 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2011/d/d4/c/solution

Put

x = e^{- t}

. Then

\int_{0}^{1} (x lo g x)^{k} d x = (- 1)^{k} \int_{0}^{\infty} t^{k} e^{- (k + 1) t} d t = \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 l o g x} = \sum_{k = 0}^{\infty} \frac{( - x l o g x ) ^{k}}{k !} .

Every term is nonnegative. By the monotone convergence theorem, integration can be interchanged with summation, and hence

\int_{0}^{1} x^{- x} d x = \sum_{k = 0}^{\infty} \frac{1}{( k + 1 ) ^{k + 1}} = j = 1 \sum \infty \frac{1}{j ^{j}} .

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