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I want to evaluate the integral $$ I = \int_0^\infty \frac{\ln x}{x^3 - 1} \, dx = \frac{4\pi^2}{27} $$ using Feynman's trick (differentiation under the integral sign).

Substitution

Let’s make the substitution $x = e^t$, so $dx = e^t dt$ and $t = \ln x $. Then: $$ I = \int_{-\infty}^\infty \frac{t \cdot e^t}{e^{3t} - 1} \, dt $$

My Approach

Define a parameter-dependent integral: $$ I(\alpha) := \int_{-\infty}^\infty \frac{t \cdot e^{\alpha t}}{e^{3t} - 1} \, dt $$

Then differentiating under the integral sign gives: $$ \frac{\partial I}{\partial \alpha} = \int_{-\infty}^\infty \frac{t^2 \cdot e^{\alpha t}}{e^{3t} - 1} \, dt $$

Idea

For $t > 0$, the function $$ \frac{1}{e^{3t} - 1} $$ can be written as a geometric series: $$ \frac{1}{e^{3t} - 1} = \sum_{n=1}^\infty e^{-3nt} $$

My question is: Is there a clean way to evaluate $I(\alpha)$ using Feynman's trick, possibly by interchanging summation and integration and then integrating back to recover $ I(1) $?
I will rigorously evaluate the convergence properties afterwards. I would like to see maybe a more intelligent choice of parameter or something that solves the integral. I know about residues, I want a solution using Feynman's trick. Thanks for any helpful answers.

Samuel
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1 Answers1

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Instead of differentiating under the integral, I find a function that when differentiated under the integral gives our desired problem. Assuming it is defined, let $$I(a)=\int_0^\infty \frac{x^a}{x^3-1}dx.$$ Our desired integral is $I'(0)$. Using $u=x^3$, we get $$I(a)=\frac13\int_0^\infty \frac{u^{\frac{a-2}3}}{1-u} du.$$ There are a few ways to evaluate the general integral $\int_0^\infty \frac{u^t}{1-u}du$ (for example this post, which uses contour integration or Ramanujan's master theorem). Or, more simply, we can use the Mellin transform for $\frac{1}{1-x}$, wherein

$$\mathcal{M}\left\{\frac{1}{x-1}\right\}(s)=\int_0^\infty\frac{x^{s-1}}{x-1}dx = -\pi \cot \pi s.$$ Thus $$I(a)=\frac13\mathcal{M}\left\{\frac{1}{x-1}\right\}\left(\frac{a+1}3\right)=-\frac{\pi}{3} \cot \pi\frac{a+1}3$$ and $$I'(a)= \frac{\pi^2}9\csc^2\pi\frac{a+1}3,$$ so $$I'(0)=\int_0^\infty\frac{\ln x}{x^3-1}dx=\frac{\pi^2}9\csc^2\frac{\pi}3 = \boxed{\frac{4\pi^2}{27}}.$$

Elias Mrug
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