Show that $\int_2^{+ \infty} \frac{\log(t)^2}{t(t-1)}dt \leq 4$

Question

I couldn't prove this inequality $$ \int_2^{+ \infty} \frac{\log(t)^2}{t(t-1)}dt \leq 4 $$ I've tried integration by parts but it doesn't work.

Answer 1

We have \begin{align*} \int_2^\infty \frac{\ln^2 t}{t(t - 1)}\,\mathrm{d} t & \overset{t = 1/x} = \int_0^{1/2} \frac{\ln^2 x}{1 - x} \,\mathrm{d} x \\ &\le \int_0^{1/2} \frac{\ln^2 x}{1 - 1/2} \,\mathrm{d} x\\ &= 2\int_0^{1/2} \ln^2 x \,\mathrm{d} x\\ &= \ln^2 2 + 2 + 2\ln 2\\ &< 4. \end{align*}

Answer 2

Here is a quick-and-dirty approach:

1. Show that $\sup_{t\geq 2} \frac{\log^2 t}{\sqrt{t}} = \frac{16}{e^2}$ (achieved at $t=e^4$). That can easily be done by differentiating the function $t\mapsto \frac{\log^2 t}{\sqrt{t}}$.

2. Compute $\int_2^\infty \frac{dt}{\sqrt{t}(t-1)} = 2\operatorname{arcsinh} 1$

3. Observe that $\frac{32}{e^2}\operatorname{arcsinh} 1 < 4$.