What's the standard way to compute this kind of integral limits?

Question

For $\alpha \in \mathbb{R}$ determine $$\lim_{x\to \infty}\int_{x}^{x+x^{\alpha}}\frac{\arctan t}{\sqrt{t}} \, dt$$

What's the standard way to proceed? If the integral converges in $[a,+\infty)$ it's okay, but what about cases like this one? Thanks in advance.

Answer 1

We have that

$$\frac{\arctan t}{\sqrt{t}}=\frac{\pi}{2\sqrt t}-\frac{1}{\sqrt t}\arctan \frac 1 t= \frac{\pi}{2\sqrt t}-\frac{1}{t\sqrt t} +o\left(\frac{1}{t\sqrt t}\right)$$

and

$$\int_{x}^{x+x^{\alpha}}\frac{\arctan t}{\sqrt{t}}= \pi(\sqrt {x^\alpha+x}-\sqrt x)-\frac{2}{\sqrt {x^\alpha+x}}+\frac{2}{\sqrt {x}} +o\left(\frac{1}{\sqrt {x^\alpha+x}}\right)=\pi\frac{x^\alpha}{\sqrt {x^\alpha+x}+\sqrt x}-\frac{2}{\sqrt {x^\alpha+x}}+\frac{2}{\sqrt {x}} +o\left(\frac{1}{\sqrt {x^\alpha+x}}\right)$$

then

$$\lim_{x\to \infty}\int_{x}^{x+x^{\alpha}}\frac{\arctan t}{\sqrt{t}}=\begin{cases}+\infty \quad \alpha>\frac12\\\\\frac \pi 2 \quad \alpha=\frac12\\\\0\quad \alpha<\frac12\end{cases}$$

Answer 2

Simple estimates on the integral give

$$\tag 1 \arctan x\cdot \frac{1}{\sqrt {x+x^\alpha}}\cdot x^\alpha\le \int_x^{x+x^\alpha}\frac{\arctan t}{\sqrt t}\,dy \le \, \frac{\pi}{2} \frac{1}{\sqrt x}\cdot x^\alpha.$$

Immediately we see from the right side of $(1)$ that if $\alpha <1/2,$ then the limit in question is $0.$ The limits for $\alpha = 1/2$ and $\alpha >1/2$ are $\pi /2$ and $\infty;$ these follow from $(1)$ with a little extra work.