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For integer $n \geq 0$, we have $\dfrac{d^n}{ds^n} x^s = (\ln x)^n \,x^s$. From this it follows, for example, that $$\int_0^{\infty}e^{-x}\ln^n x \,dx= \Gamma^{(n)}(1)$$ Question: is there a way of defining fractional derivatives in which it makes sense to say that $$\int_0^{\infty}e^{-x}\ln^q x \, dx= \Gamma^{(q)}(1) $$ for rational $q$, and are there values of $q$ for which this can be evaluated in terms of ordinary special functions?

Simply Beautiful Art
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user111187
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1 Answers1

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One way to check the answer to see if it holds by induction. Using $\frac{d^n}{dx^n}\frac{d^k}{dx^k}=\frac{d^{n+k}}{dx^{n+k}}$, and assuming it is continuous as a function of $n,k$, then you should be able to verify that your answer works.

Probably the easiest and neatest method, and it works great for the simple problems.

Second and more rigorous approach would be to use a particular definition for what $\frac{d^t}{dx^t}$ means for $t\notin\mathbb N$. One approach can be given by

$$\frac{d^t}{dx^t}f(x)=\frac1{\Gamma(1-\{t\})}\frac{d^{\lceil t\rceil}}{dx^{\lceil t\rceil}}\int_{-\infty}^x\frac{f(x)}{(x-t)^{\{t\}}}~\mathrm dt$$

In particular, this should give

$$\frac{d^t}{dx^t}e^{ax}=a^te^{ax}$$

However, as we cannot directly exchange the integrals, it appears that $\int_0^\infty t^{s-1}e^{-t}(\ln(t))^x~\mathrm dt$ is only a regularization of $\Gamma^{(x)}(s)$ in this sense.

