# Integral representation of Hermite polynomial

I'm studying Merzbacher's Quantum Mechanics. In his treatment of harmonic oscillators, he introduces Hermite polynomials and their properties. He mentions that the Hermite polynomials (in the form such that $$H_n$$ has the coefficient of the highest power term equal to $$2^n$$) can be represented as $$H_n(x) = \frac{2^n}{\sqrt{\pi}} \int_{-\infty}^{\infty} (x+iu)^n e^{-u^2}\; du,$$ for $$n=0, 1, 2, \ldots$$.
He justifies this by noting that the above satisfy, first, the relation $$H_n'(x) = 2nH_{n-1}(x)$$ for all $$n\ge1$$, and, second, that their values agree with the Hermite polynomials at $$x=0$$.

Question: I'm unsure how the above two conditions are sufficient for any family of functions (indexed by $$n=0, 1, 2, \ldots$$) to be equal to the Hermite polynomials. I'm looking for a proof.

We proceed by induction. If $$p_0(x)$$ is a polynomials such that $$p\,_0'(x)=0$$ then $$p_0(x)$$ is a constant and since $$p_0(0)=H_0(0)$$ then $$p_0(x)=H_0(x).$$ Now suppose that $$p_n(x)$$ is a polynomial such that $$p\,_n'(x) = 2nH_{n-1}(x).$$ But also $$H_n(x)$$ satisfies that equation, thus they differ by a constant. The condition that $$p_n(0)=H_n(0)$$ implies that the constant is $$0$$ and thus $$p_n(x)=H_n(x).$$ By induction this is true for all $$n$$.