I'm given the integral $$\int_0 ^{+ \infty} \frac{e ^ {-\cos t} \cdot \sin (t ^ \beta)}{t^\alpha} dt \qquad a,b \in \mathbb{R}$$
and I need to test the absolute convergence. I split it in two parts, namely $$\int_0 ^{1} \frac{e ^ {-\cos t} \cdot \sin (t ^ \beta)}{t^\alpha} dt + \int_1 ^{+ \infty} \frac{e ^ {-\cos t} \cdot \sin (t ^ \beta)}{t^\alpha} dt$$ and proved that the second part converge absolutely $\iff \alpha > 0$. So I examined the first part: the integral is bounded by $$ e \int_0 ^{1} \frac{\sin (t ^ \beta)}{t^\alpha} dt$$but now I'm facing some difficulties:
- $\alpha \ge \beta +2 \Rightarrow $ divergence
- how to test if $\alpha < \beta +2 \Rightarrow $ absolute convergence
As @Tom-Tom noticed $$e \int_0 ^{1} \frac{\sin (t ^ \beta)}{t^\alpha} dt < e \int_0 ^{1} \frac{t ^ \beta}{t^\alpha} dt = e \int_0 ^{1} t ^ {\beta - \alpha} dt = \frac{1}{\beta - \alpha} \cdot \left( 1 - \lim_{t \to 0} t ^ {\beta - \alpha + 1} \right)$$ so the integral $e \int_0 ^{1} \frac{\sin (t ^ \beta)}{t^\alpha} dt$ converge $\iff \beta - \alpha + 1 > 0 \iff \alpha < \beta + 1$