How does one compute the following integral? $$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx$$
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.
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
Using the Laplace transform of the sine function, we get $$ \begin{align} \int_{0}^{\infty} \frac{\sqrt{x}\sin x}{1+x^{2}} \, \mathrm dx &= \int_{0}^{\infty} \sqrt{x} \sin(x) \int_{0}^{\infty} \sin(t) e^{-xt} \, \mathrm dt \, \mathrm dx \\ &= \int_{0}^{\infty} \sin(t) \int_{0}^{\infty} \sqrt{x} \sin(x) e^{-tx} \, \mathrm dx \, \mathrm dx \\ &= -\Im\int_{0}^{\infty} \sin(t) \int_{0}^{\infty} \sqrt{x} e^{-(t+i)x} \, dx \, \mathrm dt \\ &= -\frac{\sqrt{\pi}}{2} \, \Im \int_{0}^{\infty} \frac{\sin(t)}{(t+i)^{3/2}} \, \mathrm dt \tag{1}\\ &= - \sqrt{\pi} \, \Im \int_{0}^{\infty} \frac{\cos (t)}{\sqrt{t+i}} \, \mathrm dt \tag{2}\\ &=- \frac{\sqrt{\pi}}{2} \, \Im \left(\int_{0}^{\infty} \frac{e^{it}}{\sqrt{t+i}} \, \mathrm dt + \int_{0}^{\infty}\frac{e^{-it}}{\sqrt{t+i}} \, \mathrm dt \right). \end{align} $$
$(1)$ $\mathcal{L} \{\sqrt{x} \} (s) = \frac{\sqrt{\pi}}{2 s^{3/2}}$ if $\Re(s) >0$
$(2)$ Integrate by parts.
Next, using the principal branch of the square root, let's integrate the function $$f(z) = \frac{e^{iz}}{\sqrt{z+i}}, $$ around an infinitely large closed quarter-circle in the first quadrant of the complex plane.
Applying Jordan's lemma, we get
$$ \begin{align} \int_{0}^{\infty} \frac{e^{it}}{\sqrt{t+i}} \, \mathrm dt&=e^{i \pi/4} \int_{0}^{\infty} \frac{e^{-t}}{\sqrt{t+1}} \, \mathrm dt \\ &= 2 e^{i \pi/4} \, e \int_{1}^{\infty} e^{-u^{2}} \, \mathrm du \tag{3} \\ &= 2 e^{i \pi/4} e \, \frac{\sqrt{\pi}}{2} \operatorname{erfc} (1) \\ &= e^{i \pi/4} e \sqrt{\pi}\operatorname{erfc} (1). \end{align}$$
And by integrating the function $$g(z) = \frac{e^{-iz}}{\sqrt{z+i}} $$ around an infinitely large closed quarter-circle in the fourth quadrant of the complex plane (indented at $z=-i$), we get $$ \begin{align} \int_{0}^{\infty}\frac{e^{-it}}{\sqrt{t+i}} \, \mathrm dt &= \int_{0}^{1} \frac{e^{-t}}{\sqrt{(1-t)e^{ i \pi /2}}} \, (-i \, \mathrm dt) + \int_{1}^{\infty} \frac{e^{-t}}{\sqrt{(t-1)e^{-i \pi /2} }} \, (-i \mathrm \, dt) \\ &= -e^{i \pi/4}\int_{0}^{1} \frac{e^{-t}}{\sqrt{1-t}} \, \mathrm dt + e^{ - i \pi /4} \int_{1}^{\infty} \frac{e^{-t}}{\sqrt{t-1}} \, \mathrm dt \\ &= -\frac{2 e^{ i \pi /4}}{e} \int_{0}^{1} e^{v^{2}} \, \mathrm dv + \frac{2e^{ - i \pi/4}}{e} \int_{0}^{\infty} e^{-w^{2}} \, \mathrm dw \tag{4} \\ &= -\frac{2e^{ i \pi /4}}{e} \frac{\sqrt{\pi}}{2} \operatorname{erfi}(1) + \frac{2 e^{ - i \pi /4}}{e}\frac{\sqrt{\pi}}{2} \\ &= \frac{\sqrt{\pi}}{e} \left(-e^{ i \pi /4} \operatorname{erfi}(1) + e^{-i \pi /4} \right). \end{align}$$
$(3)$ Let $u^{2} = t+1$.
$(4)$ Let $v^{2} = 1-t$ and let $w^{2} = t-1$.
Putting everything together, we have
$$ \begin{align} \int_{0}^{\infty} \frac{\sqrt{x}\sin x}{1+x^{2}} \, \mathrm dx &= - \frac{\sqrt{\pi}}{2} \Im \left(e^{i \pi/4} e \sqrt{\pi}\operatorname{erfc}(1) + \frac{\sqrt{\pi}}{e} \left(-e^{ i \pi /4} \operatorname{erfi}(1) + e^{-i \pi /4} \right) \right)\\ &= - \frac{\sqrt{\pi}}{2} \left(\frac{e \sqrt{\pi}}{\sqrt{2}} \operatorname{erfc}(1) - \frac{\sqrt{\pi} }{\sqrt{2}e} \operatorname{erfi}(1) - \frac{\sqrt{\pi}}{\sqrt{2}e} \right) \\ &= \frac{\pi}{2 \sqrt{2}e} \left(1+ \operatorname{\color{red}{erfi}}(1) - e^{2} \operatorname{erfc}(1) \right). \end{align}$$
My goal is to prove $$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx =\frac{\pi}{2\sqrt{2}\:e}\left(-e^2\text{erfc}(1)+\text{erf}(1) +1\right)$$ as stated in @JJacquelin's answer. My following attempt falls short of that. But it does provide an equivalent integral with much faster convergence.
Take the contour integral $$2\oint_C \frac{z^2e^{iz^2}}{1+z^4} dz$$ with contour $C$ run along the real axis from $0$ to $R$, trace the circle $Re^{i\theta}$ with $\theta$ running from $0$ to $\frac\pi 4$, then roll back to the origin along $xe^{i\frac\pi 4}$ with $x$ running from $R$ to $0$ while circumventing around the point $e^{i\frac\pi 4}$ clockwise with radius $\delta>0$. The integral on the octal circle vanishes as $R\to\infty$. We have $$\oint_C \frac{z^2e^{iz^2}}{1+z^4} dz=\int_{0}^{\infty} \frac{x^2e^{ix^2}}{1+x^4} dx-e^{i\frac34\pi}\int_0^\infty \frac{x^2e^{-x^2}}{1-x^4}dx+\cdots$$
(to be continued)