Characteristic function of $Z = X + \sigma Y$ Let $X$ and $Y$ be independent, assume that $Y$ has $\mathcal{N}(0,1)$ distribution. Let $\sigma >0$ and let $\phi$ be the characteristic function of $X: \phi(u) = \mathbb{E}[e^{iuX}]$. 
Now, we study the density function of the random variabel $Z=X+\sigma Y$. The aim is to show that the density $p(z)$ could be expressed as 
\begin{align}
&p(z) = \frac{1}{2\pi\sigma} \int_\mathbb{R} \phi\big(\frac{-y}{\sigma}\big)\exp\big(\frac{iyz}{\sigma - \frac{1}{2}y^2}\big)\ dy.\\
\textbf{Edit: this should be} \qquad &p(z)= \frac{1}{2\pi\sigma} \int_\mathbb{R} \phi\big(\frac{-y}{\sigma}\big)\exp\big(\frac{iyz}{\sigma} - \frac{1}{2}y^2\big)\ dy. \\
\textbf{Now it is very solvable with the answer below.}
\end{align}
Therefore, I first wants to show that 
\begin{align}
p(z) = \frac{1}{\sigma \sqrt{2 \pi}} \mathbb{E}\big[\exp\big(\frac{-(z-X)^2}{2\sigma^2} \big)\big]. \qquad (**)
\end{align}
To start,
\begin{align}
\mathbb{E}[e^{iuZ}] &= \mathbb{E}[e^{iu(X+\sigma Y)}]\\
&=\mathbb{E}[e^{iuX}]\mathbb{E}[e^{iu(Z-X)}], \qquad \text{where } (Z-X)\sim \mathcal{N}(0,\sigma^2)\\
(*)&= \mathbb{E}[e^{iuX}]\cdot\frac{1}{\sigma \sqrt{2 \pi}} \int_\mathbb{R} e^{uiz}\cdot e^{\frac{-(z-X)^2}{2\sigma^2}}\ dz.
\end{align}
However, I am not convinced yet that $(*)$ is valid. And I don't see if this useful for working towards one of the above statements.
Any help is appreciated!
 A: since $X$ and $Y$ are independent we have that
$$
E[f(X)\cdot g(Y)] = E[f(X)]\cdot E[g(Y)],
$$
for any measurable functions $f,g$ such that the above expressions exist. 
Since $\sigma Y \sim \mathcal{N}(0,\sigma^2)$ we can use the characteristic function of the normal distribution to obtain the characteristic function of $Z$:
$$
E[\exp(iuZ)] = E\Bigl[\exp\Bigl(iu(X+\sigma Y)\Bigr)\Bigr] = E[\exp(iuX)] \cdot E[\exp(iu\sigma Y)] = \\\phi(u) \cdot \exp\Bigl(-\frac12 u^2\sigma^2\Bigr).
$$
Now we can use Fourier Inversion to obtain the density of $Z$:
$$
p(z) = \frac 1{2\pi} \int_{\mathbb{R}} \exp(iuz) \cdot \phi(-u) \cdot \exp\Bigl(-\frac12 u^2\sigma^2\Bigr) \; du.
$$
Maybe you can continue from here.
Concerning your equation (*):
$$
E[\exp(iu(Z-X))] = E[\exp(iu\sigma Y)] = \frac{1}{\sqrt{2\pi}} 
\cdot \int_{\mathbb{R}} \exp(iu\sigma y) \exp\Bigl(-\frac12 y^2 \Bigr) \;dy =
\\ = \frac{1}{\sqrt{2\pi}\sigma} 
\cdot \int_{\mathbb{R}} \exp(iuz) \exp\Bigl(-\frac12 \frac{z^2}{\sigma^2} \Bigr) \;dz,
$$
where in the last step I used the substituion $z=\sigma\cdot y$. 
So your formula is not 100% correct: in particular, $X$ is random variable and should not be the result of an expectation, which is in general a number.
Hope that helps a little.
