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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.

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