How to find the equivalent?

Question

Sorry for my bad English ... Task: Find the equivalent to the following function $$f(t) = \int\limits_{-t}^{t}{\sqrt[x^2]{\cos(x)}dx}$$ when $t\to 0^+$. I would be happy if you explain how to do it in the general case.

Answer 1

First, it is clear that $$ f(t) = \int_{-t}^t (\cos(x))^{1/x^2}dx = \int_{-t}^t e^{\frac{1}{x^2} \ln \cos(x)}dx $$ is indeed well-defined, since $x\mapsto \frac{1}{x^2} \ln \cos(x)$ is continuous on $(-\tfrac{\pi}{2},\tfrac{\pi}{2})\setminus \{0\}$ and can be extended by continuity at $0$ as $$ \frac{1}{x^2} \ln \cos(x) = \frac{1}{x^2} \ln \left(1-\frac{x^2}{2}+o(x^2)\right) = -\frac{1}{2}+o(1) \xrightarrow[x\to0]{} -\frac{1}{2}\,. \tag{1} $$

In particular, we get that $\lim_{t\to 0^+}f(t) = 0$. Further, since the summand $$\phi(x)=e^{\frac{1}{x^2} \ln \cos(x)}$$ is even, by the fundamental theorem of calculus $$ \frac{f(t)}{t} = \frac{2\int_{0}^t \phi(x)dx}{t} \xrightarrow[t\to0]{} 2 \phi(0) = 2\lim_{t\to 0} \phi(t) = 2e^{-1/2} \tag{2} $$ the last equality by (1).

Therefore, $$ f(t) \operatorname*{\sim}_{t\to 0^+} 2e^{-1/2} t\,. \tag{3} $$