Energy of a signal with decreasing amplitude
I read that a infinite extension continuous time signal with decreasing amplitude will always be an energy signal.

That is wrong. It suffices to show a single counter example to show it's wrong: Let
$$s(t) = \frac1{\sqrt{\left\lvert t\right\rvert}},$$
which clearly is infinitely extending and always getting smaller the further you go from $t=0$. Then its energy is
\begin{align}
E_s &amp;= \int\limits_{-\infty}^{\infty} \left(\frac1{\sqrt{\left\lvert t\right\rvert}}\right)^2\,\mathrm dt\\
&amp;=2\int\limits_{0}^{\infty} \frac1{\left(\sqrt{t}\right)^2}\,\mathrm dt
\end{align}
and as &quot;$\sqrt\infty$&quot; doesn't ever exist, doesn't converge, and hence, this is clearly not an energy signal.
Your example works, too, and you get an &quot;$\log\infty$&quot; as a result. So, don't bother too much basing questions on a wrong claim.

Doesn't the saying that if a signal has finite area then it is an energy signal?

Well, &quot;area&quot; doesn't work great for signals that aren't real, or that aren't piece-wise continuous, or…
But yeah, if you can assign a finite area to it, then it's an energy signal, as that requires bounded absolute values. But thinking in &quot;areas between curve and axis&quot; really isn't the best level of abstraction for signal processing.