Consider the following shock model. The count of shocks within a certain time $t$ is a Poisson process $N(t)$ with parameter $\lambda$, while every shock brings damage $Y_i$ to the subject, which is exponentially distributed with parameter $\mu$. The subject can at most withstand damages of $a$. So the subject has a lifespan of $T$. Now how to calculate $E[T]$?
My attempt: Since $T\ge 0$, so $ \begin{align} E[T]=&\int_0^{+\infty} P(T>t)\,dt\\ =&\int_0^{+\infty} P\left(\sum_{i=1}^{N(t)}Y_i\le a\right)\,dt\\ =&\int_0^{+\infty}\left( \sum_{n=1}^{+\infty} P\left(\sum_{i=1}^{n}Y_i\le a\right)\frac{(\lambda t)^n}{n!}e^{-\lambda t}\right)\,dt\\ =&\sum_{n=1}^{+\infty} P\left(\sum_{i=1}^{n}Y_i\le a\right)\int_0^{+\infty}\frac{(\lambda t)^n}{n!}e^{-\lambda t}\,dt\\ =&\frac{1}{\lambda}\sum_{n=1}^{+\infty} P\left(\sum_{i=1}^{n}Y_i\le a\right)\\ =&\frac{1}{\lambda}\sum_{n=1}^{+\infty}\int_0^a\left(\mu e^{-\mu \xi}\cdot \mathbf{1}_{\xi\ge 0}\right)^{*n}\,d\xi\\ =&\frac{1}{\lambda}\int_0^a \mathcal{F}^{-1}\left[\sum_{n=1}^{+\infty}\mathcal{F}\left[\mu e^{-\mu\xi}\cdot \mathbf{1}_{\xi\ge 0}\right]^n\right]\,d\xi\\ =&\frac{1}{\lambda}\int_0^a \mathcal{F}^{-1}\left[\sum_{n=1}^{+\infty}\left(\frac{i\mu}{s+i\mu}\right)^n\right]\,d\xi\\ =&\frac{1}{\lambda}\int_0^a \mathcal{F}^{-1}\left[\frac{i\mu}{s}\right]\,d\xi\\ =&\frac{1}{\lambda}\int_0^a \frac{\mu}{2}\mathrm{sign}(\xi)\,d\xi\\ =&\frac{\mu a}{2\lambda} \end{align} $
But my answer is wrong, and the correct one is $(1+\mu a)/\lambda$. I don't know where I mess things up. Any hint will be appreciated, thank you.
EDIT: I am using Fourier transforms defined as $\mathcal{F}[f]=E[e^{isT}]=\int_{-\infty}^{+\infty}f(t)e^{ist}\,dt$ and $\mathcal{F}^{-1}[f]=\frac{1}{2\pi}\int_{-\infty}^{+\infty}f(t)e^{-ist}\,dt$