Uniform plus exponential exceeding one

Let $U$ be uniform on $(0,1)$ and $V$ be exponential with rate $1$, independent. The probability $P(U+V>1)$ can be written as $a-e^{b}$ with integers $a,b$. Compute $ab$.

$-1$. Condition on $U=u$: the exponential tail gives $P(V>1-u)=e^{u-1}$. Integrating this over $u\in(0,1)$ yields $1-\tfrac1e$, so $a=1$, $b=-1$ and $ab=-1$.