Can't solve Improper Integral $\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx$

Whilst checking for the existence of improper integrals, I came across this one: $$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx$$ So in order to check its existence I simply have to see if the limit: $$\lim_{a\to\infty} \int_{0}^{a} \frac{\sqrt{x}\sin(x)}{1+x^2} dx$$ Is a number or not.
However I seem unable to find a way to solve this particular Integral, and neither any online calculator can. I have tried all substitutions that I could think of, as well as partial integration and using any helpful trigonometric identities but they were all in vain.

• Do you simply want to know if the integral converges, or what it converges to? – user455343 Aug 11 '17 at 21:17
• I have tried integration by parts with $(arctan(x))'=\frac{1}{1+x^2}$ but it didn't work, any other which you would suggest? – Konstantinos Zafeiris Aug 11 '17 at 21:18
• @MatthewStonebraker Only knowing whether it converges or not would be fine – Konstantinos Zafeiris Aug 11 '17 at 21:19
• Not sure on accuracy or helpfulness, but Mathematica gives $$\int_0^\infty\frac{\sqrt{x}\sin(x)}{1+x^2}\,dx=\frac{\pi\left(-e^2 \text{erfc}(1) + \text{erfi}(1)+1\right)}{2\sqrt 2 e} \approx 0.608068$$ – Dando18 Aug 11 '17 at 21:24
• Convergence is easy, because the integrand is bounded in absolute value by $x^{-3/2}$ for large $x$. Use comparison. – user296602 Aug 11 '17 at 21:24

If $f$ is continuous in $I=(1, +\infty)$ and if $x^{1+\epsilon} f(x)$ is bounded in $I$ for some $\epsilon > 0$, then $\int_1^\infty f(x) dx$ converges.