# Improper integral criterion

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?

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