1. Nov 18, 2012

mmzaj

i'm trying to prove - or disprove ! - the following :
$$-\ln x\frac{\left \{ x^{1/n} \right \}}{2n^{3}}=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{s}{\left((ns)^{2}-1\right)^{2}} x^{s}ds$$
where $\left \{ x^{1/n} \right \}$ is the fractional part of $x^{1/n}$
for $x\in \mathbb{R}:x>1$, $n\in \mathbb{Z}^{+}$
i'm confused about where to close the contour: to the right , or to the left of the imaginary axis. because the integrand has poles at $n^{-1}$ and $-n^{-1}$. and by the reside theorem, i get two different results!!

Last edited: Nov 18, 2012
2. Nov 18, 2012

jackmell

Maybe you're getting different results because the results are different. Are you sure the contribution along the half-circle arc is zero whether you go around the left half plane or the right half plane?

Just compute it numerically to see if there's a difference, then if there is, try and show it analytically.

3. Nov 18, 2012

mmzaj

maybe i was closing the contour the wrong way!! i didn't use half circles, i closed it using straight segments parallel to the real/imaginary lines. thanks for the remark . however, i still have doubts about the 'steppy' nature of the result - if correct !! - .

4. Nov 18, 2012

haruspex

Is it that you get the opposite sign? If so, maybe you forgot to flip the bounds of the linear integral in order to go anticlockwise around the pole.