- #1
mmzaj
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i'm trying to prove - or disprove ! - the following :
[tex] -\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[/tex]
where [itex]\left \{ x^{1/n} \right \} [/itex] is the fractional part of [itex] x^{1/n}[/itex]
for [itex] x\in \mathbb{R}:x>1 [/itex], [itex] n\in \mathbb{Z}^{+}[/itex]
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 [itex]n^{-1} [/itex] and [itex] -n^{-1}[/itex]. and by the reside theorem, i get two different results!
[tex] -\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[/tex]
where [itex]\left \{ x^{1/n} \right \} [/itex] is the fractional part of [itex] x^{1/n}[/itex]
for [itex] x\in \mathbb{R}:x>1 [/itex], [itex] n\in \mathbb{Z}^{+}[/itex]
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 [itex]n^{-1} [/itex] and [itex] -n^{-1}[/itex]. and by the reside theorem, i get two different results!
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