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Year 2 of the mathematics course of the University of Oxford

Options II course of the University of Oxford

Past exam of the options II course of the University of Oxford

2011

Put $x=e^{-t}$. Then
$$
\int_0^1(x\log x)^k\,dx
=(-1)^k\int_0^\infty t^k e^{-(k+1)t}\,dt
=\boxed{\frac{(-1)^k k!}{(k+1)^{k+1}}},
$$
where the second substitution is $u=(k+1)t$ and the final <integral> is the given <gamma function> value. The formula also holds for $k=0$, with value $1$.

For $0<x<1$, the <exponential function> expansion gives
$$
x^{-x}=e^{-x\log x}=\sum_{k=0}^\infty\frac{(-x\log x)^k}{k!}.
$$
Every term is nonnegative. By the <monotone convergence theorem>, integration can be interchanged with summation, and hence
$$
\int_0^1x^{-x}\,dx
=\sum_{k=0}^\infty\frac1{(k+1)^{k+1}}
=\boxed{\sum_{j=1}^\infty\frac1{j^j}}.
$$
This identity is one form of <Sophomore's dream>.

Solution

Past exam of the options II course of the University of Oxford / 2011 / D / D4 / c / Solution

Past exam of the options II course of the University of Oxford / 2011 / D / D4 / c / Solution

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