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I am having difficulty with the case when a is real.

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But after this, I am not going anywhere.

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I think the correct expression for the power is $$ P = \lim_{T\rightarrow\infty}\frac{e^{-aT}-e^{aT}}{2aT}. $$

When $T \rightarrow \infty$, $e^{aT}$ tends to infinity "faster" than $1/T$ and the limit does not exist.

However, an interesting thing happens when $a=0$. The signal $x(t) = e^{-at}$ becomes $e^0=1$, which is a power signal (with power equal to 1 watt).

I'm sure that (using L'Hopital or some other trick) it can be shown that $P=1$ when $a=0$, but I have not tried it. In any case, I would say that the premise of the question is wrong, and in fact $x(t)$ is a power signal for at least one (real) value of $a$.

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    $\begingroup$ Thank you for triggering another answer. You came first! $\endgroup$ Apr 1, 2021 at 17:27
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[Note: this answer was triggered by @MBaz's answer and mention on the de L'Hospital rule, which can be avoided.]

Let us suppose that $T>0$. If you consider a complex $a$, the conjugate of $x(t)=\exp^{-at}$ is:

$$ \overline{\exp^{-at}}=\exp^{\overline{-at}}=\exp^{-\overline{a}t}\,.$$

Hence the integral on the interval is:

$$P_{x,T} = \frac{1}{T}\int_{-T/2}^{T/2} e^{-2a_rt} \mathrm{d}t\,,$$ where $a_r$ denotes the real part of $a$ (which is zero when $a$ is purely imaginary). We can treat both cases together, with $a_r\to 0$ to account for the imaginary case. Then:

$$P_{x,T} = -\frac{1}{2a_rT}( e^{-a_rT}- e^{a_rT})\,.$$

We can rewrite it with leading terms as:

$$P_{x,T} = \frac{e^{|a_r|T}}{2|a_r|T}(1 - e^{-2|a_r|T})\,.$$

If $|a_r|\ne 0$, $P_{x,T}$ behaves like $\frac{e^{|a_r|T}}{2|a_r|T}$ and tends to infinity, because $(1 - e^{-2|a_r|T})\to 1$ as $T\to\infty$. If $|a_r|\approx 0$ (imaginary case), at the limit,

$$(1 - e^{-2|a_r|T})\sim 2|a_r|T $$ then the numerator and denominator cancel, and $P_{x,T} \to 1\,.$

Basically, when $a_r$ is zero (or the power is imaginary), the function $x(t)$ really behaves as a constant or a "cisoid". Otherwise, that is a true exponential with divergence.

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    $\begingroup$ I'm glad you filled in the details. Thanks! $\endgroup$
    – MBaz
    Apr 1, 2021 at 17:55

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