Expected value of a Delta function times a function I have a output signal $Y(t)= \sum_{n=-\infty}^{\infty} Z_n \delta (t-n\tau) h(t)$ and $Z_n$ is a random variable with equal probability to be $+-1$ and h(t) is $sinc(\frac t {\tau})$

and the Expected value of $Y(t)$ is

$E(Y(t)) = E[\sum_{n=-\infty}^{\infty} Z_n \delta (t-n\tau) h(t)]$

$=\sum_{n=-\infty}^{\infty}h(n\tau) E[Z_n]$

$=0$

But if I calculated in other way

$\int_{-\infty}^\infty t Y(t) dt$

$=\int_{-\infty}^\infty t \sum_{n=-\infty}^{\infty} Z_n \delta (t-n\tau) h(t) dt$

$= \int_{-\infty}^\infty  \sum_{n=-\infty}^{\infty} tZ_n \delta (t-n\tau) h(t) dt$ 

$= \int_{-\infty}^\infty  \sum_{n=-\infty}^{\infty} n\tau Z_n h(n\tau) dt$

$=  \sum_{n=-\infty}^{\infty} [n\tau h(n\tau)\int_{-\infty}^\infty   Z_n  dt]$

Why I have two different result with same signal? Thanks for any help.
If $h(t)$ is continuous at $a$, then $h(t)\delta(t-a) = h(a)\delta(t-a)$. Since the 
$\operatorname{sinc}$ function is
continuous everywhere, 
$$\begin{align}
Y(t) &amp;= \sum_{n=-\infty}^{\infty} Z_n\delta(t-n\tau)h(t)\\
&amp;= \sum_{n=-\infty}^{\infty} Z_n\delta(t-n\tau)h(n\tau)\\
&amp;= \sum_{n=-\infty}^{\infty} Z_n\delta(t-n\tau)\operatorname{sinc}(n)\\
&amp;= Z_0\delta(t)
\end{align}$$
since $\operatorname{sinc}(n) = \begin{cases}1,&amp;n = 0,\\0,&amp; n\neq 0.\end{cases}$
You (or your textbook's authors) probably meant to write something else.
You sure you are not missing a $\star$ indicating a convolution between
that $\delta(t-n\tau)$ and $h(t)$?