What is $\lim_{n\to\infty}\int_{0}^\infty \exp(-x^n \arctan(\frac1x)) dx,n>1$

Question

The following integral $\lim_{n\to\infty}\int_{0}^\infty \exp(-x^n \arctan(\frac1x)) dx$ seems converges for all integer $n>1$ according to sum computation that i runs with wolfram alpha such that it's value are decreasing the max is for $ n=2$ close to $1.18$ and for $n=3$ it's close to $1$,$\cdots$ , because we have the composition of $\exp$ and $\tan^{-1}$ is decreasing function , Using the series representation of $\tan^{-1}(\frac1x)$ which is defined as ::$\sum_{k=0}^{\infty}\frac{(-1)^{1+2k}(\frac1x)^{1+2k}}{1+2k}$ and by substitution in the titled integral w'd give a complicated integral which is not easy to get it's closed form , Since this is the case i should ask about:$\lim_{n\to\infty}\int_{0}^\infty \exp(-x^n \arctan(\frac1x)) dx$ for $n >1$ without using it's closed form ?

Answer 1

Split the intergal into integral over $(0,1)$ and $(1,\infty)$. The inetegrand is increasing in the first interval and decreasing in the second. Since the intergal is finite for $n=1$ we can take the limit inside in both cases and the answer is $1$.