Question: Yeah, you probably are reading this question and thinking I'm one of those engineers who hasn't learned about the s-plane. I can assure you I'm not! I know that the transfer function of an RC circuit (Vout/Vin) has a REAL pole, with no imaginary component at s= -1/RC + j0. I am curious what happens when we excite an RC circuit at its TRUE pole, which is a decaying exponential with the expression e^(-t/RC).

Of course, the output of an RC Low-pass filter will never blow up to infinity. But what about the ratio of the output to the input, which after all, is the transfer function I originally defined?

Well, let's take this to LTSpice. Below is what I've simulated:
[![enter image description here](https://i.stack.imgur.com/GChDh.png)](https://i.stack.imgur.com/GChDh.png)

I would expect Vout/Vin to be infinity everywhere, however what I see is a ramp. I suspect I'm missing something about how to properly interpret the magnitude response of my transfer function with its true pole as an input. A time-domain view such as what I have should explain something, but I can't understand why it's a ramp.

If anyone has any intuition behind this question, please let me know.

Answer: 
[KD9PDP's answer](https://electronics.stackexchange.com/a/551541/95619) alreay explains very nice (+1) why the result is as you see it, but there might be an easier way to do it. You already have the impulse response (1) and the input (2):

$$\begin{align}
h(t)&=\dfrac{1}{RC}\mathrm{e}^{-\frac{t}{RC}}\tag{1} \\
s(t)&=\mathrm{e}^{-\frac{t}{RC}}\tag{2}
\end{align}$$

The convolution integral is now simple because \$h(t)\$ and \$s(t)\$ are simplified before applying the integral:

$$\begin{align}
s(t)\star h(t)&=\int\_{\tau}{\dfrac{1}{RC}\mathrm{e}^{-\frac{\tau}{RC}}\cdot\mathrm{e}^{-\frac{t-\tau}{RC}}\mathrm{d}\tau} \\
{}&=\int\_{\tau}{\dfrac{1}{RC}\mathrm{e}^{-\frac{t}{RC}}\mathrm{d}\tau} \\
{}&=\dfrac{1}{RC}\mathrm{e}^{-\frac{t}{RC}}\cdot t\tag{3} \\
\end{align}$$

Since the variable is \$\tau\$, the integrand is a constant, and integrating it means adding \$t\$, which is the only difference between (3) and (1). As noted in the other answer, it's also a matter of initial conditions (\$\mathrm{i.c.}\$), since LTspice assumes those to be given by the value of the behavioural expression at `t=0`, which is `1`.
It *is* ramping to infinity as you expected. The s-domain calculation is a steady-state calculation, and you have an initial transient (your input exponential was effectively a constant '1' for all t < 0). You have to wait for that to decay (i.e. become negligible) -- e.g. a long (infinite) time.

You can see that after 10 time constants, your 'gain' is already 10.