1
$\begingroup$

(Be prepared for a very long post)

I have deduced the following formula:

$$D^{-n}\ln(x)=\frac{x^n(\ln(x)-n)}{(-n)!}=\frac{x^n(\ln(x)-n)}{\Gamma(-n+1)}$$

Where $$D^{-1}f(x)=\int f(x)dx$$

$$D^{-2}f(x)=\int\int f(x)dxdx$$

$etc$.

$$D^0f(x)=f(x)$$

$$D^1f(x)=\frac d{dx}f(x)$$

$$D^nf(x)=\frac{d^n}{dx^n}f(x)$$

So the $n$th integral of $\ln(x)$ is given by my formula if $n$ is a natural number.

Since the formula is continuous for $n\in\mathbb{R}$ when $n$ is not a negative integer, I have made the assumption that it is the fractional derivative/integral formula for $\ln(x)$.

I have made a graph of the integral. You can truly see that if $n=0$, then it is the natural logarithm, and $f'(x)=\frac1x$, $f''(x)=-\frac1{x^2}$, etc.

If $n=-1$, then $f(x)=\int\ln(x)dx$, $f'(x)=\ln(x)$, $f''(x)=\frac1x$, etc.

However, some weird business occurs at $n=1,2,3,\dots$ Interestingly, you can prove that my formula works by induction, simply take the derivative of $D^{a}f(x)$ and compare it to $D^{a+1}f(x)$. It should work, right?

And since it is trivial that $D^0f(x)=f(x)$, then $D^1f(x)=\frac d{dx}f(x)=\frac1x$?

It leads me to the conclusion that my formula works for $n$ if it is not a negative whole number, but since it appears to be continuous, does it work for all other $n$?

Obviously, it shouldn't, because $D^1\ln(x)=D^{1/2}D^{1/2}\ln(x)=\frac1x$.

$$D^{1/2}\ln(x)=\frac{\sqrt{x}\left(\ln(x)+\frac12\right)}{\Gamma(1/2)}$$

And if the formula I sort of proposed holds true, then we have:

$$D^{1/2}D^{1/2}\ln(x)=D^{1/2}\frac{\sqrt{x}\left(\ln(x)+\frac12\right)}{\Gamma(1/2)}=\frac{\Gamma(3/2)}{\Gamma(1/2)}\left(\sum_{m=0}^{\infty}(D^{\frac12-m}g(x))(D^{m}h(x))\right)$$

$$=\frac32\sum_{m=0}^{\infty}(D^{\frac12-m}g(x))(D^{m}h(x))$$

where $g(x)=x^{1/2}$ and $h(x)=\ln(x)+\frac12$.

On the first link, there is a formula that solves $D^{n}x^k=\frac{\Gamma(k+1)}{\Gamma(k-n+1)}x^{k-n}$

In our case, $k=1/2$.

$$D^{1/2}D^{1/2}\ln(x)=\frac32\sum_{m=0}^{\infty}\left(\frac{\sqrt{\pi}}{2\Gamma(m+1)}x^m\right)(D^{m}h(x))$$

I see that $D^0h(x)=\ln(x)+\frac12$, $D^1h(x)=x^{-1}=H(x)$

The $n$th derivative of $H(x)$ is given as $(-1)^nn!x^{-1-n}$.

$$D^{1/2}D^{1/2}\ln(x)=\frac32\left(\frac{\sqrt{\pi}(\ln(x)+0.5)}{2}+\sum_{m=1}^{\infty}\left(\frac{\sqrt{\pi}}{2\Gamma(m+1)}x^m\right)((-1)^mm!x^{-1-m}\right)$$

  1. I need someone to prove or disprove whether or not this last line equals $\frac1x$.

  2. I need someone with more fractional Calculus to verify if my formula for the $n$th integral of $\ln(x)$ may be used for fractional values.

  3. I would like someone to view if my proposed summation for the fractional derivative of the multiple of two functions and tell me what they think on it, whether or not it is 'correct' or 'incorrect'.

  4. If my proposed formula for the $n$th derivative/integral is only valid until we reach the messed up summation at the end of my post, why? It doesn't make sense for the formula to disagree with known knowledge and where would it start failing? I can't imagine that it holds all the way up to $1/x$ and then fails right there, nor can I imagine it holding from $\ln(x)$ to all of its integrals but fail to work for fractional derivatives of $\ln(x)$.

  5. And of course, do mention if you catch any mistakes I may have made.

Thanks in advance for all of your efforts and time.

$\endgroup$
  • $\begingroup$ Your formula at the top of your post is good for the case $n=0$. $\endgroup$ – Mhenni Benghorbal Feb 28 '16 at 1:31
  • $\begingroup$ I gave many techniques and theorems in my book how to give formulas for the $n$th derivative and anti derivatives $\endgroup$ – Mhenni Benghorbal Feb 28 '16 at 1:40
  • $\begingroup$ See my answer. $\endgroup$ – Mhenni Benghorbal Feb 28 '16 at 1:46
  • $\begingroup$ See my paper. $\endgroup$ – Mhenni Benghorbal Feb 28 '16 at 2:04
  • 1
    $\begingroup$ @MhenniBenghorbal Thank you so much. I managed to find the more correct answer, but I still have the problem of making it work. I would like it to work for $n=-1$, and I did note that the "formula at the top of [my] post is good for the case $n=0$." It is rather late, so I will fix all of this on the morrow. $\endgroup$ – Simply Beautiful Art Feb 28 '16 at 2:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.