@JmsNxn,
I fed the given series
\( e^x \ln(x) \stackrel{\mathrm{supposedly}}{=} \sum_{n=0}^{\infty} \frac{\psi_0(n + 1)}{n!} x^n \)
to the Wolfram Alpha calculator. It says that it does not give the left-hand side, but instead, it gives
\( e^x (\ln(x) + \Gamma(0, x)) = \sum_{n=0}^{\infty} \frac{\psi_0(n + 1)}{n!} x^n \)
(I have no idea how it managed to derive that formula -- any suggestions?)
But this shows the problem. There is an extra term present, which is the upper incomplete gamma function. It decays to 0 as \( x \rightarrow \infty \).
So, your series is asymptotic to, but not equal to, \( e^x \ln(x) \). I.e. the first "equation" is really
\( e^x \ln(x) \sim \sum_{n=0}^{\infty} \frac{\psi_0(n + 1)}{n!} x^n \quad (x \rightarrow \infty) \).
So obviously, there must be something wrong in the derivation. I think I found it.
In the beginning, you assume \( D_t e^x = 0 \), which, given the definition of your \( D_t \) operator, would mean that \( \frac{d^t}{dx^t} e^x = e^x \). But that's a problem. There's a catch when working with fractional derivatives. Namely, that they are "non-local". This is similar to how the integral requires a "lower (or upper) bound". Another way to think of choosing the bound is "choosing a branch" of the inverse function. The fractional derivative is a continuous and real-indexed iteration of the derivative, which means it must also include all negative iterates as well -- and those are integrals, so the "non-local" property of the integrals must show up somewhere, and it shows up at every non-nonnegative-integer order of differentiation. In general, fractional-iterate functions that can do real and complex iterates will be multi-valued functions if whatever is being iterated is not injective, and derivative is definitely not injective (differentiate a constant function).
In the theory of fractional derivatives, when you took \( \frac{d^t}{dx^t} e^x = e^x \), you took the lower bound as \( -\infty \). (See how you have to take this as the lower bound for an integral of the exponential function if you want it to return the exponential function back? Mmm-hmm. Same here.) But, when you took
\( \frac{d^t}{dx^t} x^n = \frac{\Gamma(n+1)}{\Gamma(n+1-t)} x^{n-t} \)
which you used to build \( D_t x^n \), and thus the power series, you were taking the lower bound to be 0. That's a funny thing about these formulas -- and in some cases it is not said what the lower bound is. When you then took the power series, it was kind of like saying
\( \int_{-\infty}^{x} e^t dt = \int_{0}^{x} \left(\sum_{n=0}^{\infty} \frac{t^n}{n!}\right) dt \)
and here, the error becomes clear.
I fed the given series
\( e^x \ln(x) \stackrel{\mathrm{supposedly}}{=} \sum_{n=0}^{\infty} \frac{\psi_0(n + 1)}{n!} x^n \)
to the Wolfram Alpha calculator. It says that it does not give the left-hand side, but instead, it gives
\( e^x (\ln(x) + \Gamma(0, x)) = \sum_{n=0}^{\infty} \frac{\psi_0(n + 1)}{n!} x^n \)
(I have no idea how it managed to derive that formula -- any suggestions?)
But this shows the problem. There is an extra term present, which is the upper incomplete gamma function. It decays to 0 as \( x \rightarrow \infty \).
So, your series is asymptotic to, but not equal to, \( e^x \ln(x) \). I.e. the first "equation" is really
\( e^x \ln(x) \sim \sum_{n=0}^{\infty} \frac{\psi_0(n + 1)}{n!} x^n \quad (x \rightarrow \infty) \).
So obviously, there must be something wrong in the derivation. I think I found it.
In the beginning, you assume \( D_t e^x = 0 \), which, given the definition of your \( D_t \) operator, would mean that \( \frac{d^t}{dx^t} e^x = e^x \). But that's a problem. There's a catch when working with fractional derivatives. Namely, that they are "non-local". This is similar to how the integral requires a "lower (or upper) bound". Another way to think of choosing the bound is "choosing a branch" of the inverse function. The fractional derivative is a continuous and real-indexed iteration of the derivative, which means it must also include all negative iterates as well -- and those are integrals, so the "non-local" property of the integrals must show up somewhere, and it shows up at every non-nonnegative-integer order of differentiation. In general, fractional-iterate functions that can do real and complex iterates will be multi-valued functions if whatever is being iterated is not injective, and derivative is definitely not injective (differentiate a constant function).
In the theory of fractional derivatives, when you took \( \frac{d^t}{dx^t} e^x = e^x \), you took the lower bound as \( -\infty \). (See how you have to take this as the lower bound for an integral of the exponential function if you want it to return the exponential function back? Mmm-hmm. Same here.) But, when you took
\( \frac{d^t}{dx^t} x^n = \frac{\Gamma(n+1)}{\Gamma(n+1-t)} x^{n-t} \)
which you used to build \( D_t x^n \), and thus the power series, you were taking the lower bound to be 0. That's a funny thing about these formulas -- and in some cases it is not said what the lower bound is. When you then took the power series, it was kind of like saying
\( \int_{-\infty}^{x} e^t dt = \int_{0}^{x} \left(\sum_{n=0}^{\infty} \frac{t^n}{n!}\right) dt \)
and here, the error becomes clear.
