08/29/2013, 04:05 PM
Well if one way to continuum sum wasn't enough, I found another one! Let's start:
\( \mathcal{L} f(x) = \frac{1}{\Gamma(x)}\int_0^\infty e^{-t} f(t) t^{x-1}dt \)
\( \mathcal{L}^{-1} f(x) = \frac{e^x}{2 \pi i} \int_{c - i \infty}^{c+ i \infty} \Gamma(t) f(t) x^{-t} dt \)
If \( \frac{\Delta f(x)}{\Delta x} = f(x) - f(x-1) \) then:
\( \frac{\Delta}{\Delta x} \mathcal{L} f(x) = \mathcal{L} \frac{d f(x)}{dx} \)
The method for continuum sum follows from this!
\( \sum_a^b f(y) \, \Delta y = (\mathcal{L} \int \mathcal{L}^{-1} f(x))\,|_{x=a}^{x=b} \)
\( \mathcal{L} f(x) = \frac{1}{\Gamma(x)}\int_0^\infty e^{-t} f(t) t^{x-1}dt \)
\( \mathcal{L}^{-1} f(x) = \frac{e^x}{2 \pi i} \int_{c - i \infty}^{c+ i \infty} \Gamma(t) f(t) x^{-t} dt \)
If \( \frac{\Delta f(x)}{\Delta x} = f(x) - f(x-1) \) then:
\( \frac{\Delta}{\Delta x} \mathcal{L} f(x) = \mathcal{L} \frac{d f(x)}{dx} \)
The method for continuum sum follows from this!
\( \sum_a^b f(y) \, \Delta y = (\mathcal{L} \int \mathcal{L}^{-1} f(x))\,|_{x=a}^{x=b} \)