I like integrals 2

Calculus Level 5

∫ 0 1 ln ⁡ ( 1 − x ) e x d x = − A B e ∑ n = 1 ∞ 1 n ⋅ n ! \large \int_0^1 \dfrac{\ln(1-x) }{e^x} \, dx = - \dfrac A{B e} \sum_{n=1}^\infty \dfrac1{n \cdot n!}

If the equation above holds true for positive integers A A and B B , where A A and B B are coprime positive integers, find A + B A+B .


The answer is 2.

Hamza A by Hamza A , 19, USA

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1 solution

using substitution, ∫ 0 1 log ⁡ ( 1 − x ) e − x d x = e − 1 ∫ 0 1 log ⁡ ( x ) e x d x \int_0^1 \log(1-x) e^{-x} dx = e^{-1} \int_0^1 \log(x) e^x dx using integration by parts, ∫ 0 1 log ⁡ ( x ) e x d x = lim ⁡ y → 0 [ ∫ y 1 log ⁡ ( x ) e x d x ] = lim ⁡ y → 0 [ − log ⁡ ( y ) e y − ∫ y 1 e x / x d x ] = lim ⁡ y → 0 [ − log ⁡ ( y ) ( e y − 1 ) − ∫ y 1 ( e x − 1 ) / x d x ] = lim ⁡ y → 0 [ − log ⁡ ( y ) ( e y − 1 ) ] − ∫ 0 1 ( e x − 1 ) / x d x = − ∫ 0 1 ( e x − 1 ) / x d x \int_0^1 \log(x) e^x dx = \lim_{y \rightarrow 0} [\int_y^1 \log(x) e^x dx] = \lim_{y \rightarrow 0} [-\log(y) e^y - \int_y^1 e^x / x dx] = \lim_{y \rightarrow 0} [-\log(y) (e^y-1) - \int_y^1 (e^x-1) / x dx] = \lim_{y \rightarrow 0} [-\log(y) (e^y-1)] - \int_0^1 (e^x-1) / x dx = - \int_0^1 (e^x-1) / x dx and note that ∫ 0 1 ( e x − 1 ) / x = ∑ n = 1 ∞ 1 n ⋅ n ! \int_0^1 (e^x-1) / x = \sum_{n=1}^{\infty} \frac{1}{n\cdot n!} so A = B = 1 A=B=1

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