Oh! very very fascinating!
I'd read an article on the lower bounds in fractional differentiation and I had been wondering if it would affect this proof. It seems it does.
Now it all makes sense.
But still, I do believe that this is still something interesting. I would like to figure out how wolfram got that equation...
By re-substituting the original proof and keeping the lower bound at 0 we get:
\( e^x \cdot \ln(x) + K(0, x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}\psi_0(n+1) \)
if
\( K(t, x) = \frac{d}{dt}\frac{d^t}{dx^t} e^x \neq 0 \), where we take the upper zero bound.
Which, given by your wolfram result, seems to imply:
\( K(0, x) = e^x \cdot \Gamma(0, x) \)
what is the second argument for in this gamma function?
And if I may ask, what occurs with the convergence with this series:
\( \ln(x) + \Gamma(0, x)= \sum_{n=0}^{\infty} x^n (\sum_{k=0}^{n} (-1)^k \frac{\psi_0(n-k)}{k!(n-k)!}) \)
Do we have the same asymptotic development for large x?
If so perhaps we can do something similar to Tommy's 2sinh method?
I'd read an article on the lower bounds in fractional differentiation and I had been wondering if it would affect this proof. It seems it does.
Now it all makes sense.
But still, I do believe that this is still something interesting. I would like to figure out how wolfram got that equation...
By re-substituting the original proof and keeping the lower bound at 0 we get:
\( e^x \cdot \ln(x) + K(0, x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}\psi_0(n+1) \)
if
\( K(t, x) = \frac{d}{dt}\frac{d^t}{dx^t} e^x \neq 0 \), where we take the upper zero bound.
Which, given by your wolfram result, seems to imply:
\( K(0, x) = e^x \cdot \Gamma(0, x) \)
what is the second argument for in this gamma function?
And if I may ask, what occurs with the convergence with this series:
\( \ln(x) + \Gamma(0, x)= \sum_{n=0}^{\infty} x^n (\sum_{k=0}^{n} (-1)^k \frac{\psi_0(n-k)}{k!(n-k)!}) \)
Do we have the same asymptotic development for large x?
If so perhaps we can do something similar to Tommy's 2sinh method?
