How this convolution integral is solved to obtain autocorrelation function of current for shot noise?

Question

I am studying shot noise characteristics from this source:

Here the author writes that autocorrelation function is given by: $$R_I(\tau)=\bar{h}*h*R_Z(\tau)$$ where $R_Z(\tau)=q^2(\lambda^2+\lambda \delta(\tau))$ and $\bar{h}(t)=h(-t)$ now applying the value of $R_Z(\tau)$ in the integral we have $$=q^2\lambda^2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\bar{h}(\tau-s)h(s-t)\text{d}s\text{d}t+q^2\lambda\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\bar{h}(\tau-s)h(s-t)\delta(t)\text{d}s\text{d}t$$ $$=q^2\lambda^2(\int_{-\infty}^{\infty}{h}(t)\text{d}t)^2+q^2\lambda\bar{h}*h(\tau)$$ the second term I am able to derive as I know that $$q^2\lambda\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\bar{h}(\tau-s)h(s-t)\delta(t)\text{d}s\text{d}t$$ $$=q^2\lambda\int_{-\infty}^{\infty}\bar{h}(\tau-s)h(s)\text{d}s$$ however, I can not get that how: $$q^2\lambda^2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\bar{h}(\tau-s)h(s-t)\text{d}s\text{d}t$$ turns out to be $$=q^2\lambda^2(\int_{-\infty}^{\infty}{h}(t)\text{d}t)^2$$ can somebody please help me out with this?

PS: As per the answer of Matt L., as inner integral is independent of $s$ so I can do certain manipulations to get requisite output, however, I have a doubt here that I want to clear: $$q^2\lambda^2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\bar{h}(\tau-s)h(s-t)\text{d}s\text{d}t$$ is precisely $$q^2\lambda^2\int_{t=-\infty}^{\infty}\int_{s=-\infty}^{\infty}\bar{h}(\tau-s)h(s-t)\text{d}s\text{d}t$$ now repeating the steps as suggested in the answer I have: $$q^2\lambda^2\int_{s=-\infty}^{\infty}\bar{h}(\tau-s)\left[\int_{t=-\infty}^{\infty}h(s-t)\text{d}t\right]\text{d}s$$ I believe that the inner integral should be some function of $s$ let us say $f(s)$ and then we should have $$q^2\lambda^2\int_{s=-\infty}^{\infty}\bar{h}(\tau-s)f(s)\text{d}s$$ The reason I believe so is because $s$ may be constant for inner integral but outer integral needs $s$, how I can I simply replace $s$ without changing the outer integral limits and $\text{d}s$ in outer integral.

Also taking the answer's view: $$q^2\lambda^2\int_{s=-\infty}^{\infty}\bar{h}(\tau-s)\left[\int_{t=-\infty}^{\infty}h(s-t)\text{d}t\right]\text{d}s$$ we replace $s-t$ with some dummy variable $t'$, but if we do so, shouldn't the outer integral will change accordingly as it also contains $s$?

Answer 1

Note that

$$\begin{align}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\bar{h}(\tau-s)h(s-t)dsdt&=\int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty}h(s-t)dt\right]\bar{h}(\tau-s)ds\\&=\int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty}h(t)dt\right]\bar{h}(\tau-s)ds\\&=\int_{-\infty}^{\infty}h(t)dt\int_{-\infty}^{\infty}h(s-\tau)ds\\&=\int_{-\infty}^{\infty}h(t)dt\int_{-\infty}^{\infty}h(s)ds\\&=\left(\int_{-\infty}^{\infty}h(t)dt\right)^2\end{align}$$

The second equality is true because the inner integral is independent of $s$. Similarly, the integral of $h(s-\tau)$ in the third line is independent of $\tau$.