Simply Beautiful Art
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  • hmm ? you have to find a derivation of $\displaystyle\frac{d^q}{ds^q} e^{cs}$ for real $q > 0$ – reuns Jun 17 '16 at 13:04
  • @user1952009 $\frac{d^q}{ds^q}e^{cs}=c^qe^{cs}$ by induction. – Simply Beautiful Art Jun 17 '16 at 13:34
  • how do you know it ? – reuns Jun 17 '16 at 13:35
  • @user1952009 It is rather easy actually. $\frac d{dx}e^{cx}=ce^{cx}$, so for $n\in\mathbb N$, $\frac{d^n}{dx^n}e^{cx}=c^ne^{cx}$, which implies this holds for $n\in\mathbb Z$, which implies using the method my answer suggests, that this holds for $n\in\mathbb{Q}\implies n\in\mathbb{R}\implies n\in\mathbb{C}$ – Simply Beautiful Art Jun 17 '16 at 13:40
  • so to be clear you don't have any proof – reuns Jun 17 '16 at 13:41
  • @user1952009 ... ? I don't understand, that derivative is well known and what not. I'm not quite sure what you are asking for when you say you want a proof. – Simply Beautiful Art Jun 17 '16 at 13:43
  • this is a 2 years old question whose you don't have the answer : how to define the fractional derivative such that it is a well-defined linear operator and that $D^q e^{cx} = c^q e^{cx}$ – reuns Jun 17 '16 at 13:47
  • @user1952009 It is actually far beyond 2 years old and is one of the first fractional derivatives considered, like here: http://www.mathpages.com/home/kmath616/kmath616.htm I am actually slightly unsure how to handle this problem using "linear operators", but I'm sure someone else can. Sorry. – Simply Beautiful Art Jun 17 '16 at 14:02
  • in wiki/Fractional_calculus there is what is needed for proving $D^q x^a = \frac{\Gamma(a+1)}{\Gamma(a+1-q)} x^{a-q}$ but it gives $D^q e^{cx} = D^q\sum_{k=0}^\infty \frac{c^k x^k}{k!} = \sum_{k=0}^\infty D^q \frac{c^k x^k}{k!} = \sum_{k=0}^\infty \frac{c^k x^{k-q}}{\Gamma(k+1-q)}$.... maybe using the laplace transform definition of $D^q$ will be easier – reuns Jun 17 '16 at 14:31
  • @user1952009 No, actually, I do think different ways of calculating fractional derivatives may come out differently. Which is a question I want to ask. Plus, that's a bad way of doing it: http://math.stackexchange.com/questions/572828/fractional-derivative-of-a-taylor-series – Simply Beautiful Art Jun 17 '16 at 14:36
  • the Laplace/Fourier transform definition when it converges should be compatible with the $D^q x^a = \frac{\Gamma(a+1)}{\Gamma(a+1-q)} x^{a-q}$ definition, the problem is when something does not converge – reuns Jun 17 '16 at 14:40
  • @user1952009 You know, I was actually trying to take the fractional derivative of a function via Taylor series, because that tends to make sense, but it turns out to seem slightly wrong somehow. I don't actually know how to do those transforms, but I do note that if we have $c=1$ in your example above, you get false statements, as $D^qe^x\ne D^{q+1}e^x$, they actually come out different. – Simply Beautiful Art Jun 17 '16 at 14:42
  • how do you know $D^{q+1} e^x \ne D^q e^x$ ? – reuns Jun 17 '16 at 15:36
  • @user1952009 I tried their Taylor series, and $D^qe^x-D^{q+1}e^x=\frac{x^{-q-1}}{\Gamma(-q)}$, which happens when you reindex the series so you can cancel off terms. Which seems contradictory, since we should have $D^qe^x=e^x$ for any $q$. – Simply Beautiful Art Jun 17 '16 at 15:51
  • I don't get your argument, but it seems $D^q e^x\ne e^x$ (when $q$ is not an integer) so that the correct answer is that with the usual definition of the fractional derivative in term of the Fourier transform $D^q f = FT^{-1}[(2 i \pi \xi)^q FT[f]]$ which is the same as $D^q [x^a 1_{x > 0}] = \frac{\Gamma(a+1)}{\Gamma(a+1-q)} x^{a-q} 1_{x > 0}$ then $D^q [e^x] \ne e^x$ – reuns Jun 17 '16 at 16:05
  • @user1952009 Yes, and many fractional derivative formulas point out $D^qe^x=e^x$, implying that there isn't just one way to doing these things and one answer may be just as right as the next. – Simply Beautiful Art Jun 17 '16 at 16:07
  • don't know what those fractional derivative are meant to be, but they are probably not linear operators (on tempered distribution) – reuns Jun 17 '16 at 16:08
  • @user1952009 Sadly, I can't help you there. Maybe not all fractional derivative formulas aren't linear operators, which I guess would be a pain for you. :/ – Simply Beautiful Art Jun 17 '16 at 16:09
  • of course it is a problem if it is not linear ! how would you treat $D^q \sum_k c_k e^{k x}$ then ? what about the $D^q \int_0^\infty f(t,x) dt$ in the original question ? so I think the problem is to find what are the possible definition of $D^q$ such that $D^q e^{cx} = c^q e^{cx}$ and see if one of them is linear (I think they are not, but I've no proof) – reuns Jun 17 '16 at 16:11
  • @user1952009 I usually follow an induction method if the answer seems it should be obvious. If the answer isn't, then I'd probably resort to the formula book. – Simply Beautiful Art Jun 17 '16 at 16:12
  • so did you get why the fact it is linear is the most important ? and try searching in your texts if you find any clear definition of $D^q$ such that $D^q e^{cx}= c^q e^{cx}$ – reuns Jun 17 '16 at 16:15
  • @user1952009 I'm sorry, but to begin with, I don't fully understand 'linear' in this context. I haven't even studied linear algebra, so sorry. – Simply Beautiful Art Jun 17 '16 at 16:16
  • @user1952009 Oh, well, that would strongly imply that we should be able to differentiate a function via its Taylor series... I feel as though I should study up on this. Thanks for the inputs. http://math.stackexchange.com/questions/1829921/do-different-methods-of-calculating-fractional-derivatives-have-to-be-equal – Simply Beautiful Art Jun 17 '16 at 16:20
  • $D^q(a f+b g) = a D^q f + b D^q g$ (for any two functions $f,g$ and constants $a,b$) !!! this is what allows us to write (when $K$ is finite, or when $K = \infty$ and that everything converges) $D^q \sum_{k=0}^K a_k f_k = \sum_{k=0}^K a_k D^q f_k$ and yes the fact that the usual derivative is a linear operator is why we can differentiate easily so many analytic functions via their Taylor/Dirichlet/Fourier series or their integral representation (since $\int$ is a limit of $\sum$) – reuns Jun 17 '16 at 16:21
  • @user1952009 Note that if we differentiate a fractional and positive amount of times on the Taylor series, we get some $x^{-a}$ term, implying that $\lim_{x\to0}f^{(n)}(x)$ is undefined around $x=0$ for all functions? And yet converges to something everywhere else? – Simply Beautiful Art Jun 17 '16 at 16:26
  • I answered to that in your question on the integration constant .. $\frac{\Gamma(a+1)}{\Gamma(a-n-1)} = 0$ for $a-n-1$ a negative integer – reuns Jun 17 '16 at 16:30
  • @user1952009 Oh... I feel like this is overly complicated. (I really don't like the constant of differ-integration...) I'll come back to this in a few hours, thanks for the thought invoking talk. No no, but when taking fractional derivatives, $a-n-1\notin\mathbb{Z}$, which lets the problems come in. – Simply Beautiful Art Jun 17 '16 at 16:34
  • @reuns *sighs* long time no see, how's my answer look now lol. And while you are at it, this new question of mine may be of interest to you. – Simply Beautiful Art Jul 17 '17 at 19:18