Maximum of a killed telegraph integral ($\mathbb{P}({\sup }_{t\ge 0}{\int }_0^t{X}_{s}ds>c\mid X_0=1)=e^{-c\sqrt{\lambda (2q+\lambda )}}$)

Let $X$ switch between $\pm 1$ at rate $q$ and be killed into zero at rate $\lambda$. To exceed a new level it must cross it in state $+1$. Restarting there by the Strong Markov property gives the multiplicative identity ${\psi }_{+}(c+d)={\psi }_{+}(c){\psi }_{+}(d)$, hence an exponential tail. First-step conditioning gives ${\psi }_{-}(c)={\int }_0^{\infty }q{e}^{-(q+\lambda )u}{\psi }_{+}(c+u)du$ and ${\psi }_{+}(c)=e^{-(q+\lambda )c}+{\int }_0^{c}q{e}^{-(q+\lambda )u}{\psi }_{-}(c-u)du$. Substituting the exponential determines its rate as $\sqrt{\lambda (2q+\lambda )}$.