Integral $ \int_{0}^\infty (\cos(4x))^{\sin(x)} \text{ dx} =\mbox{?}$

Question

$$ \int_{0}^\infty (\cos(4x))^{\sin(x)} \text{ dx} =\mbox{?}$$

I've tried $\frac{\pi}{2} -x = t$ (integration using substitution) but had no progress.

Answer 1

There are at least two ways in which this integral does not make sense:

1. $\cos(4x)$ can be negative on the domain of integration. How are you defining raising these negative numbers to arbitrary real powers? Is there an absolute value missing somewhere?

2. The improper integral of a (nonzero) periodic function $f(x)$ never exists. To see this, notice that if the value of the improper integral converged, it must be equal to

   $$\lim_{n\to \infty} n \int_0^{2\pi} f(x) \,dx$$

and so $\int_0^{2\pi} f(x)\,dx = 0$. Therefore $\int_0^t f(x)\,dx$ is itself periodic and there is no chance it converges as $t\to \infty$.