Who's up to the challenge? 25

Calculus Level 5

∫ 0 1 { 1 x } 3 d x = − H U − M γ + M 1 U 1 ln ⁡ ( S π ) − B ln ⁡ A \int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} ^{ 3 } } dx=-\frac { H }{ U } -Mγ+\frac { { M }_{ 1 } }{ U_1 } \ln { (Sπ) } -B\ln { A }

In the equation above, A A is the Glaisher–Kinkelin constant, all other variables are positive integers, and all the fractions mentioned are coprime.

Find H + U + M + M 1 + U 1 + S + B . H+U+M+{ M }_{ 1 }+{ U }_{ 1 }+S+B.

Note : { x } \{x\} denotes the fractional part of x . x.

This is a part of " Who's up to the challenge? "


The answer is 17.

Hamza A by Hamza A , 19, USA

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1 solution

Mark Hennings
Mar 31, 2016

The integral is ∫ 0 1 { x − 1 } 3 d x = ∫ 1 ∞ { u } 3 u 2 d u = ∑ n = 1 ∞ ∫ 0 1 u 3 ( n + u ) 2 d u = ∫ 0 1 u 3 ψ ′ ( 1 + u ) d u = ψ ( 2 ) − 3 ∫ 0 1 u 2 ψ ( 1 + u ) d u = 1 − γ + 6 ∫ 0 1 u ln ⁡ Γ ( 1 + u ) d u \begin{array}{rcl}\displaystyle \int_0^1 \{x^{-1}\}^3\,dx & = & \displaystyle \int_1^\infty \frac{\{u\}^3}{u^2}\,du \; = \; \sum_{n=1}^\infty \int_0^1 \frac{u^3}{(n+u)^2}\,du \; = \; \int_0^1 u^3 \psi'(1+u)\,du \\ & = & \displaystyle \psi(2) - 3\int_0^1 u^2 \psi(1+u)\,du \; = \; 1 - \gamma + 6\int_0^1 u \ln \Gamma(1+u)\,du \end{array} This last integral can be evaluated in terms of generalized polygamma functions (integrating by parts once more) and then simplified, yielding ∫ 0 1 { x − 1 } 3 d x = − 1 2 − γ + 3 2 ln ⁡ ( 2 π ) − 6 ln ⁡ A , \int_0^1 \{x^{-1}\}^3\,dx \; = \; -\tfrac12 - \gamma + \tfrac32\ln(2\pi) - 6\ln A \;, giving the answer 17 \boxed{17} .

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