7
$\begingroup$

(The question arises from playing with translating series into integrals)

I wanted to see, what it means to have a "continuous" relative for powerseries and other series; the most simple one perhaps $$ \begin{array} {} f_1(x) = \sum _{k=0}^\infty x^k = {1 \over 1-x} &\to & g_1(x)= \int_0^\infty x^t \,dt = - { 1\over \log(x) } \end{array}$$ I couldn't get this $$ \begin{array} {} f_2(x)=\sum _{k=0}^\infty {x^k \over k!} = \exp(x) &\to& g_2(x)=\int_0^\infty { x^t\over \Gamma(1+t) } \,dt =\text{ ???} \end{array}$$ by, for instance, Wolfram alpha...
and I'd like to proceed to some more general $$ \begin{array} {} f_\varphi(x)=\sum _{k=0}^\infty \varphi(k) x^k &\to& g_\varphi(x)=\int_0^\infty \varphi(t) x^t \,dt =\text{ ???} \end{array}$$ where $\varphi(k)$ is some meaningful function producing common sets of coefficients for the power series.

Playing a bit with numerical evaluations for $f_2(x),g_2(x)$ so far did not uncover anything obvious, but I am interested, whether there are some relations known; for instance whether there are relations between $g_2(x) \cdot g_2(y)$ perhaps analoguously to $ f_2(x) \cdot f_2(y) = f_2(x+y)$ or the like.

Is there something known about it? Is there a possibly a list of sums/integrals-relations done elsewhere?

In general, I'd like to get more intuition about this; it reminds me that I should possibly re-read in the explanations for the Euler/MacLaurin-summation formula where the "dance between discrete and continuous" (as it is a title of a nice article about the work of Delabaere on Euler's divergent series) has a similar relevance. (But this is possibly too much for this Q&A-site ...)

$\endgroup$
6
  • $\begingroup$ Your $g_1(x)$ should equal $$\frac1{\log{\frac1{x}}}$$ $\endgroup$
    – Ron Gordon
    Apr 16, 2014 at 19:47
  • $\begingroup$ @RonGordon: ah, thanks; I missed it because W/A gave it for the indefinite case without a sign. I've just edited the text. $\endgroup$ Apr 16, 2014 at 19:53
  • $\begingroup$ Why is this elementary-number-theory? $\endgroup$
    – chubakueno
    Apr 16, 2014 at 23:41
  • 1
    $\begingroup$ See the question .it is related:math.stackexchange.com/questions/158527/… $\endgroup$
    – Mathlover
    Apr 17, 2014 at 0:12
  • $\begingroup$ @Mathlover: very nice - thank you. That gives a lot to chew for me... $\endgroup$ Apr 17, 2014 at 4:16

2 Answers 2

1
$\begingroup$

Here's a thought: rewrite $g_2$ as

$$\int_0^{\infty} dt \frac{e^{-t \log{(1/x)}}}{t \Gamma(t)}$$

Now, $\frac1{t \Gamma(t)}$ is an entire function and therefore has the following Taylor series representation in the whole complex plane:

$$\frac1{t \Gamma(t)} = \sum_{k=0}^{\infty} a_k t^k$$

where $a_0=1$, $a_1=\gamma$, and for $k \ge 2$, the $a_k$ satisfy

$$a_k = \frac{a_1 a_{k-1}-\sum_{j=2}^{k} (-1)^j \zeta(j) a_{k-j}}{k} $$

Integrate and get the following

$$g_2(x) = \sum_{k=0}^{\infty} a_k \frac{k!}{\left ( \log{\frac1{x}}\right )^{k+1}} $$

$\endgroup$
3
  • $\begingroup$ A bizarre expression, indeed! I'll try it numerically tomorrow (It's late today) $\endgroup$ Apr 16, 2014 at 20:47
  • $\begingroup$ @GottfriedHelms: try to deduce the behavior of $a_k k!$ for large $k$; that should be interesting. $\endgroup$
    – Ron Gordon
    Apr 16, 2014 at 20:49
  • $\begingroup$ Hmm, I seem to have half the tour. I arrive at the coefficients $a_k$ simply letting Pari/GP give the power series expansion of $\exp(u*t)/\Gamma(1+t)$ in powers of $t$ and let it then determine the formal integral, which of course has a term-by-term approach . I interpret $u$ as $log(x)$ and the first three coefficients of the two variate power-series (in u and t) are your $a_k \cdot k!$. Well, to make the relation completely clear I've to proceed a bit more and go through this on my scribblepad. I'll come back to this then. $\endgroup$ Apr 17, 2014 at 19:36
0
$\begingroup$

One more relation by W/A: $$ \begin{array} {} f_{1,1}(x) = \sum _{k=1}^\infty k \cdot x^{k-1} = {1 \over (1-x)^2} &\to & g_{1,1}(x)= \int_1^\infty t \cdot x^{t-1} \,dt = { 1 - \log(x) \over \log^2(x) } \\& & \text{for Re} (\log(x)) \lt 1 \end{array}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .