Other then as a fantastic tool to evaluate some difficult real integrals, complex integrals have many purposes.
Firstly, contour integrals are used in Laurent Series, generalizing real power series.
The argument principle can tell us the difference between the poles and roots of a function in the closed contour $C$:
$$\oint_{C} {f'(z) \over f(z)}\, dz=2\pi i (\text{Number of Roots}-\text{Number of Poles})$$
and this has been used to prove many important theorems, especially relating to the zeros of the Riemann zeta function.
Noting that the residue of $\pi \cot (\pi z)f(z)$ is $f(z)$ at all the integers. Using a square contour offset by the integers by $\frac{1}{2}$, we note the contour disappears as it gets large, and thus
$$\sum_{n=-\infty}^\infty f(n) = -\pi \sum \operatorname{Res}\, \cot (\pi z)f(z)$$
where the residues are at poles of $f$.
While I have only mentioned a few, basic uses, many, many others exist.
As already suggested, it's hard to see how we can use RT here, since $e^{-z^2/2}$ is entire; there are no poles, hence no residues.
If this came up on a complex exam, maybe this is what they had in mind: Say the integral is $I(s)$. It's an exercise to use Cauchy's Theorem to show that $I(s)=I(0)$; now if we're allowed to regard $I(0)=\sqrt{2\pi}$ as "well-known" we're done.
Comment It's actually an important integral: If you write $(x+is)^2=x^2+isx-s^2$ you see that $I(s)=\sqrt{2\pi}$ implies that the function $e^{-x^2/2}$ is its own Fourier transform.
Ok, for the benefit of readers who don't want to do any exercises:
Details: Let $f(z)=e^{-z^2/2}$ and define $$I(s)=\int_{-\infty}^\infty f(x+is)\,dx.$$
$I(s)=I(0)$ for every $s\in\Bbb C$.
Proof: It's clear that $I(s+it)=I(s)$ for $t\in\Bbb R$, so we may assume that $s\in\Bbb R$. In fact we'll assume $s>0$ just so it's clear what the picture looks like.
For $A>0$ let $$I_A(s)=\int_{-A}^Af(x+is)\,dx,$$so $I(s)=\lim_{A\to\infty}I_A(s)$.
Let $R_A$ be the rectangle with vertices $\pm A$, $\pm A+is$. Since $f$ is entire, Cauchy's Theorem shows that $$\int_{R_A}f=0.$$Parametrizing $R_A$ in the obvious way, and letting $[p,q]$ denote the line segment from $p$ to $q$, this says precisely that $$I_A(0)-I_A(s)=\int_{[-A,-A+is]}f - \int_{[A, A+is]}f.$$Now since $|e^{\alpha+i\beta}|=e^\alpha$we see that $$|f(x+iy)|=e^{(y^2-x^2)/2},$$hence $$|f(z)|\le e^{(s^2-A^2)/2}
\quad(z\in[A,A+is]).$$So $$\lim_{A\to\infty}\int_{[A,A+is]}f=0$$by uniform convergence. Similarly for $[-A,-A+is]$; hence $$\lim_{A\to\infty}(I_A(0)-I_A(s))=0,$$or $$I(0)-I(s)=0.$$
Best Answer
Cauchy's integral formula has a straight-forward generalization to polydiscs. If $f$ is holomorphic on (a neighbourhood) of the polydisc $\Omega = \mathbb{D}(z_1,r_1) \times \mathbb{D}(z_2,r_2) \times \cdots \times \mathbb{D}(z_n,r_n)$, and $a \in \Omega$, then
$$ f(a) = \frac{1}{(2\pi i)^n} \int_{|z_1|=r_1} \int_{|z_2|=r_2}\cdots\int_{|z_n|=r_n} \frac{f(z_1, z_2, \ldots, z_n)}{(z_1-a_1)(z_2-a_2)\cdots(z_n-a_n)}\,dz_1\cdots dz_n. $$
The really fascinating thing is that we are integrating just on a tiny part of the boundary of $\Omega$.
For more general domains $\Omega \subset \mathbb{C}^n$, you have the Bochner-Martinelli integral formula, and there are lots of other integral representation formulas. Keywords: Leray and Koppelman integral formulas. See for example Range: Holomorphic Functions and Integral Representations in Several Complex Variables for details.
Regarding generalizations of residues to several complex variables, see this recent question.