 Solution

For $x>0$ write $f(x)=x^{-a}I(x)$, where $I(x)={\int }_0^x{e}^{-t}{t}^{a-1}dt$. Because $a>0$, the lower endpoint is integrable. At each positive upper endpoint the fundamental theorem of calculus gives $I^{\prime }(x)=e^{-x}{x}^{a-1}$. The product rule therefore yieldsDifferentiating once more and substituting this expression for $f^{\prime }$ givesSubstitution now cancels the $e^{-x}$ terms and the terms proportional to $f$, proving $x{f}^{\prime \prime }+(1+a+x)f^{\prime }+af=0$.

The change of variables $t=xu$ identifies $f$ as the scaled lower incomplete gamma integral ${\int }_0^1{e}^{-xu}{u}^{a-1}du$. It has a smooth extension at zero, with $f(0)=1/a$, $f^{\prime }(0)=-1/(a+1)$ and $f^{\prime \prime }(0)=1/(a+2)$. Thus the apparent singularities in the derivative formulas cancel there; those formulas with denominators $x$ were derived for $x>0$.