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How to integrate

$$\int_{0}^{1} \int_{0}^{1} \ln\left(\frac{1}{\sinh^2(x) + \cosh^2(y)}\right) \,dx\,dy$$

My attempt

$$\int_{0}^{1} \int_{0}^{1} \ln\left(\frac{1}{\sinh^2(x) + \cosh^2(y)}\right) \,dx\,dy = - \int_{0}^{1} \int_{0}^{1} \ln(\sinh^2(x) + \cosh^2(y)) \,dx\,dy$$

$$= - \int_{0}^{1} \int_{0}^{1} \ln\left(\frac{e^{2x} + e^{-2x} + e^{2y} + e^{-2y}}{4}\right) \,dx\,dy$$

$$ =- \int_{0}^{1} \int_{0}^{1} \ln\left(e^{2x} + \frac{1}{e^{2y}} + e^{2y} + \frac{1}{e^{2x}}\right) \,dx\,dy + 2\ln(2)$$

$$= - \int_{0}^{1} \int_{0}^{1} \ln\left(\frac{e^{2x+2y} + 1}{e^{2y}} + \frac{e^{2x+2y} + 1}{e^{2x}}\right) \,dx\,dy + 2\ln(2)$$

$$=- \int_{0}^{1} \int_{0}^{1} \ln(e^{2(x+y)} + 1) \,dx\,dy +\int_{0}^{1} \int_{0}^{1} \ln(e^{2(x+y)}) \,dx\,dy - \int_{0}^{1} \int_{0}^{1} \ln(e^{2x} + e^{2y}) \,dx\,dy + 2\ln(2)$$

$$= - \int_{0}^{1} \int_{0}^{1} \ln(e^{2x} + e^{2y}) \,dx\,dy - \int_{0}^{1} \int_{0}^{1} \ln\left(1 + e^{2(x+y)}\right) \,dx\,dy + 2\ln(2) + 2$$

I need help with the last two integrals. Any help is appreciated.

Edit

Here is my second attempt

$$I = 2 \ln(2) \int_{0}^{1} \int_{0}^{1} \,dx\,dy + \int_{0}^{1} \int_{0}^{1} \ln\left(\frac{1}{e^{-2x} + e^{2x} + e^{-2y} + e^{2y}}\right) \,dx\,dy$$

$x \rightarrow e^{-2x}, \quad y \rightarrow e^{-2y}$

$$=2\ln(2) - \frac{1}{4} \int_{e^{-2}}^{1} \int_{e^{-2}}^{1} \frac{\ln(y(x^2 + 1) + x(y^2 + 1)) - \ln(xy)}{xy} \,dx\,dy$$
$$=2\ln(2) - \frac{1}{4} \int_{e^{-2}}^{1} \int_{e^{-2}}^{1} \frac{\ln(y(x^2 + 1) + x(y^2 + 1))}{xy} \,dx\,dy + \frac{1}{2} \int_{e^{-2}}^{1} \frac{\ln(x)}{x} \,dx \cdot \int_{e^{-2}}^{1} \frac{1}{y} \,dy$$
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  • 1
    $\begingroup$ I haven't checked your calculations so far, but looking at what you have at the end, I'd say your best bet is polylogs. Perhaps you made a mistake earlier and this isn't needed $\endgroup$
    – A P
    Feb 12 at 12:03
  • $\begingroup$ Mathematica evaluates it as$$\frac{\text{Li}_3\left(-e^4\right)}{4}+\frac{3 \zeta (3)}{16}+\frac{4}{3}+\frac{\pi ^2}{12}+2 \log (2)$$ $\endgroup$
    – Robert Lee
    Feb 13 at 1:39
  • $\begingroup$ @RobertLee any ideas how to evaluate the integral? $\endgroup$
    – user1285841
    Feb 13 at 6:14

