# Average value of complex exponential

I'm interested in evaluating the following limit $$\lim_{T\to\infty}\frac{1}{T}\int_0^T dt\;e^{i(\varepsilon'-\varepsilon)t},$$ with $$\varepsilon$$ and $$\varepsilon'$$ being real numbers. Performing the integral first yields $$\frac{-i}{\varepsilon'-\varepsilon}\left(e^{i(\varepsilon'-\varepsilon)T}-1\right).$$ Dividing by $$T$$ and taking the limit, seems to converge to zero. This does not look correct at first glance.

It is suspiciously temping to using the principle value of the exponential as $$\int_0^\infty dt\;e^{i(\varepsilon'-\varepsilon)t}=\pi\delta(\varepsilon'-\varepsilon)+\frac{i}{\varepsilon'-\varepsilon}$$ Would that be of any use? Setting $$\varepsilon=\varepsilon'$$ in the first equation, the limit evaluates to 1, hinting at the fact that zero may not be correct.

• not sure of what you want because if $\epsilon \ne \epsilon'$ fixed the limit is clearly zero, while in the second approach you also get $0$ as long as $\epsilon \ne \epsilon'$ so again not sure what you mean by "Setting $\epsilon =\epsilon'$"; maybe more context would illuminate what you are actually interested in; obviously the integral is $1$ for $\epsilon=\epsilon'$ which corresponds to what you say in the second approach, but why is that relevant in general when $\epsilon \ne \epsilon'$? Commented Oct 13, 2022 at 15:49
• Why does the answer of $0$ "not look correct"? Intuitively, as long as $\epsilon'\neq\epsilon$, the values of $e^{i(\varepsilon'-\varepsilon)t}$ oscillate around the circle so "average out" to $0$. Commented Oct 13, 2022 at 18:00
• @Conrad I was simply exploring what would happen for different values of $\varepsilon'-\varepsilon$, and checked the obvious case $\varepsilon'-\varepsilon=0$. The context is that there is an integral over $\varepsilon$ and $\varepsilon'$, so I hoped there would be a $\delta(\varepsilon'-\varepsilon)$ hidden in there. Commented Oct 13, 2022 at 18:01
• @EricWofsey Sure, but I need a general expression also valid for $\varepsilon'-\varepsilon=0$ Commented Oct 13, 2022 at 18:02
• the fact that the integral is $0$ when $\epsilon \ne \epsilon'$ and $1$ when they are equal is precisely a delta function, so if you want a general expression, $\delta(\epsilon-\epsilon')$ will do Commented Oct 13, 2022 at 18:04

## 1 Answer

Let $$\omega = \epsilon - \epsilon '$$. Then we wish to find $$\lim_{T\to \infty} \frac{1}{T}\int _0^T dt\cdot e^{i\omega t}$$

After Integrating: $$\lim_{T\to \infty} \frac{-i}{\omega T}(e^{i\omega T}-1)=\frac{e^{i\omega T /2}}{\omega T}(2\sin{\frac{\omega T}{2}})=\frac{\sin {\omega T}}{\omega T}+i\frac{2\sin^2{\omega T /2}}{\omega T}$$

If $$\omega$$ is non-zero, then the denominator dominates and the limit is zero for both the real and imaginary parts. If $$\omega$$ is zero applications of l'hopital's rule yield 1 for the real part and $$0$$ for the imaginary part.

So $$\lim(...)= \delta(\omega)$$