Showing Hermite polynomials are orthogonal

I need to show that two Hermite polynomials are orthogonal, but I'm a little confused.

I have: $H_2(x) = 4x^2-2$ and $H_3(x) = 8x^3-12x$

I know I need to integrate $$\int_{-L}^L (4x^2-2)(8x^3-12x) dx=0,$$ because it says I need to show it's orthogonal on $[-L,L]$, where $L$ is a constant.

Do I just pick random values for $L$, or is there some sort of procedure?

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I realize that I might be a bit too late the hero, but I'm leaving this comment that got too long for the comment box as an answer.

What seems to have been forgotten by the OP and the answerers is that for one to speak of orthogonality, one should always specify the associated inner product. For the Hermite polynomials $H_n(x)$, the relevant inner product is

$$\langle f,g \rangle=\int_{-\infty}^\infty f(x)g(x)\color{red}{\exp(-x^2)}\,\mathrm dx$$

Having said this, the general idea in the other two answers is correct: $H_2(x)H_3(x)$ is indeed an odd function, while $\exp(-x^2)$ is even. Their product is odd, and thus $\langle H_2,H_3\rangle$ certainly ought to be zero. As you are now dealing with an improper integral, some more finesse in proving that the integral is sensible is needed; that work is left as an exercise.

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Think of $L$ as an arbitrary, but fixed interval length. These functions are orthogonal regardless of the actual value of $L$, as the straightforward integration demonstrates:

\begin{align} &\int_{-L}^L (4x^2-2)(8x^3-12x) dx\\ &=\left[8 \left(\frac{2 x^6}{3}-2 x^4+\frac{3 x^2}{2}\right)\right]_{-L}^{L}\\ &=\left(8 \left(\frac{2 L^6}{3}-2 L^4+\frac{3 L^2}{2}\right)\right)-\left(8 \left(\frac{2 (-L)^6}{3}-2 (-L)^4+\frac{3 (-L)^2}{2}\right)\right)\\ &=0. \end{align}

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Note that the integrand is an even function--$4x^2-2$--times an odd function--$8x^3-12x$--so is an odd function. This function is a polynomial, and so is continuous. Given any $L>0$, we have that the integral over $[-L,L]$ of an odd function (that's continuous there) is $0$.

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