Asymptotics to $f(n) = \int_0^1\bigl ( \frac{ \operatorname{li}(x)}{x} \bigr)^{2n + 1} \,(x-1) \, dx $

Question

Consider

$$f(n)=\int_0^1\Bigl(\frac{\operatorname{li}(x)}{x}\Bigr)^{2n + 1}\,(x-1)\,dx $$

Where $n$ is a positive integer.

(I know that $f(1) = \zeta(3) $ but I already made a Question about proving that)

If I am not mistaken $f(6) = 123 482 $ or about that value. (I assume not exactly that integer!)

So I started to wonder. How fast does $f(n)$ grow?

What are very good asymptotics for $f(n) $?

I had the idea to investigate $ f ‘ (t) $ (with respect to $t$) and could “ express “ that simply by differentiation under the integral sign. (Feyman loved that btw) But then we are still left with integral(s) , I'm not sure how that helps. Similarly I could make a Taylor series of $f(t)$ but that would still not help me I guess? Also the behaviour at half-integers is radically different!

How to Find very good asymptotics then !?

Answer 1

Let $$\phi_n (x) = \left(\frac{\operatorname{li}(x)}{x}\right)^{2n+1} (x-1)$$ for $n \in \mathbb{N}_0$ and $x \in (0,1)$. Note that for large $n$ the function $\phi_n$ is extremely small everywhere except for one sharp peak close to $x = 1$ (at about $x \approx 1 - \mathrm{e}^{-(2n+1)}$) .

Therefore we can approximate it by its asymptotic form near $x = 1$ if $n$ is large. Using $x \sim 1$ and $\operatorname{li}(x) \sim \ln(1-x) + \gamma$ with the Euler-Mascheroni constant $\gamma$ as $x \nearrow 1$ , we obtain $$ \phi_n (x) \sim [\ln(1-x) + \gamma]^{2n+1} (x-1) \, , \, x \nearrow 1 \, . $$

Plugging this result into the integral and using the substitution $1 - x = \mathrm{e}^{-t}$ we get \begin{align} f(n) & \sim \int \limits_0^1 [\ln(1-x) + \gamma]^{2n+1} (x-1) \, \mathrm{d} x \\ &= \int \limits_0^\infty (t-\gamma)^{2n+1} \mathrm{e}^{-2 t} \, \mathrm{d} t \\ &= \mathrm{e}^{-2\gamma} \int \limits_{-\gamma}^\infty s^{2n+1} \mathrm{e}^{-2 s} \, \mathrm{d} s \\ &\sim \mathrm{e}^{-2\gamma} \frac{\Gamma(2n+2)}{2^{2n+2}} \\ &= \mathrm{e}^{-2\gamma} \frac{(2n+1)!}{4^{n+1}} \end{align} as $n \to \infty$ .

This derivation is not quite rigorous yet, as I did not prove that the corrections to these approximations are asymptotically negligible compared to the above leading term. However, numerical calculations seem to agree with this result and confirm the value of the prefactor.