Consider $X_1,\ldots, X_n$ iid random variables with $X_1\sim\mathcal{N}(0,1)$, some real numbers $\alpha_1,\ldots,\alpha_n$.
Define $Z:=\sum_{i=1}^n (X_i+\alpha_i)^2$ and $\kappa:=\sum_{i=1}^n\alpha_i^2$
I want to show by direct integration, that
$$E[e^{uZ}]=\frac{\exp({\frac{\kappa u}{1-2u}})}{(1-2u)^\frac{\kappa}{2}}$$
for any $u\in\mathbb{C}$ with $\Re(u)\le 0$. What I have done so far is the following: $$ \begin{align} E[e^{uZ}]&=E[e^{u\sum_{i=1}^n (X_i+\alpha_i)^2}]\\ &=\prod_{i=1}^nE[e^{u (X_i+\alpha_i)^2}]\text{, since }X_i\text{ are independent}\\ &=\prod_{i=1}^n \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{u (x+\alpha_i)^2}e^{-\frac{x^2}{2}}dx\text{, since }X_i\sim\mathcal{N}(0,1)\\ &=\prod_{i=1}^n \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{(2u-1) (\frac{x+\alpha_i}{\sqrt{2}})^2}e^{x\alpha_i+\frac{1}{2}\alpha_i^2}dx\\ \end{align} $$
The right hand side can be written as
$$ \begin{align} \frac{\exp({\frac{\kappa u}{1-2u}})}{(1-2u)^\frac{\kappa}{2}}&=\exp\Big({\frac{\kappa u}{1-2u}}-\frac{\kappa}{2}\ln(1-2u)\Big)\\ \end{align} $$
How can I go on from here? Maybe I can apply some Fourier Inversion formula?
Thanks in advance!