How can I calculate: $\lim \limits_{x \to 0^+} x \int_{x}^{1} $ $\frac{cost}{t^{\alpha}}dt$ for rach $\alpha >0$, I tried to think about this as an improper integral and substituting $k=\frac{1}{x}$, and then getting something like this:
$\lim \limits_{k \to \infty^+} \frac{1}{k} \int_{\frac{1}{k}}^{1} $ $\frac{cost}{t^{\alpha}}dt$, but Im not sure Im in the right direction, because I didn't know how to continue from here... any kind of help would be appreciable.
On
We have
$\vert \frac{cost}{t^{\alpha}}\vert \leq \frac{1}{t^{\alpha}}$ on $(0,1]$,
which implies that, if $0<\alpha < 2$
$x\int _{x}^{1}\frac{cost}{t^{\alpha}}dt\leq x-x\cdot \frac{x^{1-\alpha }}{1-\alpha }=x-\frac{x^{2-\alpha }}{1-\alpha }\rightarrow 0$ as $x\rightarrow 0$
On the other hand, as soon as $0<t<\pi /3$, $\frac{cost}{t^{\alpha }}\geq \frac{1}{2t^{\alpha }}$ and so in this case, if $\alpha > 2$
$x\int _{x}^{1}\frac{cost}{t^{\alpha}}dt\geq \frac{x}{2}\int _{x}^{1}\frac{1}{t^{\alpha}}dt=\frac{x}{2}-\frac{x}{2}\cdot \frac{x^{1-\alpha }}{1-\alpha }=\frac{x}{2}-\frac{x^{2-\alpha }}{2-2\alpha }\rightarrow \infty $ as $x\rightarrow 0$
Finally, if $\alpha =2$ we have, using the Maclaurin series for cosine,
$x\int_{x}^{1}\frac{1-\frac{1}{2}t^{2}+\frac{1}{24}t^{4}+O(t^{6})}{t^{2}}dt=x\left [ (-1+\frac{1}{x})+(\frac{1}{2}-\frac{1}{2}x)-(\frac{1}{72}-\frac{1}{72}x^{3})+O(x^{5}) \right ]\rightarrow 1$ as $x\rightarrow 0$
Let $F(x)=\int_x^1 \cos t/t^\alpha\,dt$. Then $$xF(x)=\frac{F(x)}{1/x}.$$
Note that $F'(x)=-\cos x/x^\alpha$ and use L'Hospital's Rule.