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After reading the MSE post on "Integral Milking", my first instinct was to try it out on one of my favorite integrals:

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

which is equivalent to

$$\int^{\ln{\phi}}_{0}\ln\left(2\sinh{x}\right)dx=-\frac{\pi^2}{20}$$

I personally prefer the first one.

After trying to milk the integral with various techniques for a while, I came up with

$$\int^{1}_{0}\ln\left(\phi^{x}-\phi^{-x}\right)\ln\left(\phi\right)dx=-\frac{\pi^2}{20}$$

I like this one a lot except that it is multiplied by the constant $\ln\left(\phi\right)$

My solution, which made use of the change of base formula was not as elegant as I had envisioned:

$$\int_{0}^{1}\log_{\sqrt[\ln\left(\phi\right)]{e}}\left(\phi^{x}-\phi^{-x}\right)dx=-\frac{\pi^2}{20}$$

My "milking" of this integral used very simple techniques. I am curious what can be done with more advanced ones so my question is this:

What are other ways to "milk" the integral $\int^{1}_{0}\ln\left(\phi^{x}-\phi^{-x}\right)\ln\left(\phi\right)dx=-\frac{\pi^2}{20}$?

Dylan Levine
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    Should this have the "soft-question" tag? – Dylan Levine Dec 16 '23 at 19:33
  • I'm not familiar with the term "integral milking". Does that mean making slight changes to an integral such that you get a nice answer out of it? For example, IDK if there is a closed form for $\int_{0}^{\frac{1}{4}}\frac{\arctan x}{20x^{2}+11x+63}dx$, but if I change the 63 into a 64, then you can prove $\int_{0}^{\frac{1}{4}}\frac{\arctan x}{20x^{2}+11x+64}dx = \frac{1}{\sqrt{4999}}\cot^{-1}\left(\frac{523}{\sqrt{4999}}\right)\tan^{-1}\left(\frac{1}{4}\right)$. Is that what milking is? – Accelerator Dec 16 '23 at 20:34
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    @Accelerator That is correct. This post is where I got the term: https://math.stackexchange.com/questions/2821112/integral-milking – Dylan Levine Dec 16 '23 at 20:45
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    Ah, I see now. You could probably make up some crazy substitution or integrate by parts and see if you get an integral that looks nice. Or if one is skilled enough, make up some contour in the complex plane that somehow involves the original integral. – Accelerator Dec 17 '23 at 06:15
  • I wonder if something could be done with Binet's formula for the fibonacci sequence – Dylan Levine Dec 17 '23 at 16:06

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Making the problem more general $$I=\int \log (a) \log \left(a^x-a^{-x}\right)\,dx$$ $$I=-\text{Li}_2\left(-a^x\right)+\text{Li}_2\left(1-a^x\right)+\frac {1}{2} \log \left(a^x\right) \log \left(a^{-x}+a^x-2\right)$$ $$\color{blue}{J=\int_0^1 \log (a) \log \left(a^x-a^{-x}\right)\,dx=}$$ $$\color{blue}{\text{Li}_2(1-a)-\text{Li}_2(-a)+\frac{1}{2} \log (a) \log \left(\frac{(a-1)^2}{a}\right)-\frac{\pi ^2}{12}}$$ and for $a=\phi$ the nice result.

Claude Leibovici
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