I found an integral that could solve p vs np. To calculate the max of a function you have to calculate the L$\infty$ norm of the function, which is $\lim_{a \to \infty}(\int_{-\infty}^{\infty}f(x)^a)^{\frac{1}{a}}$. If you can compute it in closed form for a polynomial then p=np, but it can't be computed because it has an infinite number of terms. But e^f(x) has the same maximum as f(x). The L$\infty$ norm of e^f(x) involves computing the antiderivative of (e^f(x))^k, but it simplifies to $\lim_{a \to \infty} (\int_{-\infty}^{\infty} (e^{a})^{f(x)})$ which is $\lim_{a \to \infty} (\int_{-\infty}^{\infty} a^{f(x)})$. This can be integrated unlike f(x)^a. If you can find the general antiderivative for an nth degree polynomial, then p=np.