Evaluate the $I=\frac{1}{\pi}\int_0^{\infty}\frac{e^{-xt}\sin (a\sqrt{x})}{x}\,\mathrm dx$

Question

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.

Answer 1

$$ I=\frac{1}{\pi}\int_0^{\infty}\frac{e^{-xt}\sin (a\sqrt{x})}{x}\,\mathrm dx $$ $$ \frac{dI}{da} = \frac{1}{\pi}\int_0^{\infty}\frac{e^{-xt}\cos (a\sqrt{x})}{\sqrt{x}}\,\mathrm dx \\=\frac{1}{\sqrt{\pi t}}e^{-\frac{a^2}{4t}} $$ (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 $$ \\I= \erf { (\frac {a} {2 \sqrt{t}})} $$

Answer 2

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: $$ I = \color{red}{\operatorname{Erf}\frac{a}{2\sqrt{t}}}.$$