Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / Section B / 10

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 $x>0$.

The kernel has an integrable singularity at the moving endpoint $t=x$. Direct substitution into the ordinary Leibniz integral rule would produce the undefined boundary value $h(x)/\sqrt{x-x}$ and the nonintegrable differentiated kernel $-h(t)/[2(x-t{)}^{3/2}]$. Those separately divergent expressions cannot be treated as ordinary finite terms. The derivative of an Abel integral is obtained by rewriting the integral before differentiating.

Assume $h$ is continuously differentiable on the interval concerned. For fixed $x>0$, take $u=h(t)$ and $dv=(x-t{)}^{-1/2}dt$, so $v=-2\sqrt{x-t}$. Integration by parts, interpreted first with an endpoint cutoff and then by a limit, givesThe upper endpoint term vanishes, and the lower endpoint supplies the explicit square-root term.

Differentiate the rewritten expression. Its new kernel vanishes at the moving upper endpoint, and its partial derivative is an integrable inverse square root. The derivative of an Abel integral isFor a rigorous route, rewrite $2{\int }_0^x{h}^{\prime }(t)\sqrt{x-t}dt$ as ${\int }_0^x{\int }_0^s{h}^{\prime }(t)/\sqrt{s-t}dtds$ using Fubini's theorem, and apply the fundamental theorem of calculus to the continuous inner integral. The assumption $h(0)\ne 0$ implies $g(x)=2h(0)\sqrt{x}+O(x^{3/2})$, so the extension $g(0)=0$ has no finite right derivative at zero.