Help to solve two integrals $\int_1^\infty\frac{1}{t\log^2(1+t)}$ $\int_1^\infty\frac{1}{t\log(1+t)}\,dt$

Question

If $\int_{1}^{\infty} |f(t)|^{2}\,dt < \infty$, is $\int_{1}^{\infty} \frac{|f(t)|}{\sqrt{t}}\,dt < \infty$?

https://math.stackexchange.com/a/3636380/776594

$\int_1^\infty|f(t)|^2\,dt=\int_1^\infty\frac{1}{t\log^2(1+t)}<\infty$

$\int_1^\infty \frac{|f(t)|}{\sqrt{t}}\,dt=\int_1^\infty\frac{1}{t\log(1+t)}\,dt=\infty.$

I understand the counter example Clayton offered, but I have problems to solve the integrals. I tried to solve by comparison with some other integrals. But I didn't compute properly. Anyone can help or give some ideas to solve two integrals?

$\int_1^\infty\frac{1}{t\log^2(1+t)}$

$\int_1^\infty\frac{1}{t\log(1+t)}\,dt$

Thank you and I'm sincerely thankful for all who wrote comments to my previous posting

Answer 1

For both integrals, use the substitution $s = \log(1+t)$:

$$I_1 = \int_{\log 2}^\infty \frac{e^s}{e^s-1} \frac{1}{s^2}\:ds$$

$$I_2 = \int_{\log 2}^\infty \frac{e^s}{e^s-1} \frac{1}{s}\:ds$$

We can see for both integrals the factor $$1 \leq \frac{e^s}{e^s-1} \leq 2$$

leading us to

$$ I_1 \leq 2\int_{\log 2}^\infty \frac{1}{s^2}\:ds = \frac{2}{\log 2}$$

and

$$I_2 \geq \int_{\log 2}^\infty \frac{1}{s}\:ds = \infty$$

Answer 2

The substitution $t=e^{s}-1$ makes this very easy.

Note that $\frac {e^{s}} {e^{s}-1}$ is bounded above and below by positive numbers on $(\ln 2 ,\infty)$.