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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.

Gary
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Lai
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  • This post may be useful. – NikoWielopolski Mar 23 '22 at 07:41
  • In terms of $f(x):=\int_0^x\ln(1+\sin t)dt$,$$\begin{align}\int_0^x\ln(1-\sin t)dt&=-f(-x)\\implies\int_0^x\ln(1-\cos t)dt&=f(x-\pi/2)-f(-\pi/2)\\implies\int_0^x\ln(1+\sin t-\cos t)dt&=\frac12[x\ln2+f(x)+f(x-\pi/2)-f(-\pi/2)]\\implies\int_0^{\pi/4}\ln(1+\sin t-\cos t)dt&=\frac12\left[\frac{\pi\ln2}{4}+f(\pi/4)+f(-\pi/4)-f(-\pi/2)\right].\end{align}$$ – J.G. Mar 23 '22 at 08:11

2 Answers2

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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 $n\geq 5$.

Claude Leibovici
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Inspired by Claude, I try to find its indefinite integral. $$ \begin{aligned} \int \ln (1+\sin x-\cos x) d x = & \int \ln \left(2 \sin ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}\right) d x \\ = & \int \ln 2 d x+\int \ln \left(\sin \frac{x}{2}\right) d x+\int\ln \left(\sin \frac{x}{2}+\cos \frac{x}{2}\right) d x \\ = & x \ln 2+\int \ln (\sin \frac{x}{2} ) d x+\frac{1}{2} \int \ln \left(1+\sin x\right)dx \end{aligned} $$

$$ \begin{aligned}\int \ln \left(2 \sin \frac{x}{2}\right) d x=&\int \ln \left(\frac{e^{\frac{x i}{2}}-e^{\frac{-x i}{2}}}{i}\right) d x \\ = &\int \ln \left(e^{\frac{x i}{2}}\left(1-e^{-x i}\right)\right) d x-\int \ln i d x\\ = & \frac{x^2 i}{4}+\int \ln \left(1-e^{-x i}\right) d x-\frac{\pi x}{2} i \\ = & \left(\frac{x^2}{4}-\frac{\pi x}{2}\right) i-\frac{1}{2} i \operatorname{Li}_2\left(e^{-x i}\right) \\ = & \frac{1}{2} \Im\left(\operatorname{Li}_2\left(e^{-x i}\right)\right) \textrm{ for real }x .\end{aligned} $$ Hence $$\int \ln \sin \left(\frac{x}{2}\right)d x =-x\ln 2+ \frac{1}{2} \Im\left(\operatorname{Li}_2\left(e^{-x i}\right)\right) $$

\begin{aligned} \int \ln (1+\cos x) d x&=\int \ln \left(2 \cos ^2 \frac{x}{2}\right) d x \\ = & \int \ln \left(\frac{\left(2 \cos \frac{x}{2}\right)^2}{2}\right) d x \\ = & 2 \int \ln \left(2 \cos \frac{x}{2}\right) d x-x \ln 2 \\ = & 2 \int \ln \left(e^{\frac{x}{2} i}+e^{-\frac{x}{2} i}\right) d x-x \ln 2 \\ = & 2 \int \ln e^{\frac{x}{2} i} d x+2 \int \ln \left(1+e^{-x i}\right) d x-x \ln 2 \\ = & 2 \Im\left(\operatorname{ Li} _2\left(-e^{-x i}\right)\right)-x \ln 2 \end{aligned}

Now we can conclude that

$$\int \ln (1+\sin x-\cos x) d x = -\frac{x}{2} \ln 2+\frac{1}{2}\Im\left(\operatorname{ Li} _2\left(e^{-x i}\right)\right)+\Im\left(\operatorname{ Li} _2\left(-e^{-x i}\right)\right)+C $$ and hence $$\boxed{\int_0^t \ln (1+\sin x-\cos x) d x = -\frac{t}{2} \ln 2+\frac{1}{2}\Im\left(\operatorname{ Li} _2\left(e^{-ti}\right)\right)+\Im\left(\operatorname{ Li} _2\left(-e^{-ti}\right)\right)} $$ In particular, $$ \int_0^{\frac{\pi}{4}} \ln (1+\sin x-\cos x) d x = -\frac{\pi}{8} \ln 2+\frac{1}{2} \Im\left(\operatorname{ Li} _2\left(\frac{1-i}{\sqrt{2}}\right)\right)+\Im\left(\operatorname{ Li} _2\left(\frac{i-1}{\sqrt{2}}\right)\right) $$

Lai
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