definite integrals - Compute $int_{0}^{infty} frac{sqrt{x}sin(x)}{1+x^2} dx$

How does one compute the following integral?
$$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx$$

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I have tried extending $x$ to the complex plane then evaluating the following contour integral
$$\oint_C \frac{\sqrt{x}e^{ix}}{1+x^2} dx$$
with the contour $C$ running along the whole real axis and then upper semicircle. I obtain
$$\int_{0}^{\infty} \frac{\sqrt{x}\big(\cos(x)+\sin(x)\big)}{1+x^2}\,\mathrm dx=\frac\pi{e\sqrt2}$$
but not the original integral.

Answer

For information :

$\int_{0}^{\infty} \frac{\sqrt{x}\sin(y\:x)}{1+x^2} dx$ is the Fourier Sine Transform of $\frac{\sqrt{x}}{1+x^2}\quad$ In the Harry Bateman's Tables of Integral Transforms an even more general formula can be seen on page 71, Eq.28 , the Fourier Sine Transform of $x^{2\nu}(x^2+a^2)^{-\mu-1}$ :
$$\frac{1}{2}a^{2\nu-2\mu}\frac{\Gamma(1+\nu)\Gamma(\mu-\nu)}{\Gamma(\mu+1)}y \:_1\text{F}_2(\nu+1;\nu+1-\mu,3/2;a^2y^2/4)\:+\:4^{\nu-\mu-1}\sqrt{\pi}\frac{\Gamma(\nu-\mu)}{\Gamma(\mu-\nu+3/2)}y^{2\mu-2\nu+1}\:_1\text{F}_2(\mu+1;\mu-\nu+3/2,\mu-\nu+1;a^2y^2/4)$$
With$\quad y=1\quad;\quad a=1\quad;\quad \nu=1/4\quad;\quad \mu=0\quad\to\quad \int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx =$
$$=\frac{1}{2}\Gamma(5/4)\Gamma(-1/4) \:_1\text{F}_2(5/4;5/4,3/2;1/4)\:+\:4^{-3/4}\sqrt{\pi}\frac{\Gamma(5/4)}{\Gamma(5/4)}\:_1\text{F}_2(1;5/4,3/4;1/4)$$
The Generalized Hypergeometric $_1$F$_2$ function (don't confuse with the well-known 2F1) reduces to functions of lower level in the particular cases :

$\:_1\text{F}_2(5/4;5/4,3/2;1/4)=\sinh(1)$

$\:_1\text{F}_2(1;5/4,3/4;1/4)=\frac{\sqrt{\pi}}{4e}\left(e^2\text{erfi}(1)+\text{erf}(1) \right)$

and after simplification :
$$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx = -\frac{\pi}{\sqrt{2}}\sinh(1)+\frac{\pi}{2\sqrt{2}\:e}\left(e^2\text{erfi}(1)+\text{erf}(1) \right)$$
$$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx =\frac{\pi}{2\sqrt{2}\:e}\left(e^2\text{erfi}(1)+\text{erf}(1) +1-e^2\right)$$
$$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx =\frac{\pi}{2\sqrt{2}\:e}\left(-e^2\text{erfc}(1)+\text{erf}(1) +1\right)$$
For the Hypergeometric $_1$F$_2$ function, see : http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F2/02/

About the functions erf, erfi, erfc, see : http://mathworld.wolfram.com/Erf.html , http://mathworld.wolfram.com/Erfi.html , http://mathworld.wolfram.com/Erfc.html