Evaluate this hardcore limit

There appears to be a limit as $n$ as a positive integer increases. Unfortunately, the evaluation of the expression that is the result of the integration runs a 32GiB PC running Wolfram Mathematica 12 out of memory somewhere above n greater than one billion. At n equal to 614, portions of the expression underflow (have values too small to represent accurately).

The result of integration with $n$ as a variable is: $\cos \left(\frac{\pi }{2 n}\right) \Gamma \left(\frac{n+1}{n}\right)-\frac{\pi \left(E_{\frac{n-1}{n}}\left(-i \pi ^n\right)+E_{\frac{n-1}{n}}\left(i \pi ^n\right)\right)}{2 n}$

Noting the conjecture that the upper limit of the integration is $\infty$ and not $\pi$ , $\underset{n\to \infty }{\text{lim}}\int_0^{\infty } \cos \left(x^n\right) \, dx \Rightarrow \underset{n\to \infty }{\text{lim}}\cos \left(\frac{\pi }{2 n}\right) \Gamma \left(\frac{n+1}{n}\right) \Rightarrow 1$

I'm not sure if something is missing from the question, but there's just enough there to solve.

Certainly we can't have $x^n$ getting arbitrarily large, otherwise the limit wouldn't exist.

There are two ways for this not to happen: $n$ can be negative, in which case $x^n \to 0$ , and the limit is $\cos0=1$ .

The other way is if $n=0$ , in which case $x^n \to 1$ , and the cosine in the limit is irrational. Since the only option for an answer is an integer, it must be $\boxed1$ .

The limit only exists and equal to $\boxed 1$ when $n=0$ .