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 Continuum sum = Continuum product JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 08/22/2013, 04:01 PM I was thinking about making the transformation from continuum sum to continuum product and encountered the following problem. The continuum sum can be written as a triple integral. If $f(y)$ is defined on the line $\Re(s) = \sigma$ And as its imaginary part goes to infinity it grows slower than $Ce^{-|\Im{s}|\frac{\pi}{2}}$ decays to zero. $\sum_a^b f(y)\, \sigma y = \frac{1}{2 \pi i}\Big[ \frac{1}{\Gamma(s)}\int_0^{\infty}e^{-t} \int_0^{\infty}u^{s-1} \int_{\sigma-i \infty}^{\sigma + i\infty}\Gamma(y)f(y)(t-u)^{-y}\,dydudt \,\,\Big]_{s=a}^{{s=b}$ Now the immediate problem is that I think mellin inversion will not work on functions like the logarithm, or any function that gives complex values to the logarithm. and this is how we go from the continuum sum to the continuum product. Therefore, we'd have to find a more general way of inverting the mellin transform that could work on functions like the logarithm. Or we could do some algebra and prove the result that $\frac{d}{ds} \sum_1^s f(y) \, \sigma y = \sum_1^s \frac{df(y)}{dy}\,\sigma y$ As well as the same result for the integral. With that we can get an expression for the factorial as: $C \Gamma(s) = e^{\int_1^s \psi(t) dt}$ Where psi is the di gamma function and can be shown to equal: $\psi(s) - C = \sum_1^{s-1} \frac{\sigma y}{y}$ « Next Oldest | Next Newest »

 Messages In This Thread Continuum sum = Continuum product - by tommy1729 - 08/21/2013, 11:27 PM RE: Continuum sum = Continuum product - by JmsNxn - 08/22/2013, 04:01 PM

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