The value of $\frac{1}{T}\int_{0}^{T} e^{ikt} dt$ when $T \rightarrow \infty$

Question

I have this integral $\left(\dfrac{1}{T}\int_{0}^{T} e^{ikt} dt\right)$ when $T \rightarrow \infty$, in an equation that I am studying, and it seems that this integration gives a value of $\dfrac{\delta(k)}{2}$. I would like to know how it computes to this result. $T$ is a quantization time period that goes to infinity.

Thank you

Answer 1

If $k\neq 0$, note that $$\int_0^Te^{ikt}dt = \frac{1}{ik}e^{ikt} \bigg|_0^T = \frac{e^{ikT} - 1}{ik}. $$ Taking the absolute values, note that \begin{align*} \lim_{T\to \infty}\left|\frac{1}{T}\int_0^Te^{ikt}dt\right| &= \lim_{T\to \infty}\frac{1}{kT}\left|e^{ikT}-1\right|\\ &\leq \lim_{T\to \infty}\frac{2}{kT} \\&= 0. \end{align*} In the case when $k=0$, the integrand is constant (it takes the value 1 everywhere) and $$ \frac{1}{T}\int_{0}^T dt = 1. $$