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Let denote the fractional part of . Evaluate the integral
By definition it holds that
We note however that
Evaluate the integral
First of all we note that for
Hence by Parseval we get that
Consider the branch of which is defined outside the segment and which coincides with the positive square root for . Let then evaluate the contour integral:
It is a classic case of residue at infinity. Subbing the counterclockwise contour integral rotates the northern pole of the Riemannian sphere to the southern one and the contour integral is transformed to a clockwise one. Hence:
The equality does hold for all if we take the standard branch , otherwise it is not that obvious why this holds, since we are dealing with a multi-valued function.
Lemma 1: Let . It holds that
Lemma 2: Let . It holds that
We begin by squaring the identity of lemma 2. Hence,
Integrating the last equation we get,
Expanding the LHS we get that
Let . Prove that
We are basing the whole solution on the Beta function and its derivative. We recall that
Setting and back at we get that
Differentiating with respect to we get that
where we made use of the reflection formulae of both the Gamma and the digamma function; and .
Now for our integral we have successively: