About a sequence of (improper) integrals.

Question

One considers the following sequence $$ I_n=\int_0^1 \log^2(t).(1-t)^n dt\ , $$ using elementary methods (integration by parts, twice), one can show that $$ I_n=2\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{(k+1)^3} $$ Question : Is there a simple way to prove or disprove that this (decreasing, non-negative) sequence tends to zero ?

Motivation : In the theory of star algebras, one considers the sub-star algebra of $\mathcal{C}(]0,1],\mathbb{C})$ generated by $\{\log(t),t\}$ and the positive linear form $\varphi : P\to \int_0^1 P(t) dt$ (where $P(t)=Q(\log(t),t),\ Q\in \mathbb{C}[X,Y]$) which generates a hilbertian form $$ \langle u(t)\mid v(t)\rangle:=\varphi(v^*u) \qquad (*) $$ the target being to prove that the operator $f\to \log(t)f(t)$ is discontinuous for the norm induced by (*).

Late edit (Thanks to all contributors) We see here that, with $P_n(t)=\frac{n}{\log^2(n)}(1-t)^n$, we have $\lim_{n\to \infty}P_n=0$ whereas $\lim_{n\to \infty}\log(t)P_n(t)=1$ for this norm.

Answer 1

You can obtain the antiderivative using hypergeometric functions and the definite integral is "just" $$I_n=\int_0^1 \log^2(t)\,(1-t)^n\, dt=\frac{6 \left(H_{n+1}\right){}^2-6 \psi ^{(1)}(n+2)+\pi ^2}{6 (n+1)}$$ which is asymptotic to $$\frac{6 \log ^2(n)+12 \gamma \log (n)+(\pi ^2+6 \gamma ^2)}{6 n}+O\left(\frac{1}{n^2}\right)$$

Answer 2

By the Cauchy–Schwarz inequality and Euler's integral $n!=\int_0^1 (-\log (t))^n\mathrm{d}t$, $$ \int_0^1 {\log ^2 (t)(1 - t)^n {\rm d}t} \le \sqrt {\int_0^1 {\log ^4 (t){\rm d}t} \int_0^1 {(1 - t)^{2n} {\rm d}t} } = \sqrt {4!\frac{1}{{2n + 1}}} = \frac{{2\sqrt 6 }}{{\sqrt {2n + 1} }} \to 0. $$ For a more precise asymptotics, we can use the dominated convergence theorem: \begin{align*} \frac{n}{{\log ^2 (n)}}\int_0^1 {\log ^2 (t)(1 - t)^n {\rm d}t} = \int_0^n {\left[ {1 - \frac{{\log (s)}}{{\log (n)}} } \right]^2\left( {1 - \frac{s}{n}} \right)^n {\rm d}s} \to \int_0^{ + \infty } {{\rm e}^{ - s} {\rm d}s} = 1, \end{align*} i.e., $$ \int_0^1 {\log ^2 (t)(1 - t)^n {\rm d}t} \sim \frac{{\log ^2 (n)}}{n} $$ as $n\to+\infty$.