∫ 0 1 { 2 0 1 7 2 0 1 7 x 1 } d x
If the closed form of the integral above can be represented as b 1 − d e ζ ( c ) , where a and b are co-prime integers, then evaluate b + c + d + e .
Notations
Inspiration Tapas Mazumdar . Try this first.
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@Mark Hennings sir . Can we both make a wiki on fractional integration?
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∫ 0 1 { 2 0 1 7 x 2 0 1 7 1 1 } d x = ∫ 0 1 2 0 1 7 x 2 0 1 7 1 1 d x − ∫ 0 1 ⌊ 2 0 1 7 x 2 0 1 7 1 1 ⌋ d x =
The below summation is created from seeing where the floor function increments, with the integral interval preserved.
2 0 1 6 1 − n = 1 ∑ ∞ ∫ ( n ∗ 2 0 1 7 ) − 2 0 1 7 ( ( n + 1 ) ∗ 2 0 1 7 ) − 2 0 1 7 n d x =
2 0 1 6 1 − 2 0 1 7 2 0 1 7 1 n = 1 ∑ ∞ n ∗ ( ( n + 1 ) 2 0 1 7 1 − n 2 0 1 7 1 ) = 2 0 1 6 1 − 2 0 1 7 2 0 1 7 ζ ( 2 0 1 7 )
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As shown in the inspiration question, we have ∫ 0 1 { n x 1 } d x = N ∫ 1 ∞ { t } t − N − 1 d t = N − 1 N − ζ ( N ) In this problem, the substitutions y = N N x and t = y − 1 give ∫ 0 1 { N N x 1 } d x = N N − 1 1 ∫ 0 N { y − 1 } y N − 1 d y = N N − 1 1 ∫ N 1 ∞ { t } t − N − 1 d t and hence the integral is equal to N N − 1 1 ( ∫ N 1 1 { t } t − N − 1 d t + N − 1 1 − N ζ ( N ) ) = N N − 1 1 ( ∫ N 1 1 t − N d t + N − 1 1 − N ζ ( N ) ) which evaluates as N − 1 1 − N N ζ ( N ) With N = 2 0 1 7 , this makes the answer 2 0 1 6 + 2 0 1 7 + 2 0 1 7 + 2 0 1 7 = 8 0 6 7 .