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Calculus Level 5

0 2016 H x ( 2 ) d x \large \int_0^{2016} H_x^{(2)} \, dx

Let H n ( 2 ) { H }_{ n }^{ (2) } be the generalized harmonic number of order n n of 2, H n ( 2 ) = k = 1 n 1 k 2 \displaystyle H_n ^{(2)} = \sum_{k=1}^n \dfrac 1{k^2} .

If the integral above is equal to a ζ ( b ) H c a\zeta(b) - H_c , where a , b a,b and c c are positive integers, find a + b + c a+b+c .

Notations :

  • ζ ( ) \zeta(\cdot) denotes the Riemann zeta function .

  • H n H_n denotes the n th n^\text{th} harmonic number , H n = 1 + 1 2 + 1 3 + + 1 n H_n = 1 + \dfrac12 + \dfrac13 + \cdots + \dfrac1n .


The answer is 4034.

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1 solution

By the identity 0 a H x ( 2 ) d x = a ζ ( 2 ) H a a = c = 2016 , b = 2 a + b + c = 4034 \displaystyle \int_0^a H_x^{(2)} dx = a\zeta (2) - H_a \implies a = c = 2016, \, b = 2 \implies a + b + c = \boxed{4034} .

Is there a way to prove this identity?

adam madni - 11 months ago

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There should be a proof but I don't know it.

Chew-Seong Cheong - 10 months, 4 weeks ago

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