1 Answer 1

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$$ \begin{split} J &= \int \frac{\ln(y(x^2+1)+x(y^2+1))}{xy}\, dx\\ & = \frac{1}{y} \int \frac{\ln(y(x^2+1)+x(y^2+1))}{x}\, dx \\ & = \frac{1}{y} \int \frac{\ln(x+y)+\ln(x+\frac{1}{y})+\ln(y)}{x}\, dx \\ & = \frac{1}{y} (K + L + \ln(x)\ln(y)) \end{split} $$

$$ \begin{split} K & = \int \frac{\ln(x+y)}{x}\, dx \\ & = \int \frac{\ln(\frac{x}{y}+1)}{x}\, dx + \ln(y)\int \frac{1}{x}\, dx \\ & \qquad \textrm{Substitution} \quad \boxed{\begin{aligned} u&=-\frac{x}{y},\\ du&=-\frac{1}{y}dx \end{aligned}} \\ & = -\int - \frac{\ln(1-u)}{u}\, du + \ln(y)\int\frac{1}{x}\, dx \\ & = -\textrm{Li}_2(-\frac{x}{y}) +\ln(x)\ln(y) \end{split} $$

$$ \begin{split} L & = \int \frac{\ln(x+\frac{1}{y})}{x}\, dx \\ & = \int \frac{\ln(xy+1)}{x}\, dx - \ln(y)\int \frac{1}{x}\, dx \\ & \qquad \textrm{Substitution} \quad \boxed{\begin{aligned} u&=-xy,\\ du&=-ydx \end{aligned}} \\ & = -\textrm{Li}_2(-xy)-\ln(x)\ln(y) \end{split} $$

$$ J = \frac{-\textrm{Li}_2(-xy)-\textrm{Li}_2\left(-\frac{x}{y}\right)+\ln(y)\ln|x|}{y} $$

$$ [J]^1_{x=e^{-2}} = \frac{2\ln(y)+\textrm{Li}_2\left(-\frac{1}{e^2y}\right)+\textrm{Li}_2\left(-\frac{y}{e^2}\right)+\frac{1}{6}(3\ln^2(y)+\pi^2)}{y} $$

$$ \begin{split} M & = \int \frac{2\ln(y)+\textrm{Li}_2\left(-\frac{1}{e^2y}\right)+\textrm{Li}_2\left(-\frac{y}{e^2}\right)+\frac{1}{6}(3\ln^2(y)+\pi^2)}{y}\, dy\\ & = \frac{1}{6}\int\frac{6\textrm{Li}_2(-e^{-2}y)+3\ln^2(y)+12\ln(y)+6\textrm{Li}_2\left(-\frac{e^{-2}}{y}\right)+\pi^2}{y}\, dy\\ & \qquad \textrm{Substitution} \quad \boxed{\begin{aligned} u&=e^{-2}y,\\ du&=e^{-2}dy,\\ \frac{1}{y}&=\frac{e^{-2}}{u} \end{aligned}} \\ &= \frac{1}{6}\int\frac{3\left((\ln(u)+2)^2+4\ln(u)+2\textrm{Li}_2(-u)+2\textrm{Li}_2\left(-\frac{e^{-4}}{u}\right)\right)+\pi^2+24}{u}\, du\\ &= \frac{1}{6}\left(3\int\frac{(\ln(u)+2)^2+4\ln(u)+2\textrm{Li}_2(-u)+2\textrm{Li}_2\left(-\frac{e^{-4}}{u}\right)}{u}\, du +(\pi^2+24)\int\frac{1}{u}\, du\right)\\ &= \frac{1}{6}\left(3N+(\pi^2+24)\ln(u)\right) \end{split} $$

