Can the following integrals be solved analytically?

Question

Question

• Can the following integrals be solved analytically?

\begin{eqnarray} \frac{\sigma^2 }{b^2} \int^T_t \left( e^{ -b(T- s)} - 1 \right)^2 ds \end{eqnarray}

• where $\sigma, b > 0$, and $T \ge t \ge 0$.

Answer 1

\begin{eqnarray} \frac{\sigma^2 }{b^2} \int^T_t \left( e^{ -b(T- s)} - 1 \right)^2 ds = \frac{\sigma^2 }{b^2}\Big( e^{ -2bT}\int^T_t e^{ 2bs} ds - 2e^{ -bT} \int^T_te^{ bs} ds + \int^T_t ds \Big) \end{eqnarray} where $$\int^T_t e^{ 2bs} ds = \frac{1}{2b}(e^{2bT}-e^{2bt})$$ $$\int^T_t e^{ bs} ds = \frac{1}{2b}(e^{bT}-e^{bt})$$ $$ \int^T_t ds = T-t$$

Answer 2

The substitution $u:=T-s$ converts the problem to $$\frac{\sigma^2}{b^2}\int_0^{T-t}(e^{-2bu}-2e^{-bu}+1)ds=\frac{\sigma^2}{b^2}\left[\frac{-3+4e^{-b(T-t)}-e^{-2b(T-t)}}{2b}+T-t\right].$$