Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 2 / 19

The Leibniz integral rule accounts for both the moving upper endpoint and the integrand's dependence on $a$:For the specified integrand this gives $I^{\prime }(a)=e^{-a^2}-{\int }_0^{a}x{e}^{-ax}dx$. When $a\ne 0$, integration by parts givessoFor the requested independent check, direct integration gives $I(a)=(1-e^{-a^2})/a$ at $a\ne 0$. Its derivative is $[2a^2{e}^{-a^2}-(1-e^{-a^2})]/a^2$, agreeing with the result above. At zero the apparent singularity is removable: $I(a)=a-a^3/2+O(a^5)$, so $\boxed{I(0)=0,I^{\prime }(0)=1}$.

For $0<x<1$,In particular its absolute value is bounded by $|b|x^{\min (b,0)}$, proving convergence when $b>-1$; the apparent endpoint singularity at one is removable, with value $b$. On any compact parameter interval with lower bound $b_0>-1$, the differentiated integrand is $x^b$, dominated by $x^{b_0}$. The dominated convergence theorem therefore justifies differentiation under the integral sign:Hence $J(b)=\ln (b+1)+C$. Since the integrand is identically zero at $b=0$, $J(0)=0$ fixes $C=0$. ThusThis proves the logarithmic integral representation of the natural logarithm, including its parameter range.