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Some context: over the past two weeks, I have been solving an integral. I completed it last night. However, I just realized that I have essentially shown $$\sum_{k=0}^nf(n,k) = \int_{\mathbf{R}}f(n,x) \, \mathrm{d}x$$ Where $n$ is a fixed non-negative integer.

My question: let $f(t,x)$ be an everywhere-differentiable real-valued function of $x$ where $t$ is a non-negative integer. Further, let it be such that $f(t,x)$ cannot be written as $g(t)h(x)$. Is there any general way to go about determining whether $$\sum_{k=0}^nf(n,k) = \int_{\mathbf{R}}f(n,x) \, \mathrm{d}x$$ For all $n=0,1,2 \dots$? I ask because I can find quite a few functions which both do and do not satisfy the above, and am wondering whether the the work I just did was unnecessary.

Descartes Before the Horse
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2 Answers2

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Your formula looks to encompass arithmetic formulas like this one :

$$\sum_{k=1}^n f(k)=\dfrac{n(n+1)}{2}=\int_{\tfrac12}^{n+\tfrac12} f(x)\mathrm{d}x \ \ \ \ \text{where} \ \ \ \ f(x):=x. \tag{1}$$

(more exactly $f(x)=x$ for $\tfrac12 \leq x \leq n+\tfrac12$ and $f(x)=0$ elsewhere).

More generally formulas like (1) can be found by adapting formulas using Bernoulli polynomials $B_n(x)$ (like one can find in this reference):

$$\sum_{k=1}^{n} k^{p}=\int_0^{n+1} B_{p}(x)\mathrm{d}x$$

Besides, could you say which function $f$ you have found fulfilling your equation ?

Jean Marie
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  • The function I am considering is $f(n,x) = {n \choose x}{{x+n} \choose n}$ where $n=0,1,2 \dots$. As an integral, we consider its continuous analog using the gamma function. – Descartes Before the Horse Feb 25 '20 at 06:37
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    A connected issue, with interesting comments : https://math.stackexchange.com/q/507696 – Jean Marie Feb 25 '20 at 10:05
  • Could you give explicitly the integral you are using ?. It is possible that it has a convolution interpretation. – Jean Marie Feb 25 '20 at 11:04
  • ...or be expressed as a Mellin-Barnes integral. – Jean Marie Feb 25 '20 at 11:14
  • I have shown: $$\sum_{k=0}^n {n \choose k}{{n+k} \choose k} = \int_{\mathbf{R}}{n \choose x}{{n+x} \choose x} , dx$$ For $n=0,1,2 \dots$ and have done so by entirely real methods. A professor of mine recommended I send this to the Monthly, though I'm worried it is a consequence of a more general fact. – Descartes Before the Horse Feb 25 '20 at 14:34
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    You can send it ; the referees are there for answering you something like "this has already been done there / there, etc..., but the approach is new " ; Nevertheless, you are more likely to be selected if your bibliography is solid from the start. My experience on this site is that it happens frequently that 2 or 3 days after posting a question, somebody gives a very accurate answer. – Jean Marie Feb 25 '20 at 16:12
  • Good news : In fact your sum has a combinatorial interpretation as "Delannoy numbers": (see (https://oeis.org/A001850)) with a big number of tracks. – Jean Marie Feb 25 '20 at 17:13
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    Yes, that is the appeal of my integral :) AFAIK, I have not found the integral representation of the central Delannoy numbers I just stated anywhere in literature. My worry is that it is a trivial consequence of another fact, though my proof of it certainly is not trivial. – Descartes Before the Horse Feb 25 '20 at 17:16
  • Indeed, up to my knowledge, function $f_n(x):=\dfrac{\Gamma(n+1+x)}{\Gamma(x+1)^2 \Gamma(n+1-x)}$ is not present in the litterature. Its curve is - rather surprisingly - almost gaussian with maximum at $x_0=(7/10)n$. Stirling approximation should give an explanation... – Jean Marie Feb 25 '20 at 18:03
  • See p. 118 of this 2014 high level Ph.D. thesis (https://www.lptmc.jussieu.fr/files/ABDELAZIZ/LairezThese.pdf) where the "residue" is a classical concept in complex analysis you may have already met. If you have some problem understanding french, tell me. – Jean Marie Feb 25 '20 at 18:22
  • An interesting (but not at all evident) 2017 article (https://www.researchgate.net/publication/318259032_A_Continuous_Analogue_of_Lattice_Path_Enumeration) – Jean Marie Feb 25 '20 at 21:06
  • @Raymond Manzoni Here is a beautiful question about a somewhat mysterious function $f_n$ for which maybe you have some answers (almost all information is to be found into the comments). – Jean Marie Feb 26 '20 at 06:31
  • Connected : https://math.stackexchange.com/q/634339 – Jean Marie Feb 26 '20 at 19:14
  • Then my result should follow directly from the fact that $\sum_{k=0}^n{n \choose k}{{n+k} \choose k} = \sum_{k=-\infty}^\infty{n \choose k}{{n+k} \choose k}$...right? – Descartes Before the Horse Feb 26 '20 at 19:46
  • As I have understood rapidly in the reference just above : converting the LHS and the RHS into their "integral of gammas" counterpart gives a good approximation value for bounds such as $0,n$ and exact value when you take bounds $-\infty,+\infty$ (which is your result). The interest of this reference is also that they cope with the gaussian aspects. – Jean Marie Feb 26 '20 at 20:13
  • It is not generally the case that $\sum_{k = -\infty}^\infty f(k) = \int_{\mathbb{R}}f(x)dx$. I have a question: does it require any complex-analytic methods to show that, in the problem you linked, that the doubly infinite sum is equal to the integral over the real line? – Descartes Before the Horse Feb 26 '20 at 22:33
  • "in the problem you linked" : which one do you mean ? – Jean Marie Feb 26 '20 at 22:44
  • The link to the answer from Noam Elkies. To clarify my question: he notes that $\sum_{k=0}^n {n \choose k} = \sum_{-\infty}^\infty{n \choose k}$. However, it is not always the case that $\sum_{-\infty}^\infty f(k) = \int_{\mathbb{R}}f(x)dx$. So how does he deduce that the doubly infinite sum is indeed equal to the integral over the real line? – Descartes Before the Horse Feb 27 '20 at 01:07
  • I must take the time for a closer look at this document. Besides, to answer to one of your previous question, I think that even if your method uses real analysis, you will have to understand at least a little why and how complex analysis methods (in particular involving residues) play a rôle. – Jean Marie Feb 27 '20 at 09:09
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    Another recent reference to Delannoy numbers. – Jean Marie Mar 01 '20 at 23:59
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Identities of this kind are well known and they are consequences of a general theorem for band limited functions. For another example with binomial coefficients see How to prove $\int_{-\infty}^{\infty}\binom{n}{x}^2\binom{2n+x}{x} dx=\binom{2n}{n}^2$ .

Martin Nicholson
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