Will the integral still be 1? We can show that $\int_{\mathbb{R}}\frac{e^{-\frac{(x-m)^2}{2}}}{\sqrt{2\pi}}=1$. The easiest way is by observing that this is the probability density of a normal random variable, with mean m, and variance 1.
But what happens if we change m with im, where i is the imaginary number $\sqrt{-1}$. Is the integral still always 1 then?
 A: Yes, the integral will still be $1$. One can see that $f\colon \mathbb{C}\to \mathbb{C}$ defined by
$$f(z) := \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \exp \biggl(-\frac{(x-z)^2}{2}\biggr)dx$$
is constant in various ways.
First of course one has to check that $f$ is well-defined. For $\lvert z\rvert \leqslant K$, one can bound the modulus of the integrand by a function $C_K\cdot e^{-\frac{x^2}{4}}$ (see below), so not only is $f$ well-defined, by the dominated convergence theorem it is also continuous.
One can also use the dominated convergence theorem to see that $f$ is (arbitrarily often) differentiable and differentiation under the integral is legitimate, but the more idiomatic way to show that $f$ is actually holomorphic is using Morera's theorem. For any triangle $\Delta \subset \mathbb{C}$, we have
$$\int_{\partial \Delta} f(z)\,dz = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} \int_{\partial \Delta} \exp \biggl(-\frac{(x-z)^2}{2}\biggr)\,dz \,dx = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} 0\,dx = 0,$$
where the change of order of integration is justified by Fubini's theorem together with the bound from above, and the inner integral over $\partial \Delta$ vanishes by Cauchy's integral theorem. Hence, by Morera's theorem $f$ is holomorphic.
Now the identity theorem together with $f\lvert_{\mathbb{R}} \equiv 1$ shows that $f(z) = 1$ for all $z\in \mathbb{C}$.
One can also use the Fourier transformation to show $f \equiv 1$ by real methods. The key is that $g(x) := e^{-\frac{x^2}{2}}$ is a fixed point of the Fourier transform - when one uses the
$$\mathscr{F}[h](\omega) := \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} h(x) e^{-ix\omega}\,dx$$
variant of the Fourier transform. For other variants the Fourier transform is still a function of the form $e^{-\alpha x^2}$ for some $\alpha > 0$ and some details in the computation change, but the basic structure is the same.
We recall that for the translates $\tau_y h \colon x \mapsto h(x-y)$, we have
$$\mathscr{F}[\tau_y h](\omega) = e^{-iy\omega}\mathscr{F}[h](\omega),$$
and thus with $z = u+iv$ we have
\begin{align}
f(z) &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} \exp\biggl(-\frac{\bigl((x-u) - iv\bigr)^2}{2}\biggr) dx\\
&= \frac{e^{\frac{v^2}{2}}}{\sqrt{2\pi}} \int_{\mathbb{R}} \exp\biggl(-\frac{(x-u)^2}{2}\biggr) e^{i(x-u)v}\,dx\\
&= e^{\frac{v^2}{2}-iuv}\mathscr{F}[\tau_u g](-v)\\
&= e^{\frac{v^2}{2}-iuv}\cdot e^{-iu(-v)}e^{-\frac{(-v)^2}{2}}\\
&= 1.
\end{align}

To show that for every $K\in (0,+\infty)$ there is a $C_K\in (0,+\infty)$ such that
$$\biggl\lvert \exp \biggl(-\frac{(x-z)^2}{2}\biggr)\biggr\rvert \leqslant C_k\cdot e^{-\frac{x^2}{4}}$$
for all $z$ with $\lvert z\rvert \leqslant K$, we use the elementary inequality $\lvert e^w\rvert \leqslant e^{\lvert w\rvert}$ for $w\in \mathbb{C}$ and the addition theorem of the exponential function. For $\lvert x\rvert \geqslant 4K$ we have $\lvert z\rvert \leqslant K \leqslant \frac{1}{4}\lvert x\rvert$, and hence $\lvert xz\rvert \leqslant \frac{x^2}{4}$, so
$$\biggl\lvert \exp \biggl(-\frac{(x-z)^2}{2}\biggr)\biggr\rvert \leqslant e^{-\frac{x^2}{2}}\cdot e^{\lvert zx\rvert}\cdot e^{\frac{\lvert z\rvert^2}{2}} \leqslant e^{\frac{K^2}{2}}\cdot e^{-\frac{x^2}{4}}.$$
On the compact set $[-4K,4K]$ the continuous function
$$e^{\frac{x^2}{4}}\biggl\lvert \exp \biggl(-\frac{(x-z)^2}{2}\biggr)\biggr\rvert$$
is bounded, say by $B_K$, and then we can choose $C_K = \max \bigl\{B_K, e^{\frac{K^2}{2}}\bigr\}$.
