I started by using the error function:
$$ \begin{align} \text{erf}(y)=\frac{2}{\sqrt{\pi}}\int_{0}^{y} e^{-u^{2}} du, \end{align} $$
where I let $\tau=u^{2}$ then $d\tau=2u\;du$, $du=\frac{d\tau}{2u}=\frac{d\tau}{2\sqrt{\tau}}$. Then substituting and letting $y=\sqrt{t}$:
$$ \begin{align} \text{erf}(\sqrt{t})=\frac{2}{\sqrt{\pi}}\int_{0}^{\sqrt{t}} e^{-\tau} \frac{d\tau}{2\sqrt{\tau}}=\frac{1}{\sqrt{\pi}} \int_{0}^{\sqrt{t}} \frac{e^{-\tau}}{\tau} \;d\tau, \end{align} $$
the only problem is with $\sqrt{t}$ in the upper limit when it should be $t$, any idea how can I arrive to this result?
The substitution $\tau=u^{2}$ maps $[0, y]$ onto $[0, y^2]$, so the boundaries of the integral must be substituted as well: $$ \operatorname{erf}(y)=\frac{2}{\sqrt{\pi}}\int_{0}^{y} e^{-u^{2}} du =\frac{1}{\sqrt{\pi}}\int_{0}^{y^2} \frac{e^{-\tau}}{\sqrt \tau} d\tau $$ Now set $y=\sqrt t$ to get the desired result.