integral involves positive definite function and Bessel function

Question

Could the following integral be $0$? \begin{eqnarray*} \lim_{x \to 0^+} \int_0^\pi \vartheta^{\alpha+\frac{1}{2}} \frac{J_{\alpha-\frac{1}{2}} ( x \vartheta )}{x^{\alpha-\frac{1}{2}}} g(\vartheta) d \vartheta \end{eqnarray*} where $g(\vartheta)$ is a continuous function on $[0,\pi]$, $J_\alpha (x)$ is the Bessel function, and $$\int_0^\pi \vartheta^{\alpha+\frac{1}{2}} J_{\alpha-\frac{1}{2}} ( x \vartheta ) g(\vartheta) d \vartheta >0$$ for $x>0$.

Answer 1

With your assumption, and with asymptotic behavior as $ x\rightarrow 0$: $$J_a(x)\sim \dfrac{1}{\Gamma(a+1)}\left(\dfrac{x}{2}\right)^a$$ then \begin{eqnarray*} \lim_{x \to 0^+} \int_0^\pi \vartheta^{\alpha+\frac{1}{2}} \frac{J_{\alpha-\frac{1}{2}} ( x \vartheta )}{x^{\alpha-\frac{1}{2}}} g(\vartheta) d \vartheta = \frac{1}{2^{\alpha-\frac{1}{2}}\Gamma(\alpha+\frac{1}{2})}\int_0^\pi \vartheta^{2\alpha} g(\vartheta) d \vartheta \end{eqnarray*} which depends on the last inteegral.