Integrate $I=\int_0^{\pi/2}x^2\ln(\sinh x)\ln(\cosh x)dx$

Question

Hi I am trying to evaluate the integral $$I=\int_0^{\pi/2}x^2\ln(\sinh x)\ln(\cosh x)dx.$$ Note we can write the integrand as $$ x^2 \ln\big(\frac{e^x-e^{-x}}{2}\big) \ln\big(\frac{e^x+e^{-x}}{2}\big)=x^2\left( \big(\ln(e^x-e^{-x})-\ln 2\big)\big(\ln(e^x+e^{-x})-\ln 2\big)\right). $$ I am unsure of where to go from here. Thanks

Answer 1

Sinilar to Integrate $\int_0^\pi \theta^2 \ln^2\big(2\cosh\frac{\theta}{2}\big)d \theta$,

$$\int_0^\frac{\pi}{2}x^2\ln(\sinh x)\ln(\cosh x)~dx$$

$$=\int_0^\frac{\pi}{2}x^2\ln\dfrac{e^x-e^{-x}}{2}\ln\dfrac{e^x+e^{-x}}{2}~dx$$

$$=\int_0^\frac{\pi}{2}x^2(\ln(e^x-e^{-x})-\ln2)(\ln(e^x+e^{-x})-\ln2)~dx$$

$$=\int_0^\frac{\pi}{2}x^2(\ln^22-(\ln(e^x-e^{-x})+\ln(e^x+e^{-x}))\ln2+\ln(e^x-e^{-x})\ln(e^x+e^{-x}))~dx$$

$$=\int_0^\frac{\pi}{2}x^2(\ln^22-\ln(e^{2x}-e^{-2x})\ln2+\ln(e^x-e^{-x})\ln(e^x+e^{-x}))~dx$$

$$=\int_0^\frac{\pi}{2}x^2(\ln^22-(2x+\ln(1-e^{-4x}))\ln2+(x+\ln(1-e^{-2x}))(x+\ln(1+e^{-2x})))~dx$$

$$=\int_0^\frac{\pi}{2}x^2(\ln^22-(2x+\ln(1-e^{-4x}))\ln2+x^2+(\ln(1-e^{-2x})+\ln(1+e^{-2x}))x+\ln(1-e^{-2x})\ln(1+e^{-2x}))~dx$$

$$=\int_0^\frac{\pi}{2}(x^4-2x^3\ln2+x^2\ln^22+x^3\ln(1-e^{-4x})-x^2\ln(1-e^{-4x})\ln2+x^2\ln(1-e^{-2x})\ln(1+e^{-2x}))~dx$$

$$=\int_0^\frac{\pi}{2}\left(x^4-2x^3\ln2+x^2\ln^22+\sum\limits_{n=1}^\infty\dfrac{x^3e^{-4nx}}{n}-\sum\limits_{n=1}^\infty\dfrac{x^2e^{-4nx}\ln2}{n}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^\infty\dfrac{(-1)^nx^2e^{-2(n+k)x}}{nk}\right)~dx$$

$$=\left[\dfrac{x^5}{5}-\dfrac{x^4\ln2}{2}+\dfrac{x^3\ln^22}{3}-\sum\limits_{n=1}^\infty\dfrac{(32n^3x^3+24n^2x^2+12nx+3)e^{-4nx}}{128n^5}+\sum\limits_{n=1}^\infty\dfrac{(8n^2x^2+4nx+1)e^{-4nx}\ln2}{32n^4}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^\infty\dfrac{(-1)^n(2(n+k)^2x^2+2(n+k)x+1)e^{-2(n+k)x}}{4nk(n+k)^3}\right]_0^\frac{\pi}{2}$$

$$=\dfrac{\pi^5}{160}-\dfrac{\pi^4\ln2}{32}+\dfrac{\pi^3\ln^22}{24}-\sum\limits_{n=1}^\infty\dfrac{(4\pi^3n^3+6\pi^2n^2+6\pi n+3)e^{-2\pi n}-3}{128n^5}+\sum\limits_{n=1}^\infty\dfrac{((2\pi^2n^2+2\pi n+1)e^{-2\pi n}-1)\ln2}{32n^4}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^\infty\dfrac{(-1)^n((\pi^2(n+k)^2+2\pi(n+k)+2)e^{-\pi(n+k)}-2)}{8nk(n+k)^3}$$

Answer 2

The only way I found is based on the infinite Taylor expansion of the integrand. For the first terms, we have $$x^2\ln(\sinh x)\ln(\cosh x)=\frac{1}{2} x^4 \log (x)+x^6 \left(\frac{1}{12}-\frac{\log (x)}{12}\right)+x^8 \left(\frac{\log (x)}{45}-\frac{1}{60}\right)+x^{10} \left(\frac{197}{45360}-\frac{17 \log (x)}{2520}\right)+x^{12} \left(\frac{31 \log (x)}{14175}-\frac{439}{340200}\right)+x^{14} \left(\frac{53}{128304}-\frac{691 \log (x)}{935550}\right)+O\left(x^{16}\right)$$ This shows a polynomial in $x$ as well as $x^n \log(x)$ terms. The integration of polynomials does not make any problem and, for the other terms, we can easily prove (integration by parts) that $$I=\int x^n \ln(x)dx=\frac{x^{n+1} ((n+1) \log (x)-1)}{(n+1)^2}$$ As a function of terms used for the Taylor series, the value of the integral is oscillating around $0.392634$.

According to RIES, this last number seems to be quite close to $$\frac{1}{\sqrt{e^2-\frac{1}{4}}-\frac{1}{8}}$$