Joint Laplace transform of Brownian square and its time integral ($\mathbb{E}{e}^{-\alpha B_t^2-({\beta }^2/2){\int }_0^t{B}_s^{2}ds}=[\cosh (\beta t)+(2\alpha /\beta )\sinh (\beta t){]}^{-1/2}$)

For standard Brownian motion and $\alpha ,t\ge 0$, $\beta >0$, the title gives the joint Laplace transform of the terminal square and the integrated square of Brownian motion. At $\beta =0$ the answer is $(1+2\alpha t{)}^{-1/2}$. To prove the formula, use the stochastic exponentialOn every finite interval $[0,T]$, $0<Z_t\le e^{\beta T/2}$, so the bounded local martingale criterion makes it a martingale. The Girsanov theorem changes $B$ into the zero-started Ornstein-Uhlenbeck process $dB=dW-\beta Bdt$ under the terminal density $Z_t$. Its terminal variance is $v=(1-e^{-2\beta t})/(2\beta )$. Multiplying $Z_t$ by $\exp [(\beta /2-\alpha )B_t^2-\beta t/2]$ recovers the integrand on the left of the title. The Gaussian quadratic exponential moment then gives $e^{-\beta t/2}[1-(\beta -2\alpha )v{]}^{-1/2}$, which simplifies to the title formula. At $\alpha =0$ this reduces to the Laplace transform of the integrated square of Brownian motion after Brownian scaling.