∫ 0 1 { x 1 } 1 − x x d x = A γ + B
The equation above holds true for integers A and B . Find A + B .
Notations :
{ ⋅ } denotes the fractional part function .
γ denotes the Euler-Mascheroni constant , γ ≈ 0 . 5 7 7 2 .
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How do you know that k = 1 ∑ ∞ k + 1 ζ ( k + 1 ) − 1 = 1 − γ ?
∫ 0 1 { x 1 } 1 − x x d x = n = 1 ∑ ∞ ∫ n + 1 1 n 1 ( x 1 − n ) 1 − x x d x = n = 1 ∑ ∞ ∫ n + 1 1 n 1 ( n − 1 − x n − 1 ) d x
= 2 1 + n = 2 ∑ ∞ ( n ( n 1 − n + 1 1 ) + ( n − 1 ) ( ln ( 1 − n 1 ) − ln ( 1 − n + 1 1 ) ) )
= 2 1 + n = 2 ∑ ∞ ( n + 1 1 + ( n − 1 ) ( ln ( n n − 1 ) − ln ( n + 1 n ) ) )
= 2 1 + n = 2 ∑ ∞ ( n + 1 1 + ( n − 1 ) ( ln ( n − 1 ) − 2 ln ( n ) + ln ( n + 1 ) ) )
= 2 1 + N → ∞ lim n = 2 ∑ N ( n + 1 1 + ( n − 1 ) ( ln ( n − 1 ) − 2 ln ( n ) + ln ( n + 1 ) ) )
= N → ∞ lim n = 1 ∑ N ( n + 1 1 ) + n = 2 ∑ N ( n − 1 ) ln ( n − 1 ) − 2 n = 2 ∑ N ( n − 1 ) ln ( n ) + n = 2 ∑ N ( n − 1 ) ln ( n + 1 )
= N → ∞ lim n = 1 ∑ N ( n + 1 1 ) + n = 1 ∑ N − 1 n ln ( n ) − 2 n = 2 ∑ N ( n − 1 ) ln ( n ) + n = 3 ∑ N + 1 ( n − 2 ) ln ( n )
= N → ∞ lim n = 1 ∑ N ( n + 1 1 ) + n = 3 ∑ N − 1 n ln ( n ) − 2 n = 3 ∑ N − 1 ( n − 1 ) ln ( n ) + n = 3 ∑ N − 1 ( n − 2 ) ln ( n ) − N ln ( N ) + ( N − 1 ) ln ( N + 1 )
= N → ∞ lim n = 1 ∑ N ( n + 1 1 ) − N ln ( N ) + ( N − 1 ) ln ( N + 1 )
= N → ∞ lim n = 2 ∑ N + 1 ( n 1 ) + ln ( N N ( N + 1 ) N − 1 )
= N → ∞ lim − 1 + n = 1 ∑ N + 1 ( n 1 ) + ln ( N N ( N + 1 ) N ) − ln ( N + 1 )
= N → ∞ lim n = 1 ∑ N + 1 ( n 1 ) − ln ( N + 1 ) + N → ∞ lim ln ( ( 1 + N 1 ) N ) − 1
= γ + ln ( e ) − 1 = γ
See my solution...below...
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I = ∫ 0 1 { x 1 } 1 − x x d x
By Maclaurin series I'll write it as: I = k = 1 ∑ ∞ ∫ 0 1 { x 1 } x k d x
Let J = ∫ 0 1 { x 1 } x k d x
Substitute: x 1 = x
J = i = 1 ∑ ∞ ∫ 1 ∞ y k + 2 y − i d y
On solving it, we get: J = k 1 − k + 1 ζ ( k + 1 )
Now, coming back to our main sum: I = k = 1 ∑ ∞ ( k 1 − k + 1 ζ ( k + 1 ) )
I'll rearrange the terms as:
I = k = 1 ∑ ∞ ( k 1 − k + 1 1 ) − k = 1 ∑ ∞ k + 1 ζ ( k + 1 ) − 1
Hence, I = γ