Evaluate a complex exponential at negative infinity [duplicate]

I am learning about the properties of the Fourier Series (FS), which is defined by:

$$x(t) = \sum_{k=-\infty}^{\infty}c_{x}[k]e^{j2\pi kt/T}\tag{1}$$

where $$c_{x}[k] = \frac{1}{T}\int_{T}x(t)e^{-j2\pi kt/T}\,dt.\tag{2}$$

One of the properties of FS is the time integration property, which is that:

$$\int_{-\infty}^{t}x(\tau)\,d\tau \overset{\mathscr{FS}}{\underset{T}\longleftrightarrow}\frac{c_{x}[k]}{j2\pi k/T},\quad k\neq 0 \ \text{ if }\ c_{x}[0] = 0 \tag{3}.$$ To prove this property, I integrated $$(1)$$ from $$-\infty$$ to $$t$$ in terms of $$\tau$$. However, when I am evaluating the indefinite integral, I need to find the value of $$e^{\frac{j2\pi k(-\infty)}{T}}$$.

I almost jumped to the conclusion that an exponential raised to $$-\infty$$ must be $$0$$, which is what the proof requires in order for the property to hold true. However, $$k$$ can be negative, and, more importantly, a complex exponential is periodic, so I don't know how to evaluate it at $$-\infty$$.

I would like some insight into why we can treat this evaluation as zero.

• – MBaz
Nov 17, 2020 at 22:34