A limit property of the integral of $\sin(x) / x^{1 + b}$ and little $o$ notation

Question

Suppose $b > 1$ it is clear that the integral $$\int_{y}^1 \frac{\sin(x)}{x^{1 + b} } \ dx \to \infty, \ y \to 0^+ \ .$$ However I am interested in showing that there exists a constant $C$ such that $$ \int_{y}^1 \frac{\sin(x)}{x^{1 + b} } \ dx = C y^{1 - b} + o (1) \ .$$ I am not that used to little $o$-notation but to my understanding I wish to show that there exists a constant $C$ such that $$ \lim_{y \to 0^+} \left( \int_{y}^1 \frac{\sin(x)}{x^{1 + b} } \ dx - C y^{1 - b} \right) = 0 \ .$$ My problem is that this seems to be an "exact" rate of the divergence of the integral so I have no idea how to approach this limit.

Answer 1

Hint. One may use, as $y \to 0^+$, $$ x-\frac{x^3}6\le\sin x\le x-\frac{x^3}6+\frac{x^5}{120},\quad x \in [y,1]. $$ giving $$ \int_{y}^1 \frac{dx}{x^{b} }-\frac16\int_{y}^1 \frac{dx}{x^{b-2} }\le\int_{y}^1 \frac{\sin x}{x^{1 + b} } \ dx\le \int_{y}^1 \frac{dx}{x^{b} }-\frac16\int_{y}^1 \frac{dx}{x^{b-2} }x+\frac1{120}\int_{y}^1 \frac{dx}{x^{b-5} }x $$ or, as $y \to 0^+$, $$ \int_{y}^1 \frac{\sin x}{x^{1 + b} } \ dx= \frac1{b-1}y^{1-b} + o(1). $$