$$ \begin{split} N &= \int\frac{(\ln(u)+2)^2+4\ln(u)+2\textrm{Li}_2(-u)+2\textrm{Li}_2\left(-\frac{e^{-4}}{u}\right)}{u}\, du\\ &=\int\frac{(\ln(u)+2)^2}{u}\, du +4\int\frac{\ln(u)}{u}\, du+2\int\frac{\textrm{Li}_2(-u)}{u}\, du+2\int\frac{\textrm{Li}_2\left(-\frac{e^{-4}}{u}\right)}{u}\, du\\ & \qquad \textrm{Substitution} \quad \boxed{\begin{aligned} v&=\ln(u)+2,\\ dv&=\frac{1}{u}du \end{aligned}} \\ &=\int v^2\, dv+4\int\frac{\ln(u)}{u}\, du+2\int\frac{\textrm{Li}_2(-u)}{u}\, du+2\int\frac{\textrm{Li}_2\left(-\frac{e^{-4}}{u}\right)}{u}\, du\\ &=\frac{(\ln(u)+2)^2}{3}+4\int\frac{\ln(u)}{u}\, du+2\int\frac{\textrm{Li}_2(-u)}{u}\, du+2\int\frac{\textrm{Li}_2\left(-\frac{e^{-4}}{u}\right)}{u}\, du\\ &=\frac{(\ln(u)+2)^2}{3}+2\ln^2(u)+2\int\frac{\textrm{Li}_2(-u)}{u}\, du+2\int\frac{\textrm{Li}_2\left(-\frac{e^{-4}}{u}\right)}{u}\, du\\ &=\frac{(\ln(u)+2)^2}{3}+2\ln^2(u)+2\textrm{Li}_3(-u)+2\int\frac{\textrm{Li}_2\left(-\frac{e^{-4}}{u}\right)}{u}\, du\\ &=\frac{(\ln(u)+2)^2}{3}+2\ln^2(u)+2\textrm{Li}_3(-u)-2\textrm{Li}_3\left(-\frac{e^{-4}}{u}\right) \end{split} $$

$$ \begin{split} M &= \frac{1}{6}\left(3\left(\frac{(\ln(u)+2)^2}{3}+2\ln^2(u)+2\textrm{Li}_3(-u)-2\textrm{Li}_3\left(-\frac{e^{-4}}{u}\right)\right)+(\pi^2+24)\ln(u)\right) \\ &= \frac{1}{6}\left((\ln(u)+2)^3+6\ln^2(u)+(\pi^2+24)\ln(u)+6\textrm{Li}_3(-u)-6\textrm{Li}_3\left(-\frac{e^{-4}}{u}\right)\right)\\ &= \textrm{Li}_3(-e^{-2}y)+\frac{\ln^3(y)}{6}+\frac{(\pi^2+24)(\ln(y)-2)}{6}+(\ln(y)-2)^2-\textrm{Li}_3\left(-\frac{e^{-2}}{y}\right)\\ &= \textrm{Li}_3(e^{-2}y)+\frac{\ln(y)(\ln(y)(\ln(y)+6)+\pi^2)}{6}-\textrm{Li}_3\left(-\frac{e^{-2}}{y}\right)\\ \end{split} $$

$$ [M]^1_{x=e^{-2}} = \frac{1}{3}\left(-3\textrm{Li}_3\left(-\frac{1}{e^4}\right)-8+\pi^2\right)-\frac{3\zeta(3)}{4} $$

Bringing all this back to your work,

$$ \begin{split} I&=2\ln(2)-\frac{1}{4}\left(\frac{1}{3}\left(-3\textrm{Li}_3\left(-\frac{1}{e^4}\right)-8+\pi^2\right)-\frac{3\zeta(3)}{4}\right)-1(2)\\ &=\frac{\text{Li}_3\left(-e^4\right)}{4}+\frac{3 \zeta (3)}{16}+\frac{4}{3}+\frac{\pi ^2}{12}+2 \ln (2)\\ \end{split} $$

which is the answer by Robert.

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  • $\begingroup$ More than elegant!! $\endgroup$
    – user1285841
    Feb 14 at 17:38

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