# Given $$f(x) = \int_{1}^{\sqrt{x}} \mathrm{e}^{-t^{2}} dt$$ Find $$\int_{0}^{1} \frac{f(x)}{\sqrt{x}} dx$$

This question appeared on Mathematics Stack Exchange. My solution is an alternative to the better solution here.

f(x) = \int\limits_{1}^{\sqrt{x}} \mathrm{e}^{-t^{2}} dt
\label{eq:161108-1}
\tag{1}

\int\limits_{0}^{1} \frac{f(x)}{\sqrt{x}} dx
\label{eq:161108-2}
\tag{2}

We need the following result

\int \mathrm{erf}(x) dx = x\,\mathrm{erf}(x) + \frac{\mathrm{e}^{-x^{2}}}{\sqrt{\pi}}
\label{eq:161108-3}
\tag{3}

Proof: Integrate by parts
\begin{align}
\int \mathrm{erf}(x) dx &= x\,\mathrm{erf}(x) -\frac{2}{\sqrt{\pi}} \int x\,\mathrm{e}^{-x^{2}} dx \\
&= x\,\mathrm{erf}(x) + \frac{\mathrm{e}^{-x^{2}}}{\sqrt{\pi}}
\end{align}
we used the substitution $$u=x^{2}$$.

Now we evaluate equation \eqref{eq:161108-1}

f(x) = \frac{\sqrt{\pi}}{2} \mathrm{erf}(x) \Big|_{1}^{\sqrt{x}} = \frac{\sqrt{\pi}}{2} [\mathrm{erf}(\sqrt{x}) – \mathrm{erf}(1)]
\label{eq:161108-4}
\tag{4}

Substitute equation \eqref{eq:161108-4} into equation \eqref{eq:161108-2} and evaluate the following two integrals.

\begin{align}
I_{1} &= \int\limits_{0}^{1} \frac{\mathrm{erf}(\sqrt{x})}{\sqrt{x}} dx \\
&= 2\int\limits_{0}^{1} \mathrm{erf}(z) dz \\
&= 2\,\mathrm{erf}(1) + \frac{2}{\mathrm{e}\sqrt{\pi}} – \frac{2}{\sqrt{\pi}}
\end{align}
we used the substitution $$z=\sqrt{x}$$.

I_{2} = \int\limits_{0}^{1} \frac{1}{\sqrt{x}} dx = 2

Putting all of the pieces together yields our final result
\begin{align}
\int\limits_{0}^{1} \frac{f(x)}{\sqrt{x}} dx
&= \frac{\sqrt{\pi}}{2} \left(2\,\mathrm{erf}(1) + \frac{2}{\mathrm{e}\sqrt{\pi}} – \frac{2}{\sqrt{\pi}} – 2\,\mathrm{erf}(1) \right) \\
&= \frac{1}{\mathrm{e}} – 1
\end{align}