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!!
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 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.
 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 !! - .

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 Quote by mmzaj 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!!