A calculus problem by Hamza A

Calculus Level 3

∫ 0 1 ln ⁡ ( x ) 1 − x d x = − ζ ( a ) \large \displaystyle\int _{ 0 }^{ 1 }{ \frac { \ln { (x) } }{ 1-x } \, dx } =-\zeta(a)

The equation above holds true for some constant a a . Find a a .

Notation : ζ ( ⋅ ) \zeta(\cdot) denotes the Riemann zeta function .


The answer is 2.

Hamza A by Hamza A , 19, USA

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1 solution

I = ∫ 0 1 ln ⁡ x 1 − x d x Let u = 1 − x ⇒ x = 1 − u ⇒ d x = − d u = ∫ 0 1 ln ⁡ ( 1 − u ) u d u By Maclaurin series... = ∫ 0 1 − ∑ k = 1 ∞ u k k u d u = − ∫ 0 1 ∑ k = 1 ∞ ( u k − 1 k ) d u = − ∑ k = 1 ∞ ∫ 0 1 u k − 1 k d u = − ∑ k = 1 ∞ [ u k k 2 ] 0 1 = − ∑ k = 1 ∞ 1 k 2 = − ζ ( 2 ) \begin{aligned} I & = \int_0^1 \frac{\ln x}{1-x} dx \quad \quad \small \color{#3D99F6}{\text{Let }u = 1-x \space \Rightarrow x = 1-u \space \Rightarrow dx = - du} \\ & = \int_0^1 \frac{\color{#3D99F6}{\ln (1-u)}}{u} du \quad \quad \small \color{#3D99F6}{\text{By Maclaurin series...}} \\ & = \int_0^1 \frac{\color{#3D99F6}{\small \displaystyle - \sum_{k=1}^\infty \frac{u^k}{k}}}{u} du \\ & = - \int_0^1 \sum_{k=1}^\infty \left( \frac{u^{k-1}}{k}\right) du \\ & = - \sum_{k=1} ^\infty \int_0^1 \frac{u^{k-1}}{k} du \\ & = - \sum_{k=1}^\infty \left[ \frac{u^{k}}{k^2} \right]_0^1 \\ & = - \sum_{k=1}^\infty \frac{1}{k^2} \\ & = - \zeta(2) \end{aligned}

⇒ a = 2 \Rightarrow a = \boxed{2}

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