Recent content by mmzaj

it boils down to finding the discontinuities of ##\log f(x)##

http://www.wolframalpha.com/input/?i=log(((1%2Bix)%5E(3)-1)%2F((1-ix)%5E(3)-1)*((1-ix)%2F(1%2Bix))%5E(3%2F2))%2B(3*i*arctan(x)-2*i*sum%5Barctan(x%2F(1-e%5E(2*pi*i*k%2F3))),%7Bk,1,2%7D%5D)-i*pi there you go

where should i expect the jumps to happen ? that's where i am stuck. and we can just forget about the graphical discrepancy and correct my analytic calculation.

have you ever used WF ? it doesn't return results for general n ! and posting one example won't be of help if it doesn't say where the jumps are ! thanks for the very helpful and constructive post anyways !

graphically, there seems to be a difference between what I've calculated and the plot of ##\log f(x)## by multiples of ##2\pi## . but i am not able to locate the exact locations of the jumps.

We have the rational function : $$f(x)=\frac{(1+ix)^{n}-1}{(1-ix)^{n}-1}\left(\frac{1-ix}{1+ix}\right)^{n/2}\;\;\;,\;\;n\in \mathbb{Z}^{+}$$ It's not hard to prove that ...

We have the integral: I(s)=\int_{0}^{\infty}\log\left(1+\frac{s^{2}}{4\pi^{2}} \log^{2}(1+ix)\right ) e^{-2\pi nx}dx Where s is a complex parameter, and n is a positive integer. Things i tried: Set \log(1+ix)=y , so that I(s)=-i\int_{c}\log\left(1+\frac{s^{2}}{4\pi^{2}}y^{2} \right...

the limit is better stated this way, i guess. 2\lim_{N\rightarrow \infty}\sum_{n=1}^{N} \left[\int_{\log n}^{\log N}\frac{x}{x^{2}+\frac{4\pi^{2}}{s^{2}}}dx-\left(\frac{\log N}{(\log N)^{2}+\frac{4\pi^{2}}{s^{2}}} \right ) \right ]

this is not a homework the third term is outside the sum i tried your suggestion, but wasn't helpful thanks for the remarks though

We have the following limit: \lim _{N\rightarrow \infty}N\log\left(1+\frac{(s\log N)^{2}}{4\pi^{2}} \right )-\sum_{n=1}^{N}\log\left(1+\frac{(s\log n)^{2}}{4\pi^{2}} \right )-N\left(\frac{2\log N}{(\log N)^{2}+\frac{4\pi^{2}}{s^{2}}} \right ) Where is a complex parameter. any thoughts are...

And it doesn't converge . i kinda knew it , thanks . i was trying to evaluate a limiting case of the integral : \int_{0}^{\infty}\frac{d}{dx}\left(\frac{\zeta(x)}{x}+\frac{1}{2x}-\frac{1}{x-1} \right )\ln\left(-e^{-zx}\right)dx For z\in \mathbb{C}

Does this integral converge !? \int_{0}^{\infty}x\frac{d}{dx}\left(\frac{\zeta(x)}{x}+\frac{1}{2x}-\frac{1}{x-1} \right )dx Where \zeta(x) is the Riemann zeta function. Of course, we can integrate by parts to obtain ...

you missed the fact that the inverse of the complex exponential - the complex \log function- is multivalued. namely : \frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}}=-e^{-2\pi i x}=e^{-2\pi i \left(x-1/2\right)}=e^{-2\pi i \left(\left \{ x \right \}-1/2\right)} Where \left \{ x \right \} is the...

The integral above is equivalent to : \int_{1}^{\infty}\left(\frac{1}{2}-x+\left \lfloor x \right \rfloor \right )\left(\frac{1}{x\left(\ln x+z\right)}\right)dx And \int_{1}^{\infty}\sum_{n=1}^{\infty}\frac{\sin(2 \pi n x)}{n\pi}\left(\frac{1}{x\left(\ln x+z\right)}\right)dx

you would think ! but no, it doesn't ...