Hello everyone,i'm trying to solve this problem:
For what values $a$ and $b$ is $$ \int_{\frac{1}{\pi}}^{\infty} x^{a}[\sin\frac{1}{x}]^{b}dx $$ convergent???
So i tried like this:
using the $ x=1/t$, integral becomes $$ \int_{0}^{\pi} \frac{[\sin(t)]^{b}}{t^{a+2}}dt $$
However,$[\sin(t)]^{b}$ ~ $t^b$,as t approaches $0$,so sub-integral function behaves like $t^{b-a-2}$,when $t \rightarrow 0$ ,which converges for $a+2-b<1$,or $a<b-1$ using the p-test.
On the other hand,at $x=\pi$,i'm not sure what's happening.It should be the similar to previous one,using the $\pi-t=u$,$u \rightarrow 0$,but again i'm not sure,and i would be very grateful if someone provide me a hint.
This should be solutions for positive $a$ and $b$.Now if $b<0$,i get $$ \int_{\frac{1}{\pi}}^{\infty} \frac{x^{a}}{[\sin\frac{1}{x}]^{b}}dx = \int_{0}^{\pi} \frac{1}{t^{a+2}[\sin(t)]^{b}}dt $$ using the very same change of variable,and at $0$ should converges for $a<-b-1$.
And this is what i got so far.I should also do for $a<0$ and both negative,but i thought to ask you first is this ok so far?thanks in advance
When $t$ is close to $\pi$ the function $\frac{1}{t^{a+2}}$ behaves like a constant, while the function $$ (\sin t)^b = \left(\sin(\pi-t)\right)^b $$ behaves like $(\pi-t)^b$, hence the conditions for the integrability of $\frac{(\sin x)^b}{x^{a+2}}$ over $(0,\pi)$ are $b>(a+1)$ (like you correctly stated) and $b>-1$, or just: $$ b>\max(a+1,-1).$$