Integration $\frac{\sin(x)p(x)}{q(x)}$

Question

Can we integrate this function: $$\int^{\infty}_{0} \frac{x^2 \sin(xr)}{x^2+a^2-ca^2x}\,dx$$ ?

Answer 1

The fraction goes to $1$ at $+\infty$. If it were Improper Riemann integrable, then we would have that $\forall \varepsilon > 0$, $\exists c>$ so that $\lvert \int_a^b f \rvert < \varepsilon \forall c<a<b$. Let $1>\delta>0$ be given.We can find $c$ big enough to have that $$g(x):=\frac{x^2}{x^2+a^2-ca^2x}\geqslant 1-\delta$$ Let $n \in \mathbb{N}$ be given with $2n\pi \geqslant c$. Then, we have that $$\int_{2n\pi}^{2n \pi +\pi} \sin(x)g(x)\mathrm{d}x \geqslant \int_{2n\pi}^{2n \pi+\pi} \sin(x)(1-\delta) \mathrm{d}x=2(1-\delta)$$ But this contradicts the criterion mentioned at the beggining, so it's not integrable.