# A cool integral: $\int^{\ln{\phi}}_{0}\ln\left(e^{x}-e^{-x}\right)dx=-\frac{\pi^2}{20}$

I was looking at the equation $$\ln{e^{x}-e^{-x}}$$ and found that the zero was at $$x=\ln{\phi}$$ where $$\phi$$ is the golden ratio. I thought that was pretty cool so I attempted to find the integral. I am currently at the beginning of Calculus BC so it is definitely over my skill level. Here was my attempt:

$$\int^{\ln{\phi}}_{0}\ln\left(e^x-e^{-x}\right)dx$$

$$\int^{\ln{\phi}}_{0}\ln\left(\frac{e^{2x}-1}{e^x}\right)dx$$

$$\int^{\ln{\phi}}_{0}\left(\ln\left(e^{2x}-1\right)-\ln\left(e^x\right)\right)dx$$

$$\int^{\ln{\phi}}_{0}\left(\ln\left(e^{2x}-1\right)-x\right)dx$$

$$\int^{\ln{\phi}}_{0}\ln\left(e^{2x}-1\right)dx-\frac{\phi^2}{2}$$

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

I can't figure out how to solve these two integrals I am left with (integration by parts doesn't seem to work). Wolfram Alpha says to use polylogarithms but I was hoping there would be a method that there would be an approach to solving it that I could understand.

• It's more cool writing down like this: $\displaystyle \int_0^{\ln\varphi}\ln(2\sinh x)dx$
– FDP
Oct 16, 2023 at 23:49

\begin{align}J&=\int^{\ln{\phi}}_{0}\ln\left(\text{e}^{x}-\text{e}^{-x}\right)dx\\ &\overset{u=\text{e}^x}=\int_1^\varphi \frac{\ln\left(u-\frac{1}{u}\right)}{u}du\\ &=\underbrace{\int_1^\varphi \frac{\ln(u^2-1)}{u}du}_{z=u^2-1}-\int_1^\varphi \frac{\ln u }{u}du\\ &=\boxed{\frac{1}{2}\underbrace{\int_0^{\varphi}\frac{\ln z}{z+1}dz}_{=A}-\frac{1}{2}\ln^2\varphi}\end{align} On the other hand, \begin{align}J&=\int_1^\varphi \frac{\ln(u^2-1)}{u}du-\frac{1}{2} \ln^2\varphi\\ &=\underbrace{\int_1^\varphi \frac{\ln(u-1)}{u}du}_{z=u-1}+\underbrace{\int_1^\varphi \frac{\ln(u+1)}{u}du}_{\text{IBP}}-\frac{1}{2} \ln^2\varphi\\ &=\int_0^{\frac{1}{\varphi}}\frac{\ln z}{1+z}dz-\int_1^{\varphi}\frac{\ln z}{1+z}dz+\frac{3}{2}\ln^2\varphi\\ &=\int_0^1 \frac{\ln z}{1+z}dz-\underbrace{\int_{\frac{1}{\varphi}}^1\frac{\ln z}{1+z}dz}_{w=\frac{1}{z}}-\int_1^{\varphi}\frac{\ln z}{1+z}dz+\frac{3}{2}\ln^2\varphi\\ &=\int_0^1 \frac{\ln z}{1+z}dz+\int_1^{\varphi}\frac{\ln z}{z(1+z)}dz-\int_1^{\varphi}\frac{\ln z}{1+z}dz+\frac{3}{2}\ln^2\varphi\\ &=\int_0^1 \frac{\ln z}{1+z}dz-2\int_1^{\varphi}\frac{\ln z}{1+z}dz+2\ln^2\varphi\\ &=3\underbrace{\int_0^1 \frac{\ln z}{1+z}dz}_{=-\frac{\pi^2}{12}}-2\int_0^{\varphi}\frac{\ln z}{1+z}dz+2\ln^2\varphi\\ &=\boxed{2\ln^2\varphi-\frac{\pi^2}{4}-2A}\\ \end{align} Therefore, \begin{align}4J+J&=\left(2A-2\ln^2\varphi p\right)+\left(2\ln^2\varphi-\frac{\pi^2}{4}-2A\right)=-\frac{\pi^2}{4}\\ J&=\boxed{-\frac{\pi^2}{20}} \end{align}