Estimating integral of periodic function

To explain my problem, let me first start with an easier version:

• Let $\Sigma: \mathbb R \to \mathbb C$ be a periodic (continuous) function, $\Sigma(t+T)=\Sigma(t)$.
• Then the integral $$I_1 = \int_0^T \mathrm dt \int_0^T \mathrm d\tau\, \Re\left\{ \Sigma(t-\tau)^\ast \Sigma(t) \right\}$$ is non-negative.
• This can be easily seen by substituting $\tau \rightarrow x=t-\tau$, then we get $$\int_0^T \mathrm dt \int_{t-T}^t \mathrm dx\, \Sigma(x)^\ast \Sigma(t) = \int_0^T \mathrm dt \int_0^T \mathrm dx\, \Sigma(x)^\ast \Sigma(t) \geq 0 \;.$$

I claim: Also the integral $$I_2 = \int_0^T \mathrm dt \int_0^T \mathrm d\tau\, \Re\left\{ \mathrm e^{-\alpha\tau} \Sigma(t-\tau)^\ast \Sigma(t) \right\}$$ is non-negative for $\alpha > 0$. My reasoning:

• For $\alpha \to 0$, $I_2 \to I_1 \geq 0$.
• For $\alpha \to \infty$, $I_2 \to 0$ from above (because the $\mathrm e^{-\alpha\tau}$ gives the strongest weight to $\tau \sim 0$ where the $\Sigma^\ast \Sigma$-part is non-negative)
• I would be surprised if the integral became negative for some intermediate $\alpha$, but I can't prove that it doesn't. Also I have done some numerics and not found a counterexample.

Does someone either have a counterexample, or know how to prove $I_2 \geq 0$?
In the best case, can the whole thing be extended to $\int_0^\infty \mathrm d\tau$?

Found an answer myself, leaving it for posterity. Everything becomes relatively easy once we write $\Sigma(t)$ as a Fourier series, $$\Sigma(t) = \sum_k s_k\, \mathrm{exp}\left(\mathrm i k \frac{2\pi t}{T} \right) \;.$$ (Sorry for calling that function "$\Sigma$"...)
I'll write down the $\int_0^\infty \mathrm d\tau$ case, but $\int_0^T$ works just as well: \begin{align} I_2 &= \sum_{k,\ell} s_k^\ast s_\ell \int_0^T \mathrm dt \int_0^\infty \mathrm d\tau\, \Re\left\{ \mathrm e^{-\alpha\tau} \mathrm e^{-\mathrm i k \frac{2\pi(t-\tau)}{T}} \mathrm e^{\mathrm i \ell \frac{2\pi t}{T}} \right\} \\ &= \sum_{k,\ell} s_k^\ast s_\ell \int_0^T \mathrm dt \cos\left( (\ell-k) \frac{2\pi t}{T} \right) \int_0^\infty \mathrm d\tau\, \mathrm e^{-\alpha\tau} \cos\left( k \frac{2\pi\tau}{T} \right) + 0 \\ &= T^3 \alpha \sum_k \frac{|s_k|^2}{4k^2\pi^2 + T^2\alpha^2} \geq 0 \end{align}