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How to calculate $$ \int_0^{\pi/2}\log(1+\sin(x))\log(\cos(x)) \,dx \,\,?$$ I tried to use the Fourier series of log sine and log cos and I got that the integral is equal to : $$ \frac{\pi^2}{24}-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{n+k}}{k(4k^2-(2n-1)^2)}$$ has anyone a idea to how to find the closed-form of the last series or how to start out differently with the integral?

DMcMor
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Ahmad Albaw
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    Some relevant integrals: https://math.stackexchange.com/questions/1299459/which-function-can-be-used-for-substitution , https://math.stackexchange.com/questions/890373/log-trig-integral-with-sin-cos-and-tan , https://math.stackexchange.com/questions/1783743/a-beautiful-integral-int-0-pi-2-log-sin-x-log-cos-x-dx?noredirect=1 – VIVID Feb 08 '21 at 14:56
  • With some help from Mathematica, I obtained $$ \frac{\pi }{8}\sum\limits_{k = 1}^\infty {\frac{1}{{4^k k}}\binom{2k}{k}(H_k + \log 4)} - \frac{1}{2}\sum\limits_{k = 1}^\infty {\frac{{4^k }}{{(2k + 1)^2 }}\frac{1}{{\binom{2k}{k}}}(H_{2k + 1} - H_k )} . $$ I do not know if this helps. – Gary Feb 08 '21 at 15:28
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    If you make change $t=\frac{\pi}{2}-x$ you will come to the integral type $\int_0^{\pi/2}\log(1+\sin(x))\log(\cos(x)) ,dx=\int_0^{\pi/2}\log(1+\cos(x))\log(\sin(x)) ,dx\sim\int_0^{\pi}\log(2\cos^2(x/2))\log(2\sin(x/2)\cos(x/2)) ,dx$ Then you will have integrands type $\log(\cos(x))\log(\sin(x))$ - third link provided by @VIVID – Svyatoslav Feb 08 '21 at 15:37
  • I do apologize - it is not as easy as it looks at first glance. If we change the integrand to $\sim\cos(x/2)\sin(x/2)$ we get very uncomfortable integration limits $(0, \pi/4)$ – Svyatoslav Feb 08 '21 at 17:21

2 Answers2

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Substitute $t=\tan\frac x2$

\begin{align} &\int_0^{\pi/2}\ln(1+\sin x)\ln(\cos x) \,dx \\ =&\>2\int_0^{1}\frac{\ln\frac{(1+t)^2}{1+t^2}\ln \frac{1-t^2}{1+t^2} }{1+t^2}\,dt =\>4I_1 +4 I_2 -2I_3- 6I_4+2I_5 \end{align} where, per the results

\begin{align} I_1 &= \int_0^1 \frac{\ln (1+t)\ln(1-t)}{1+t^2} dt = -G \ln 2-K+\frac{3 \pi ^3}{128}+\frac{3\pi}{32} \ln ^22\\ I_2 &= \int_0^1 \frac{\ln^2(1+t)}{1+t^2} dt = -2 G \ln 2-4 K+\frac{7 \pi ^3}{64}+\frac{3\pi}{16} \ln ^22 \\ I_3 &= \int_0^1 \frac{\ln (1+t^2)\ln(1-t)}{1+t^2} dt = -\frac{1}{2} G \ln 2+4 K -\frac{5 \pi ^3}{64}+\frac{\pi}{8} \ln ^22 \\ I_4 &= \int_0^1 \frac{\ln (1+t^2)\ln(1+t)}{1+t^2} dt = -\frac{5}{2} G \ln 2-4 K+\frac{7 \pi ^3}{64}+\frac{3\pi}{8} \ln ^22\\ I_5 &= \int_0^1 \frac{\ln^2(1+t^2)}{1+t^2} dt = -2 G \ln 2+4 K-\frac{7 \pi ^3}{96}+\frac{7\pi}{8} \ln ^22 \end{align}

with $K= \Im\text{Li}_3\left(\frac{1+i}{2}\right)$. Together $$ \int_0^{\pi/2}\ln(1+\sin x)\ln(\cos x) \,dx =4\Im\text{Li}_3\left(\frac{1+i}{2}\right)-\frac{11\pi^3}{96}+\frac{3\pi}8\ln^22 $$

Quanto
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  • The numerically value is not correct – Ahmad Albaw Feb 08 '21 at 16:27
  • @AhmadAlbaw - fixed a couple of typos; should be good now – Quanto Feb 08 '21 at 17:22
  • Instead $t=\tan(\frac\pi4-\frac x2)$ – Quanto Apr 29 '22 at 21:58
  • Quanto i think you better edit your solution by setting x=(1-x)/(1+x) after you did the half tan subbing. Will look much nicer and simpler – Ali Shadhar Jun 15 '22 at 08:44
  • @AliShadhar - Appreciate the suggestion, which simplifies the integrand considerably and also requires whole-sale changes. It feels almost a different solution. Rather someone posts it as a new answer. – Quanto Jun 15 '22 at 12:25
  • I agree but no need to make any changes to your solution as you can create addendum section and add your second solution there but its your call of course. – Ali Shadhar Jun 15 '22 at 16:23
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We have a closed form for the inner sum $$S_k=\sum_{n=1}^{\infty}\frac{(-1)^{n+k}}{k(4k^2-(2n-1)^2)}$$ $$S_k=(-1)^k \frac{\Phi \left(-1,1,\frac{1}{2}-k\right)-\Phi \left(-1,1,\frac{1}{2}+k\right)}{8 k^2}$$ where appears the Hurwitz-Lerch transcendent function. This can rewrite $$S_k=(-1)^k \frac{-\psi \left(\frac{1}{4}-\frac{k}{2}\right)+\psi \left(\frac{3}{4}-\frac{k}{2}\right)+\psi \left(\frac{k}{2}+\frac{1}{4}\right)-\psi \left(\frac{k}{2}+\frac{3}{4}\right) }{16 k^2 }$$

Claude Leibovici
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