Someone could help me with this exercise, that is giving me hard times. I've already tried some variable changes, but they didn't work so far.
Consider the following integral: $$\int_{0}^{\sqrt{x}}\frac{\ln(\cos(t))\tan(t)}{t^{\alpha}}\,dt.$$
- for $x \in (0,1)$, find $\alpha \in \mathbb{R}$ s.t. the integral converges
- for such $\alpha$ values find the order of the infinitesimal function: $$F(x)=\int_{0}^{\sqrt{x}}\frac{\ln(\cos(t))\tan(t)}{t^{\alpha}}\,dt.$$
You just need some asymptotic comparisons near $t =0$. Since $\cos(t) \sim 1 -t^2/2$ and $\tan(t) \sim t$ near $t=0$, you have $$f(t) := \frac{\ln(\cos(t))\tan(t)}{t^\alpha} \sim \frac{t\ln(1-t^2/2)}{t^{\alpha}}.$$ Next since $ln(1+X) \sim X$ near $X=0$, you have $\ln(1-t^2/2) \sim -t^2/2$ near $t = 0$ so $$f(t) \sim \frac{t\cdot(-t^2/2)}{t^{\alpha}} = - \frac{1}{2t^{\alpha-3}}.$$ In order for $$\int^1_0 \frac{dt}{t^p}$$ to converge, you need $p < 1$, and thus here we need $\alpha - 3 < 1$ or $\boxed{\alpha < 4}$. The $x$ doesn't matter in the first part. In the second part, integrating, we see that $$F(x) \sim -\int^{\sqrt x}_0 \frac{dt}{2t^{\alpha - 3}} = \frac{1}{2(\alpha-4)t^{\alpha -4}}\bigg]^{t=\sqrt x}_{t=0} = \frac{1}{2(\alpha -4)x^{\alpha/2- 2}}.$$