72. $f(x)$ given by
\[
f(x)=\begin{cases}
\frac{1}{x},&x\geq1\\
\frac{1}{\sqrt{x}},&0<x<1
\end{cases}
\]
has the required property.
73. Suppose $f(x):=(\ln x)^{-2}\in L^1$, i.e$.$ the integral $\int_1^\infty f(x)\,dx<\infty$. Consider a sequence of function
\[
f_\varepsilon(x)=\frac{1}{(\ln x+\varepsilon)^2},\qquad0<\varepsilon<1.
\]
We have $f_\varepsilon(x)\to f(x)$ for all $x\in\mathbb{R}$, as $\varepsilon\to0$. By Lebesgue dominated convergence theorem, we get
\[
\lim_{\varepsilon\to0}\int f_\varepsilon(x)\,dx=\int f(x)\,dx<\infty.
\]
But it's easy to check that $\displaystyle\int f_\varepsilon(x)\,dx\geq\int_1^\infty\frac{1}{(x-1+\varepsilon)^2}\,dx=\frac{1}{\varepsilon}$. This gives a contradiction, hence the integral is infinite.
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