Can I shift this integral along the complex plane? Does
$$\int_{-\infty}^\infty \text{e}^{\ a\ (x+b)^2}\ \text dx=\int_{-\infty}^\infty \text{e}^{\ a\ x^2}\ \text dx\ \ \ \ \ ?$$
hold, even if the imaginary part of $b$ is nonzero?

What I really want to understand is what the phrase &quot;By analogy with the previous integrals&quot; means in that link. There, the expression $\frac{J}{a}$ is complex but they seem to imply the integral can be solved like above anyway.
The reusult tells us that the integral is really independend of $J$, which is assumed to be real here. I wonder if we can also generalize this integral to include complex $J$. In case that the shift above is possible, this should work out.
But even if the idea is here to perform that substitution, how to get rid of the complex $a$ to obtain the result. If everything is purely real or imaginary, then this solves the rest of the problem.
Let us write $b= r+it$. The real part of $b$ does not matter as you have already proven yourself. So wlog $r=0$.

For shifting along the imaginary axis, we have to employ the residue theorem. We have
$$
\begin{align}
\int_{-\infty}^\infty f(x+i t) \,dx&amp;- \int_{-\infty}^\infty f(x)\, dx\\
&amp;=\int_{-\infty-it}^{\infty-it} f(x) \,dx- \int_{-\infty}^\infty f(x)\, dx \\
&amp;= 2\pi i \sum \text{Res}(f)+ \int_{\infty-it}^{\infty} f(x) \,dx
- \int_{-\infty-it}^{-\infty} f(x) \,dx
\end{align},$$
where $\sum \text{Res}(f)$ is the sum over the residues of $f$ in the area $z\in \mathbb{C}$ with $-t&lt;\text{Im}\, z&lt;0$.

So the two integrals are the same if there are no residues and if the two integral at $\pm \infty$ vanish (both of which is the case for your example as long as $\text{Re}\,a &lt;0$).
Put 
$$(x+b)^2=u^2\Longrightarrow 2(x+b)dx=2udu\Longrightarrow dx= du$$ so we get

$$\int_{-\infty}^\infty e^{a(x+b)^2}dx=\int_{-\infty}^\infty e^{au^2}du$$
and your question's answered in the affirmative.

Added: You may want to divide the integral in two rays $\,(-\infty,0)\,,\,(0,\infty)\,$ and then do the above, to avoid problems with the signs after taking the square root.