= Solution
Use the <power series> of the <exponential function>:
$$
\frac{e^{-x^2}}{\sqrt x}=\sum_{n=0}^\infty a_n(x),\qquad
a_n(x)=\frac{(-1)^n x^{2n-1/2}}{n!},\quad0<x<1.
$$
Each term is a <Lebesgue integrable function>, and
$$
\int_0^1|a_n(x)|\,dx
=\frac{1}{n!(2n+1/2)}=\frac{2}{(4n+1)n!}.
$$
The sum of these nonnegative integrals is at most $2\sum_{n\geq0}1/n!=2e<\infty$. Applying <integration of a series with summable L1 norms>, just proved in part (ii), therefore justifies exchanging the series and the integral:
$$
\boxed{\int_0^1\frac{e^{-x^2}}{\sqrt x}\,dx
=2\sum_{n=0}^\infty\frac{(-1)^n}{(4n+1)n!}.}
$$
The absolute-integral bound is the justification; pointwise convergence of the exponential series alone would not suffice.
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