# Integration by parts for fractional Ornstein-Uhlenbeck process

So I have encountered a problem in a paper called Volatility is rough by Jim Gatheral et al.

A stationary fractional Ornstein–Uhlenbeck process ($$X_t$$) is defined as the stationary solution of the stochastic differential equation $$dX_t = ν dW^H_t− α (X_t − m)dt$$ where $$m ∈ \mathbb{R}$$ and $$ν$$ and $$α$$ are positive parameters (see Cheridito et al. 2003). As for usual Ornstein–Uhlenbeck processes, there is an explicit form for the solution which is given by $$X_t = ν\int^t_{−\infty}e^{−α(t−s)}dW^H_t+ m \qquad \quad (3.3)$$ Here, the stochastic integral with respect to fBm is simply a pathwise Riemann–Stieltjes integral, see again Cheridito et al. (2003).

Proposition 3.1 Let $$W_H$$ be a fBm and $$X^α$$ defined by (3.3) for a given $$α > 0$$. As α tends to zero, $$E[sup_{t∈[0,T]}|X^α_t− X^α_0− νW^H_t|]→ 0$$

Then the proof goes like this:

Starting from equation (3.3) and applying integration by parts, we get $$X^α_t = νW^H_t −\int_t^{−∞}ναe^{−α(t−s)}W^H_s ds + m$$

and some proofs that I know, just this part that I don't understand ^

So my question is: How do I even perform integration by parts for the fractional ornstein-uhlenbeck process in this proof. I don't see anyway how to obtain the answer by integration by parts. I'd appreciate if someone can answer or link me to somewhere that has methods/answers!

Essentially, we apply the integration by part to $$(e^{\alpha t}W_t^H)$$ $$\begin{equation} d(e^{\alpha t}W_t^H) = \alpha e^{\alpha t}W_t^H +e^{\alpha t}dW_t^H \end{equation}$$ Then integrate $$\begin{equation} e^{\alpha t}W_t^H - e^{\alpha b}W_b^H = \int_b^t \alpha e^{\alpha s}W_s^Hds + \int_b^te^{\alpha s}dW_s^H\end{equation}$$ So we have : $$\int_b^te^{\alpha s}dW_s^H = e^{\alpha t}W_t^H - e^{\alpha b}W_b^H - \int_b^t \alpha e^{\alpha s}W_s^Hds$$. Replacing the latter in $$(3.3)$$ by taking $$b\to-\infty$$, you will find the answer.