How to find all values for $\alpha$ and $\beta$ such that $\int _0^{\infty }f\left(x\right)$ converge

Question

$f(x) = \begin{cases} x^{\alpha }\left(1-cos\left(1-x\right)\right)^{\beta } & \text{if $\;\;\;0<x<1$} \\ \frac{1}{x^{\alpha }+x^{\beta }} & \text{if $\;\;\;1\ge x$} \end{cases}$

I split into two integrals:

$\int _0^1x^{\alpha }\left(1-cos\left(1-x\right)\right)^{\beta }dx\:+\:\int _1^{\infty }\left(\frac{1}{x^{\alpha }+x^{\beta }}\right)dx$

I want to use the comparison test. but I find it hard to choose $g(x)$ that will give me an answer for $\alpha$ and $\beta$

Answer 1

The convergence is obtained for

$$ \alpha>-1 \quad \text{and} \quad \beta>-1/2 \quad \text{and} \quad \max(\alpha,\beta)>1. $$

Hint. Potential issues are as $x \to 0^+$, as $x \to 1^-$ and as $x \to \infty.$

• As $x \to 0^+$, one has $$ f(x)=x^\alpha(1-\cos1)^\beta+O(x^{\alpha+1}) $$ and $ \displaystyle \int_0^\epsilon\!\!f(x)dx$ converges ($0<\epsilon<1$) iff $ \displaystyle \int_0^\epsilon\!\!x^\alpha dx$ converges, that is it converges iff $\alpha>-1.$

• As $x \to 1^-$, one has $$ f(x)=(1-x)^{2\beta}2^{-\beta}+O((x-1)^{2\beta+1}) $$ and $ \displaystyle \int_{1-\epsilon}^1\!\!f(x)dx$ converges ($0<\epsilon<1$) iff $ \displaystyle \int_{1-\epsilon}^1\!\!(1-x)^{2\beta} dx$ converges, that is it converges iff $\beta>-1/2.$

• As $x \to \infty$, one has $$ f(x)=\frac{1}{x^{\max(\alpha,\beta)}}+O\left(\frac{1}{x^{\max(\alpha,\beta)+(\max(\alpha,\beta)-\min(\alpha,\beta))}}\right) $$ and $ \displaystyle \int_1^\infty\!\!f(x)dx$ converges iff $ \displaystyle \int_0^1\!\!\frac{1}{x^{\max(\alpha,\beta)}} dx$ converges, that is it converges iff $\max(\alpha,\beta)>1.$