I am reading this mathematical paper, and it states the following on the second page:
"This is the classical solution if we note that
$$\frac{1}{\pi} \int_0^\infty\ \frac{1}{\alpha} e^{-\alpha t}\sin\left(\sqrt{\alpha}y\right)d\alpha = \operatorname{erf}\left(\frac{y}{2\sqrt{t}}\right)."$$
How is it possible to derive the RHS starting with the LHS? As I cannot seem to find anything that helps with this.
Recalling that $\int_0^{\infty} e^{-x^2} \cos( a x) \ \mathrm{d}x = \frac{\sqrt{\pi}}{2} e^{-\frac{a^2}{4} } $, we get
\begin{align} \frac{2}{\pi} \int_0^\infty e^{-\left(\frac{x}{2\xi} \right)^2}\frac{\sin(x)}{x} \, \mathrm{d}x & \overset{\color{blue}{x/(2\xi) \to x}}{=} \frac{2}{\pi }\int_{0}^{\infty}e^{-x^2}\frac{\sin(2\xi x)}{ x} \,\mathrm{d}x\\ & =\frac{2}{\pi }\int_{0}^{\infty}e^{-x^2}\int_{0}^{\xi}2 \cos(2 tx)\,\mathrm{d}t\, \mathrm{d}x\\ & =\frac{4}{\pi}\int_{0}^{\xi}\int_{0}^{\infty}e^{-x^2}\cos(2tx)\mathrm{d}x\, \mathrm{d}t\\ & =\frac{4}{\pi}\int_{0}^{\xi} \frac{\sqrt{\pi}}{2} e^{-t^2}\,\mathrm{d}t\\ & = \frac{2}{\sqrt{\pi}}\int_{0}^{\xi} e^{-t^2}\,\mathrm{d}t\\ & = \mathrm{erf}(\xi) \end{align} where on the last step we used that $\mathrm{erf}(z) =\frac{2}{\sqrt{\pi}} \int_{0}^{z}e^{-t^2}\mathrm{d}t$ by definition of the error function. We can thus conclude that
$$ \frac{2}{\pi} \int_0^\infty e^{-\left(\frac{x}{2\xi} \right)^2}\frac{\sin(x)}{x} \, \mathrm{d}x = \mathrm{erf}(\xi) \qquad \text{for} \quad \xi \in \mathbb{R}$$
Applying this to your question, we see that $\require{cancel}$ \begin{align*} \frac{1}{\pi} \int_{0}^{\infty} \frac{1}{\alpha}e^{-\alpha t}\sin(y\sqrt{\alpha})\mathrm{d}\alpha &\overset{\color{blue}{x = y\sqrt{\alpha}}}{=} \frac{1}{\pi} \int_{0}^{\infty} \frac{\cancel{y^2}}{x^{\cancel{2}}} e^{-\frac{x^2}{y^2}t} \sin(x) \, \frac{2\cancel{x} \, \mathrm{d}x}{\cancel{y^2}}\\ & = \frac{2}{\pi}\int_{0}^{\infty} e^{-\left(\frac{x}{y}\sqrt{t}\right)^2} \frac{\sin(x)}{x}\mathrm{d}x\\ & = \frac{2}{\pi} \int_0^{\infty}e^{-\left(\frac{x}{2\left(\color{green}{\frac{y}{2\sqrt{t}}}\right)}\right)^2}\frac{\sin(x)}{x}\mathrm{d}x\\ & = \mathrm{erf}\left(\color{green}{\frac{y}{2\sqrt{t}}}\right) \end{align*} where we used the identity previously proven with $\xi = \frac{y}{2\sqrt{t}}$.
