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. 
 A: Hint
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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{\int_{0}^{\infty}{\root{x}\sin\pars{x} \over 1 + x^{2}}\,\dd x}:\ {\Large ?}}$.

The integration is "closed" along a
$\ds{45^{\large\circ}}$-pizza slice contour in the complex plane first quadrant. Namely, 
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}
{\root{x}\sin\pars{x} \over 1 + x^{2}}\,\dd x}
\,\,\,\stackrel{x\ \mapsto\ x^{2}}{=}\,\,\,
2\int_{0}^{\infty}
{x^{2}\sin\pars{x^{2}} \over 1 + x^{4}}\,\dd x
\\[5mm] = &\
2\,\Im\int_{0}^{\infty}{x^{2}\expo{\ic x^{2}} \over 1 + x^{4}}\,\dd x
\\[5mm] = &
\lim_{\epsilon \to 0^{\large +}}\left[-2\,\Im\int_{\infty}^{1 + \epsilon}{r^{2}\,\ic\,\exp\pars{-r^{2}}
\over 1 - r^{4}}\,\expo{\ic\pi/4}\,\dd r\right.
\\[2mm] &\ \phantom{\lim_{\epsilon \to 0^{\large +}}\left[\right.}
- 2\,\Im\int_{\pi/4}^{-3\pi/4}{\ic\expo{-1} \over
1 - \pars{\expo{\ic\pi/4}
+ \epsilon\expo{\ic\theta}}^{4}}\,\epsilon\expo{\ic\theta}\ic
\,\dd\theta
\\[2mm] & \phantom{\lim_{\epsilon \to 0^{\large +}}\left[\right.}
\left. -2\,\Im\int_{1 - \epsilon}^{0}{r^{2}\,\ic\,\exp\pars{-r^{2}}
\over 1 - r^{4}}\,\expo{\ic\pi/4}\,\dd r\right]
\\[5mm] = &\
2\mrm{P.V.}\Im\int_{0}^{\infty}{r^{2}\,\ic\,\exp\pars{-r^{2}}
\over 1 - r^{4}}\,\expo{\ic\pi/4}\,\dd r
\\[2mm] &\ +
2\lim_{\epsilon \to 0^{+}}\Im\int_{-3\pi/4}^{\pi/4}
{\ic\,\expo{-1} \over
-4\expo{-\ic\pi/4}\epsilon\expo{\ic\theta}}\,\epsilon\expo{\ic\theta}\ic\,\dd\theta
\\[5mm] = &\
\root{2}\mrm{P.V.}\int_{0}^{\infty}
{r^{2}\exp\pars{-r^{2}} \over 1 - r^{4}}\,\dd r +
{1 \over 4}\,\root{2}\pi\expo{-1}
\\[5mm] = &\
{\root{2} \over 2}\,\
\underbrace{\mrm{P.V.}\int_{0}^{\infty}
{\exp\pars{-r^{2}} \over 1 - r^{2}}\,\dd r}
_{\ds{\pi\,\mrm{erfi}\pars{1} \over 2\expo{}}}
- {\root{2} \over 2}\
\underbrace{\int_{0}^{\infty}{\exp\pars{-r^{2}} \over
1 + r^{2}}\,\dd r}_{\ds{{\pi \over 2}\expo{}\,\mrm{erfc}\pars{1}}}\
\\ &\
+ {1 \over 4}\,\root{2}\pi\expo{-1}
\\[5mm] = &\
\bbx{{\root{2} \over 4}\,\pi\expo{-1}\bracks{\mrm{erfi}\pars{1} + 1}
-{\root{2} \over 4}\,\pi\expo{}\,\mrm{erfc}\pars{1}}\
\approx\ 0.6081 \\ &
\end{align}
$\ds{\mrm{erfi}\ \mbox{and}\ \mrm{erfc}}$ are
Error Functions.
