Variance of a stochastic process Problem
Consider the classical brownian $W_t$, such as $var\left(W_t-W_u\right)=\left|t-u\right|$.
Consider the process defined by
$$X(t)=e^{-t/\tau}\left(X_0+\sqrt{Q_B}\int_0^t e^{u/\tau}dW_u\right)$$
With $Q_B$ being a real positive scaling constant.
I want to study
$$Y(t,L)=\frac{X(t+L)-X(t)}{L}={\left(e^{-L/\tau}-1\right)X(t)+e^{-L/\tau}\sqrt{Q_B}\int_{t}^{t+L} e^{(u-t)/\tau}dW_u \over L}$$
Ideally for any $t$, and at least for $t\to\infty$. (I'll refer to $\lim_{t\to\infty}Y(t,L)$ as $Y_\infty(L)$ from now on).
Since $Y(t,L)$ is gaussian, with mean ${\left(e^{-L/\tau}-1\right)e^{-t/\tau} \over L}X_0$, we just have to compute its variance.

First approach
Considering that the two terms of the sum are independant (do you agree that $\int_0^t e^{u/\tau}dW_u$ and $\int_{t}^{t+L} e^{(u-t)/\tau}dW_u$ are independant ?), I simply sum the variances and get
$$var\left(Y(t,L)\right)={\left(e^{-L/\tau}-1\right)^2var(X(t))+\left(1-e^{-2L/\tau}\right)\frac{Q_B\tau}{2} \over L^2}$$
Computing $var(X(t))$ :


*

*I can either calculate directly
$$var(X(t))=Q_Be^{-2t/\tau}\int_0^t e^{2u/\tau}du=\frac{Q_B\tau}{2}\left(1-e^{-2t/\tau}\right)$$

*or, at least for $t\to \infty$, use the Laplace transform $\frac{1}{1+\tau s}$ and compute
$$Q_B\int_{-\infty}^{+\infty}\left|\frac{1}{1+2i\pi\tau f}\right|^2df=Q_B\int_{-\infty}^{+\infty}\frac{1}{1+4\pi^2\tau^2 f^2}df=\frac{Q_B}{2\tau}$$
First remark : when $t\to\infty$, the first converges to $\frac{Q_B\tau}{2}$, so the two methods don't match. I guess the second one is incorrect, but why ?
Considering $var(X(t))=\frac{Q_B\tau}{2}\left(1-e^{-2t/\tau}\right)$ (first result), we would get
$$var\left(Y(t,L)\right)=\left[\left(e^{-L/\tau}-1\right)^2\left(1-e^{-2t/\tau}\right)+\left(1-e^{-2L/\tau}\right)\right]\frac{Q_B\tau}{2L^2}$$
and
$$var\left(Y_\infty(L)\right)=\frac{Q_B\tau}{L^2}\left(1-e^{-L/\tau}\right)$$
First, is this right ?

Second approach
Another way to compute $var\left(Y_{\infty}(L)\right)$ would be to consider the Laplace transform
$$\frac{Y}{X}=\frac{e^{sL}-1}{L}$$
and evaluate
$$var\left(Y_{\infty}(L)\right)=Q_B\int_{-\infty}^{+\infty}\left|\frac{e^{2i \pi Lf}-1}{L(1+2i\pi\tau f)}\right|^2df=Q_B\int_{-\infty}^{+\infty}\frac{4sin^2\pi Lf}{L^2(1+4\pi^2\tau^2 f^2)}df$$
Let $x=\pi L f$, and we get
$$var\left(Y_{\infty}(L)\right)=\frac{4Q_B}{\pi L}\int_{-\infty}^{+\infty}\frac{sin^2x}{L^2+4\tau^2 x^2}dx$$
Since Wolfram gives us
$$\int_{-\infty}^{+\infty}\frac{sin^2x}{L^2+4\tau^2 x^2}dx=\frac{\pi}{4\tau L}(1-e^{-L/\tau})$$
we conclude
$$var\left(Y_{\infty}(L)\right)=\frac{Q_B}{\tau L^2}(1-e^{-L/\tau})$$

Now, this differs from what we got in the first approach. Which one is right (if at least one is right...) ? And why is the other (or both) wrong ?
 A: Ok, this was me being stupid and making a mistake in my solving of the system I was studying.
The integral expression of $X(t)$ was missing a $\frac{1}{\tau}$, and should have been like this :
$$X(t)=e^{-t/\tau}\left(X_0+\frac{\sqrt{Q_B}}{\tau}\int_0^t e^{u/\tau}dW_u\right)$$
Which in turn gives
$$Y(t,L)=\frac{X(t+L)-X(t)}{L}={\left(e^{-L/\tau}-1\right)X(t)+e^{-L/\tau}\frac{\sqrt{Q_B}}{\tau}\int_{t}^{t+L} e^{(u-t)/\tau}dW_u \over L}$$
and Ito's isometry now returns
$$var(X(t))=\frac{Q_B}{2\tau}\left(1-e^{-2t/\tau}\right)$$
and there is no discrepancy anymore with the integration of the PSD I was doing in parallel. Thus we also get
$$var\left(Y(t,L)\right)=\left[\left(e^{-L/\tau}-1\right)^2\left(1-e^{-2t/\tau}\right)+\left(1-e^{-2L/\tau}\right)\right]\frac{Q_B}{2\tau L^2}$$
and finally, we do get
$$var\left(Y_{\infty}(L)\right)=\frac{Q_B}{\tau L^2}(1-e^{-L/\tau})$$
Which also matches the result obtained through direct integration of the PSD, and all's right with the world again.
Sorry everyone for the trouble !
