int_{0}^{\frac{\pi}{2}} \ln (1+\cos x) d x+\int_{0}^{\frac{\pi}{2}} \ln (1-\cos x) d x \\&=\left.\int_{0}^{\frac{\pi}{2}} \ln \left(1-\cos ^{2} x\right) d x\right.\\&=2 \int_{0}^{\frac{\pi}{2}} \ln (\sin x) d x \quad \textrm{ (By my post in Footnote )}\\&=2\left(-\frac{\pi}{2} \ln 2\right)\\&=-\pi \ln 2\end{aligned} \tag*{} $ We can now conclude that $\displaystyle \boxed{\int_{0}^{\frac{\pi}{2}} \ln (1+\sin x-\cos x) d x=\frac{1}{2}\left(\frac{\pi}{2} \ln 2-\pi \ln 2\right)=-\frac{\pi}{4} \ln 2 } \tag*{} $
Then I want to find the integral with narrow width $0$ to $\frac{\pi}{4}$.
Similarly, we have $$\int_{0}^{\frac{\pi}{4}} \ln (1+\sin x-\cos x) d x =\frac{1}{2}\left[\frac{\pi}{4}\ln 2 + \int_{0}^{\frac{\pi}{4}} \ln [(1+\sin x)(1-\cos x)] d x\right] $$
$$\int_{0}^{\frac{\pi}{4}} \ln [(1+\sin x)(1-\cos x)] d x= \int_{0}^{\frac{\pi}{4}} \ln (1+\sin x) d x+\int_{0}^{\frac{\pi}{4}} \ln (1-\cos x) d x $$
\begin{aligned}\int_{0}^{\frac{\pi}{4}} \ln (1-\cos x) d x&=\int_{0}^{\frac{\pi}{4}} \ln \left(2 \sin ^{2} \frac{x}{2} \right) d x \\ &=\int_{0}^{\frac{\pi}{4}}[\ln 2+2 \ln (\sin \frac{x}{2})] d x \\ &=\frac{\pi}{4} \ln 2+4 \int_{0}^{\frac{\pi}{8}} \ln (\sin x) d x \\ \end{aligned}
However, I am stuck in this integral and $ \displaystyle \int_{0}^{\frac{\pi}{4}} \ln (1+\sin x) d x$. Your help and suggestion is highly appreciated.
Footnote: My post.
Starting from the beginning, using the tangent half-angle substitution $$I=\int\log \big[1+\sin (x)-\cos (x)\big]\, d x=2\int \frac{\log \left(\frac{2 t (t+1)}{t^2+1}\right)}{t^2+1}\,dt$$ Expanding the numerator $$\frac{\log \left(\frac{2 t (t+1)}{t^2+1}\right)}{t^2+1}=\frac{\log(2)}{t^2+1}+\frac{\log(t)+\log(t+1)-\log(t+i)-\log(t-i) }{(t+i)(t-i)}$$ Use also $$\frac 1{(t+i)(t-i)}=\frac i 2 \left(\frac{1}{t+i}-\frac{1}{t-i}\right)$$ So, we face a bunch of integrals looking like $$\int \frac {\log (t+a)}{t+b}\,dt=\text{Li}_2\left(\frac{a+t}{a-b}\right)+\log (a+t) \log \left(\frac{b+t}{b-a}\right)$$
Back to $x$,the antiderivative is $$I=i \left(\text{Li}_2\left(i e^{i x}\right)+\text{Li}_2\left(e^{i x}\right)+\frac{1 }{2}x^2\right)+x \log \left(\left(\frac{1}{2}+\frac{i}{2}\right) (\sin (x)+i \cos (x))\right)$$
Now, for $$J(t)=\int_0^t\log \big[1+\sin (x)-\cos (x)\big]\, d x$$ we have $$J(0)=-C+\frac{7 \pi ^2}{48}\,i$$
Now, making $t=\frac \pi n$
$$K_n=\int_0^{\frac \pi n }\log \big[1+\sin (x)-\cos (x)\big]\, d x$$ gives $$K_1=2 C-\frac{1}{2} \pi \log (2) \qquad \qquad K_2=-\frac{1}{4} \pi \log (2)$$ $$K_4=C-\frac{1}{8} \pi \log (2)-\frac{\psi ^{(1)}\left(\frac{1}{8}\right)+\psi ^{(1)}\left(\frac{3}{8}\right)-\psi ^{(1)}\left(\frac{5}{8}\right)-\psi ^{(1)}\left(\frac{7}{8}\right)}{32 \sqrt{2}}$$
I did not find any other looking nice.
Expanded as series
$$K_n=k \log(k)+\sum_{m=1}^\infty (-1)^n\, a_n\, k^n \qquad \text{with} \qquad k=\frac \pi n$$ and the first coefficients are $$\left\{1,\frac{1}{4},\frac{7}{72},\frac{1}{48},\frac{121}{14400},\frac{1}{288},\cdots\right\}$$
Limited to these terms, the relative error is less than $0.001$% as soon as $n\geq 5$.