Integral of $\ln(\cosh x)$?

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$$\int \ln(\cosh x) dx$$

Tried for some time, couldn't get anywhere.

This IS an elementary function, isn't it?

Thanks in advance.

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2 Answers

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The integral $\int \log(\cosh(x))\,dx$ cannot be expressed in terms of elementary functions. To see this, we enforce the substitution $x=\log(u)$ to obtain

$$\begin{align} \int \log(\cosh(x))\,dx&=\int \frac{\log(1+u^2)-\log(u)}{u}\,du-\log(2)x \\\\ &=\int \frac{\log(1+u^2)}{u}\,du -\frac12 x^2-\log(2)x \tag 1 \end{align}$$

Next, we let $u=\sqrt{v}$ in the integral on the right-hand side of $(1)$ to reveal

$$\begin{align} \int \log(\cosh(x))\,dx&=\int \frac{\log(1+v)}{2v}\,dv -\frac12 x^2-\log(2)x \\\\ &=\bbox[5px,border:2px solid #C0A000]{-\frac12\text{Li}_2(-e^{2x})-\frac12x^2-\log(2)x+C}\tag 2 \end{align}$$

which expresses the primitive in terms of the dilogarithm function, a special function given by $\text{Li}_2(x)=-\int_0^x \frac{\log(1-t)}{t}\,dt$.

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$$\ln(\cosh(x))=\ln\left(\frac{e^x+e^{-x}}2\right)=x-\ln(2)+\ln(1+e^{-2x})= x-\ln(2)+\sum_{k=1}^\infty(-1)^{k+1}\frac{e^{-2kx}}k.$$

You can integrate term-wise and get a rapidly converging series (for positive $x$), which is asymptotic to the parabola $\dfrac{x^2}2-\ln(2)\,x+C$.

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