Prove $$\mathrm{erf}(x) = \frac{1}{\pi} \int_{0}^{\infty} \mathrm{e}^{-t} \frac{\sin(2x\sqrt{t})}{t} dt$$

How to prove

\mathrm{erf}(x) = \frac{1}{\pi} \int\limits_{0}^{\infty} \mathrm{e}^{-t} \frac{\sin(2x\sqrt{t})}{t} dt

was a question posed at Mathematics Stack Exchange. The poster did not want a solution using contour integration. Here is my solution.

\begin{align}
\tag{1a}
\int\limits_{0}^{\infty} \mathrm{e}^{-t} \frac{\sin(2x\sqrt{t})}{t} dt
&= x \int\limits_{-1}^{1} \int\limits_{0}^{\infty} \mathrm{e}^{-t} \mathrm{e}^{i2x\theta \sqrt{t}} \frac{1}{\sqrt{t}} dt d\theta \\
\tag{1b}
&= 2x \int\limits_{-1}^{1} \int\limits_{0}^{\infty} \mathrm{e}^{-y^{2}+i2x\theta y} dy d\theta \\
\tag{1c}
&= 2x \int\limits_{-1}^{1} \mathrm{e}^{-x^{2} \theta ^{2}} \int\limits_{0}^{\infty} \mathrm{e}^{-(y-ix\theta)^{2}} dy d\theta \\
\end{align}

Notes:

a. Use

\frac{\sin(\phi)}{\phi} = \frac{1}{2}\int\limits_{-1}^{1} \mathrm{e}^{i\phi w} dw

b. $$\sqrt{t} = y$$

c. Complete the square.

Evaluate
\begin{align}
\int \mathrm{e}^{-(y-ix\theta)^{2}} dy &= \int \mathrm{e}^{-z^{2}} dz \\
&= \frac{\sqrt{\pi}}{2} \mathrm{erf}(z) = \frac{\sqrt{\pi}}{2} \mathrm{erf}(y-ix\theta)
\end{align}
here we let $$z = y-ix\theta$$. Applying limits to the integral yields
\begin{align}
\int\limits_{0}^{\infty} \mathrm{e}^{-(y-ix\theta)^{2}} dy
&= \frac{\sqrt{\pi}}{2} \mathrm{erf}(y-ix\theta) \Big|_{0}^{\infty} \\
&= \frac{\sqrt{\pi}}{2} [1 + i \mathrm{erfi}(x\theta)]
\tag{2}
\end{align}

Substituting equation 2 into equation 1c yields two integrals. Dropping constants we have first
\begin{align}
\int\limits_{-1}^{1} \mathrm{e}^{-x^{2} \theta ^{2}} d\theta &= \frac{1}{x} \int \mathrm{e}^{-w^{2}} dw \\
&= \frac{\sqrt{\pi}}{2x} \mathrm{erf}(w) \\
&= \frac{\sqrt{\pi}}{2x} \mathrm{erf}(x\theta) \Big|_{-1}^{1} \\
&= \frac{\sqrt{\pi}}{x} \mathrm{erf}(x)
\end{align}
secondly, we have

\int\limits_{-1}^{1} \mathrm{e}^{-x^{2} \theta ^{2}} \mathrm{erfi}(x\theta) d\theta = 0

noting that the imaginary error function is an odd function.

Putting the pieces together yields
\begin{align}
\int\limits_{0}^{\infty} \mathrm{e}^{-t} \frac{\sin(2x\sqrt{t})}{t} dt
&= 2x \int\limits_{-1}^{1} \mathrm{e}^{-x^{2} \theta ^{2}} \int\limits_{0}^{\infty} \mathrm{e}^{-(y-ix\theta)^{2}} dy d\theta \\
&= 2x \frac{\sqrt{\pi}}{2} \frac{\sqrt{\pi}}{x} \mathrm{erf}(x) \\
&= \pi\, \mathrm{erf}(x)
\end{align}