# Evaluate $\int {e^{(x!)}}^2dx$

How would I evaluate this integral? $$\int {e^{(x!)}}^2dx$$ Obviously elementary functions will not work. The structure of the integrand looks similar to erfi, but there is an $$x!$$. As someone pointed out in one of my previous questions, the factorial is only defined on integers, so this integral wouldn't be integrable for real numbers. However, if you used a gamma function, would it be possible to integrate this? And if I didn't use a gamma function and instead considered $$x!$$ as a step function over positive integers, would it be possible to integrate this?

• Also, I apologize ahead of time for any confusion caused by my question, as I am just starting AP Calculus BC at school. Aug 2, 2021 at 3:45
• "most" integrals aren't doable in closed form, so usually it isn't worth trying a random integral; wolfram can't find a solution for the gamma variant wolframalpha.com/input/… . In the case of a step function that's a sum which also looks very hard to me Aug 2, 2021 at 3:56
• To continue with Calvin's point, the integrals we give you in class are very special cases that can actually be done in a simple analytic closed form. It's like when you first learn polynomials, we only give you ones that can be factored over the integers, even though most can't.
– Alan
Aug 2, 2021 at 4:14
• Unsurprisingly, wolfram alpha doesn't find a closed form Aug 2, 2021 at 4:15
• @Alan That is so true! We are always taught about the "nice" subset of objects, but usually the vast majority are pathological. Some examples of this: (1) Almost all functions are nowhere continuous. (2) Almost all continuous functions are nowhere differentiable. (3) Almost all functions that obey linearity, $f(x+y) = f(x)+f(y)$, are nowhere continuous. (4) Almost all real numbers are incomputable (this one is the most troubling, to me). Aug 2, 2021 at 4:26