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Evaluate :$I=\int_0^{\frac{\pi}{2}} \ln^2(1+\sin(x)) dx $

Using WolframAlpha:$$I \approx 0.4162...$$

I tried to change variables $y=\tan(\frac{x}{2})$

\begin{aligned} I &= 2\int^1_0 \frac{\ln^2\left({\frac{(1+y)^2}{1+y^2}}\right)}{1+y^2} dy \\ &= 2(I_1+I_2+I_3) \end{aligned}

Where : \begin{aligned} I_1 &= 4\int^1_0 \frac{\ln^2(1+y)}{1+y^2} dy \\ I_2 &= \int^1_0\frac{\ln^2(1+y^2)}{1+y^2} dy \\ I_3 &= -4\int^1_0 \frac{\ln(1+y)\ln(1+y^2)}{1+y^2} dy \end{aligned}

Delta
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3 Answers3

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Not a full answer but evaluation of $I_1$:

According to the reference below,

$$I=\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx=4\Im\operatorname{Li}_3(1+i)-\frac{7\pi^3}{64}-\frac{3\pi}{16}\ln^22-2\ln2\ G$$ (See: Compute $ \int_0^1\frac{\ln^2(1+x)}{1+x^2}\, dx$)

Hence $I_1= 4 (4\Im\operatorname{Li}_3(1+i)-\frac{7\pi^3}{64}-\frac{3\pi}{16}\ln^22-2\ln2\ G) $

Samuel
  • 1,051
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Let $$I=\int_0^{{\pi}/{2}} \ln^2(1+\sin x) dx=2\int_0^{\pi/4} \ln^2(2\cos^2x) dx \\ J=\int_0^{{\pi}/{2}} \ln^2(1-\sin x) dx =2\int_0^{\pi/4} \ln^2(2\sin^2 x) dx $$ Note that \begin{align} J+I &=2 \int_0^{\pi/4} \ln^2(2\cos^2 x) + \ln^2(2\sin^2 x) \ dx\\ &= 2\int_0^{\pi/2} \ln^2(2\sin^2 x) dx = \pi \ln^22+\frac{\pi^3}{3}\\ J-I& =2\int_0^{\pi/4} \ln^2(2\sin^2 x) -\ln^2(2\cos^2 x) \ dx\\ &=8 \int_0^{\pi/4}\ln(\sin 2x)\ln{\tan x} \overset{t=\tan x}{dx}\\ &= 8\int_0^1 \frac{\ln^2 t -\ln t\ln\frac{1+t^2}2}{1+t^2}\ dt \\ &= \frac{5\pi^3}{16} +\frac\pi{4}\ln^22- 8\text{Im}\>\text{Li}_3\left(\frac{1+i}2\right) \end{align} where $\int_0^1 \frac{\ln^2 t}{1+t^2}dt=\frac{\pi^3}{16}$ and $ \int_0^1 \frac{\ln t\ln\frac{1+t^2}2}{1+t^2}dt =2\text{Im}\>\text{Li}_3\left(\frac{1+i}2\right) -\frac{\pi}{16}\ln^22-\frac{\pi^3}{64}$. Thus $$I=-\frac{7\pi^3}{48} +\frac{\pi}{4}\ln^22+8\text{Im}\>\text{Li}_3\left(\frac{1+i}2\right)\\ J=\frac{23\pi^3}{48} +\frac{3\pi}{4}\ln^22- 8\text{Im}\>\text{Li}_3\left(\frac{1+i}2\right)\\ $$

Quanto
  • 120,125
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The answer is $$I=\frac{7\pi^3}{24}+\pi \log^2(2)+8\Im\mathrm{Li}_3(1-i)$$

We will use the Fourier series $$\ln^2(2\cos x)=x^2+2\sum_{n=1}^{\infty}\frac{(-1)^n}{n}H_{n-1}\cos(2nx)\quad ,\quad -\frac{\pi}{2}<x<\frac{\pi}{2}$$

Start by writing

$$I=\int_0^{\pi/2}\log^2(1+\cos x)\, dx= 2\int_0^{\pi/4}(2\log(2\cos x)-\log 2)^2\, dx$$ $$=8\int_0^{\pi/4}\log^2(2\cos x)\, dx-8\log(2)\int_0^{\pi/4}\log(2\cos x)\, dx+\frac{\pi}{2}\ln^2(2)$$ It is easy to show that $$\int_0^{\pi/4}\log(2\cos x)\, dx=\frac{G}{2}$$ where $G$ is the Catalan constant. The crux is the first integral: I will leave the details but using the Fourier series given above you should be able to obtain: $$\int_0^{\pi/4}\log^2(2\cos x)\, dx=\frac{\pi^3}{192}+\sum_{n\geq 1}\frac{H_{2n}}{(2n+1)^2}(-1)^{n+1}$$ Where $$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}H_{2n+1}=-\Im\text{Li}_3(1-i)-\frac{\pi}{16}\log^2(2)-\frac{G}{2}\log 2$$ solves the integral. Gathering the results solves $I$

  • This is beyond impressive. How did you educate yourself in order to solve integrals like this? – Samuel Jun 07 '25 at 13:59