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I conjecture that the following statement holds for large values of $N$ $$ {}_3F_1\left(-N+1,1,1;2;-\frac{1}{N}\right)\to\frac{1}{2}\bigg({}_2F_1(1,1;2;1-\frac{1}{N})+\log 2+\gamma\bigg) $$ where $\gamma$ is the Euler-Mascheroni constant. PLugging in big values for $N$ in the above formula it looks like the conjecture holds, but I am unable to prove it. Anybody can do better?

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  • $\begingroup$ How did the conjecture arise? $\endgroup$ – Giovanni De Gaetano Jun 13 '16 at 11:52
  • $\begingroup$ @GiovanniDeGaetano arxiv.org/pdf/hep-ph/0505034.pdf trying to prove eq.69 there. I have proven that the integral gives ${}_3F_1$. The ${}_2F_1$ would give the log in the big $N$ limit. $\endgroup$ – PhoenixPerson Jun 13 '16 at 12:13
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As the OP remarked in a comment, the ultimate purpose of the question is to derive the asymptotic large-$N$ expansion of the integral

$$I_N=\int_0^1 dq\, \frac{q^2}{\log q}\left[\left(1-\frac{3}{N}\log q\right)^N-1\right]$$

According to equation (69) of this reference, $I_N$ should converge to $$I_N=\tfrac{1}{2}\left(\log N + \log 2 +\gamma_E\right)+{\cal O}(N^{-1/2})\qquad\qquad(*)$$

Noting the expansion $$(1+a/N)^N=\exp\left[a-\tfrac{1}{2}a^2/N+{\cal O}(N^{-2})\right]$$ I first transform from $q$ to $x=\log q$ and then take the large-$N$ limit as follows:

$$I_N=\int_{-\infty}^0 dx\, \frac{e^{3x}}{x}\left[\left(1-\frac{3x}{N}\right)^N-1\right]=\int_{-\infty}^0 dx\, \frac{e^{3x}}{x}\left[\exp\left(-3x-\tfrac{9}{2}x^2/N+{\cal O}(N^{-2})\right)-1\right]$$ $$\quad\rightarrow \int_{-\infty}^0 dx\, \frac{1}{x}\left[\exp(-\tfrac{9}{2}x^2/N)-e^{3x}\right]$$ $$\quad=-\tfrac{1}{2}\left(\log N+\log 2+\gamma_E\right).\qquad\qquad(**)$$

This differs by a minus sign from the result $(*)$ in the reference. I checked it numerically, and it's pretty clear that (**) and not (*) is the correct asymptotics.

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  • $\begingroup$ and the one half? $\endgroup$ – PhoenixPerson Jun 13 '16 at 10:33
  • $\begingroup$ fixed the one-half $\endgroup$ – Carlo Beenakker Jun 16 '16 at 15:01
  • $\begingroup$ greatly appreciated $\endgroup$ – PhoenixPerson Jun 16 '16 at 15:27
  • $\begingroup$ I like what you do but I see two pathological steps in your rationale. The first is when you write $e^{3x}=\lim(1-3x/N)$. You shouldn't use the letter $N$. Also, the last big $N$ limit you take to get $(1-9x^2/N^2)^N\to{}e^{-9x^2/N}$ is a very dubious step. In any case, thanks for all the effort you are putting $\endgroup$ – PhoenixPerson Jun 16 '16 at 16:06
  • $\begingroup$ fixed both the "pathologies" you mentioned, and fixed the missing log 2. $\endgroup$ – Carlo Beenakker Jun 16 '16 at 19:51

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