Evaluate the $I=\frac{1}{\pi}\int_0^{\infty}\frac{e^{-xt}\sin (a\sqrt{x})}{x}\,\mathrm dx$ I want to evaluate
$$I=\frac{1}{\pi}\int_0^{\infty}\frac{e^{-xt}\sin (a\sqrt{x})}{x}\,\mathrm dx$$
It seems that the solution should be in the form of the error function and also it involves contour integration.
 A: Assuming $a,t\in\mathbb{R}^+$ we have:
$$ I = \frac{2}{\pi}\int_{0}^{+\infty}e^{-tx^2}\sin(ax)\frac{dx}{x}$$
where:
$$\begin{eqnarray*}\frac{\partial I}{\partial a}&=&\frac{2}{\pi}\int_{0}^{+\infty}e^{-tx^2}\cos(ax)\,dx = \frac{2}{\pi}\Re\int_{0}^{+\infty}\exp\left(-tx^2+iax\right)\,dx\\&=&\frac{2}{\pi\sqrt{t}}\Re\int_{0}^{+\infty}\exp\left(-x^2+\frac{ia}{\sqrt{t}}x\right)\,dx=\frac{2e^{-a^2/(4t)}}{\pi\sqrt{t}}\Re\int_{0}^{+\infty}\exp\left(-\left(x-\frac{ia}{2\sqrt{t}}\right)^2\right)\,dx\\&=&\frac{1}{\sqrt{\pi t}}e^{-a^2/(4t)}\end{eqnarray*}$$
hence:
$$ I = \color{red}{\operatorname{Erf}\frac{a}{2\sqrt{t}}}.$$
A: $$
I=\frac{1}{\pi}\int_0^{\infty}\frac{e^{-xt}\sin (a\sqrt{x})}{x}\,\mathrm dx
$$
$$
\frac{dI}{da} = \frac{1}{\pi}\int_0^{\infty}\frac{e^{-xt}\cos (a\sqrt{x})}{\sqrt{x}}\,\mathrm dx
\\=\frac{1}{\sqrt{\pi t}}e^{-\frac{a^2}{4t}}
$$
(Abramowitz & Stegun, p 1026).
Integrating:
$$
\DeclareMathOperator\erf{erf}
I(a,t) = \erf { (\frac {a} {2 \sqrt{t}})} + C(t)
$$
$C(t) = 0$ since $I(0,t) = 0$, and therefore
$$
\\I= \erf { (\frac {a} {2 \sqrt{t}})}
$$
