# Can you prove a definite integral has no closed form?

It is a well known fact that some functions posses no closed form antiderivative yet still they have definite integrals that have a closed form. A classic example is the Gaussian integral $$\int_{-\infty}^\infty e^{-x^2}\mathrm dx=\sqrt\pi$$ Now look at the functions that do not possess anitderivatives in a closed form. We do this because finding the antiderivative is the most trivial way of evaluating a definite integral. Is it possible to prove that a functions definite integral simply doesn't exist in a closed form?

I imagine to do such a thing one would need to look at the nature of the integrand with respect to many different techniques for evaluating definite integrals. It looks to me that this is an extremely tedious and difficult task however is it possible?

This is similar in a sense to proving that some polynomial equations cannot have a closed form by radicals however radicals and functions considered to be "closed form" are radically different. This means that the mechanics of proof may not be very well understood.

I am specifically talking about Wikipedia's definition for closed form. And if possible the extra work that would be needed if we want to allow special functions. Beta, Gamma, Erf, PolyGamma etc.