Let $Z$ denote the lifetime of an electrical device. $Z$ is exponentially distributed with parameter $\lambda$. The time it is used daily is denoted by $X_n$ where $X_n$ are iid random variables which are uniformly distributed on $[0,1]$. Determine the probability that the device works for at least $n$ days.
What we have to calculate is $\mathbb{P}(\sum_{i=1}^n X_i \leq Z)$. The left hand side $S_n:= \sum_{i=1}^n X_i$ would be the convolution of $n$ iid random variables which I approximated by the normal distribution $\mathcal{N}(\frac{n}{2},\frac{n}{4})$. I thus get
\begin{eqnarray*} \mathbb{P}(Z \leq S_n) &=& \int_0^\infty \mathbb{P}(Z\leq S_n|S_n = s)f_{S_n}(s)ds\\ &=& \int_0^\infty \mathbb{P}(Z\leq s)f_{S_n}(s)ds\\ &=& \int_0^\infty F_{Z}(s)f_{S_n}(s)ds\\ &\approx& \int_0^\infty \frac{1}{\sqrt{2\pi}\sqrt{n/4}}\exp{\left[-\frac{(s-n/2)^2}{2n/4}\right]}(1-\exp{\lambda s})ds\\ &\approx& 1-\int_0^\infty \frac{1}{\sqrt{2\pi}\sqrt{n/4}}\exp{\left[-\frac{(s-n/2)^2}{2n/4}+\lambda s\right]}ds. \end{eqnarray*}
In the last step I approximated one of the two integrals with 1, as it is really close to 1. But whatever I do here, I am not able to calculate the integral. I tried, for example, to get that integral to be some normal distribution again, but such approaches always failed as this integral is over $(0,\infty)$. Is there any way of calculating that integral or any other approach to that problem?