0
$\begingroup$

im trying to solve the following by the limit comparison theorem.

Problem: $\int_{0}^{1} \frac{\sin(x)}{x^{3/2}(1-x)^{2/3}}dx$ is convergent or divergent?

since its a type II with 2 indeterminations I can split them and test them individually. Now, by the limit comparison test its easy to show that $\int_{1/2}^{1} \frac{\sin(x)}{x^{3/2}(1-x)^{2/3}}\,dx$ is convergent cause we can compare with $\int_{1/2}^{1} \frac{1}{(1-x)^{2/3}}\,dx$ which is convergent and $\lim_{x \to 1} \frac{\tfrac{\sin(x)}{x^{3/2}(1-x)^{2/3}}}{\tfrac{1}{(1-x)^{2/3}}}=\frac{\sin(1)}{x^{3/2}}=1$ then $\int_{1/2}^{1} \frac{\sin(x)}{x^{3/2}(1-x)^{2/3}}\,dx$ converges.

Now, for the other one, $\int_{0}^{1/2} \frac{\sin(x)}{x^{3/2}(1-x)^{2/3}}\,dx$, I know I can pick $g(x)=\frac{1}{x^{3/2}}$ and it will diverge, but if I take $h(x)=\frac{1}{x^{1/2}}$ it will converge. Which one should I pick and why?

$\endgroup$
  • $\begingroup$ Do you know the value of $\lim_{x \rightarrow 0} \frac{\sin x}{x}$? $\endgroup$ – Eric Towers Aug 13 at 16:01
  • $\begingroup$ Near $x=0$ the integrand is $\sim x^{-1/2}$. $\endgroup$ – Lord Shark the Unknown Aug 13 at 16:03
  • 1
    $\begingroup$ Taking your $g$, the ratio limit is $0$ -- which is inconclusive as the integral of $g$ over $(0,1/2]$ diverges. The better choice is $h$. $\endgroup$ – RRL Aug 13 at 16:05
  • $\begingroup$ MMA gives the following result: $$\frac{\sqrt{\pi } \Gamma \left(\frac{1}{3}\right) \, _2F_3\left(\frac{1}{4},\frac{3}{4};\frac{5}{12}, \frac{11}{12},\frac{3}{2};-\frac{1}{4}\right)}{\Gamma \left(\frac{5}{6}\right)}$$ $\endgroup$ – Dr. Sonnhard Graubner Aug 13 at 16:45
  • $\begingroup$ Which is the Taylor expansion of $\sin x\ /\ (1-x)^{2/3}$ near zero?! $\endgroup$ – dan_fulea Aug 13 at 16:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.