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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2021

Paper 3

For fixed $x>0$, $x\ne1$,
$$
\int_0^1x^y\,dy=\frac{x-1}{\log x},
$$
with limiting value one at $x=1$. Integrating over $0<x<1$ in the opposite order gives
$$
\int_0^1\int_0^1x^y\,dy\,dx
=\int_0^1\frac{dy}{y+1}
=\log2.
$$
On the other hand, after setting $x=e^{-u}$,
$$
\int_0^1\frac{x-1}{\log x}\,dx
=\int_0^\infty\frac{e^{-u}-e^{-2u}}u\,du.
$$
Consequently
$$
\boxed{
\int_0^\infty\frac{e^{-u}-e^{-2u}}u\,du=\log2}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 3 / 10B / b / Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 3 / 10B / b / Solution

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Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 3 / 10B / b / Solution

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