8
$\begingroup$

I'm interested in finding the asymptotic at $n\to\infty$ of $$b_n:= \frac{e^{-n}}{(n-1)!}\int_0^\infty\prod_{k=1}^{n-1}(x+k)\,e^{-x}dx=e^{-n}\int_0^\infty\frac{e^{-x}}{x\,B(n;x)}dx$$ Using a consecutive application of Laplace' method, I managed to get (here) $$b_n\sim(e-1)^{-n}$$ but this approach is not rigorous, and I cannot find even next asymptotic term, let alone a full asymptotic series.

So, my questions are:

  • how we can handle beta-function in this (and similar) expressions at $n\to\infty$
  • whether we can get asymptotic in a rigorous way ?
$\endgroup$
1
  • 1
    $\begingroup$ The integral is strikingly similar to the integrals that appears in a proof that $e$ is transcendental $\endgroup$
    – FShrike
    Feb 14, 2023 at 15:26

2 Answers 2

10
$\begingroup$

First approach. We have \begin{align*} b_n & = \frac{{{\rm e}^{ - n} }}{{\Gamma (n)}}\int_0^{ + \infty } {\frac{{\Gamma (x + n)}}{{\Gamma (x + 1)}}{\rm e}^{ - x} {\rm d}x} \\ & = \frac{{{\rm e}^{ - n} }}{{\Gamma (n)}}\int_0^{ + \infty } {\frac{1}{{\Gamma (x + 1)}}{\rm e}^{ - x} \left( {\int_0^{ + \infty } {s^{x + n - 1} {\rm e}^{ - s} {\rm d}s} } \right)\!{\rm d}x} \\ & = \frac{1}{{\Gamma (n)}}\int_0^{ + \infty } {\frac{1}{{\Gamma (x + 1)}}\left( {\int_0^{ + \infty } {t^{x + n - 1} {\rm e}^{ - {\rm e}t} {\rm d}t} } \right)\!{\rm d}x} \\ & = \frac{1}{{\Gamma (n)}}\int_0^{ + \infty } {t^{n - 1} {\rm e}^{ - {\rm e}t} \left( {\int_0^{ + \infty } {\frac{{t^x }}{{\Gamma (x + 1)}}{\rm d}x} } \right)\!{\rm d}t} . \end{align*} Employing Ramanujan's formula $$ \int_0^{ + \infty } {\frac{{t^x }}{{\Gamma (1 + x)}}{\rm d}x} = {\rm e}^t - \int_{ - \infty }^{ + \infty } {\frac{{{\rm e}^{ - t{\rm e}^y } }}{{y^2 + \pi ^2 }}{\rm d}y} , $$ yields the exact expression \begin{align*} b_n & = \frac{1}{{\Gamma (n)}}\int_0^{ + \infty } {t^{n - 1} {\rm e}^{ - ({\rm e} - 1)t} {\rm d}t} - \frac{1}{{\Gamma (n)}}\int_0^{ + \infty } {t^{n - 1} {\rm e}^{ - {\rm e}t} \int_{ - \infty }^{ + \infty } {\frac{{{\rm e}^{ - t{\rm e}^y } }}{{y^2 + \pi ^2 }}{\rm d}y}\, {\rm d}t} \\ & = \frac{1}{{({\rm e} - 1)^n }} - \int_{ - \infty }^{ + \infty } {\frac{1}{{({\rm e} + {\rm e}^y )^n }}\frac{1}{{y^2 + \pi ^2 }}{\rm d}y} . \end{align*} Since $$ \int_{ - \infty }^{ + \infty } {\frac{1}{{({\rm e} + {\rm e}^y )^n }}\frac{1}{{y^2 + \pi ^2 }}{\rm d}y} \le \frac{1}{{{\rm e}^n }}\int_{ - \infty }^{ + \infty } {\frac{{{\rm d}y}}{{y^2 + \pi ^2 }}} = \frac{1}{{{\rm e}^n }}, $$ we indeed have $$ b_n \sim \frac{1}{{({\rm e} - 1)^n }} $$ as $n\to +\infty$.

Second approach. Changing the order of summation and integration yields $$ \sum\limits_{n = 1}^\infty {b_n z^n } = z\int_0^{ + \infty } {\frac{{{\rm d}x}}{{({\rm e} - z)^{x + 1} }}} = \frac{z}{{({\rm e} - z)\log ({\rm e} - z)}} $$ for sufficiently small $z$. Now note that $$ \frac{z}{{({\rm e} - z)\log ({\rm e} - z)}} = \frac{{\rm e} - 1}{{({\rm e} - 1) - z}} + H(z) $$ where $H(z)$ is holomorphic in the disc $|z|<\mathrm{e}$. The first term may be expanded as $$ \frac{{\rm e} - 1}{{({\rm e} - 1) - z}} = \sum\limits_{n = 0}^\infty {\frac{1}{{({\rm e} - 1)^{n} }}z^n } . $$ On the other hand, the $n$th Maclaurin series coefficient of $H(z)$ is $\mathcal{O}((\mathrm{e}-\varepsilon)^{-n})$ by the Cauchy–Hadamard theorem for any $\varepsilon>0$ as $n\to+\infty$. Thus $$ b_n \sim \frac{1}{{({\rm e} - 1)^n }} $$ as $n\to +\infty$.

$\endgroup$
5
  • 3
    $\begingroup$ (+1) It is interesting that, for $t>0$, the integral of $t^x/\Gamma(x+1)$ always underestimate its summation counterpart! Nice work :) $\endgroup$ Feb 15, 2023 at 0:35
  • 1
    $\begingroup$ Thank you very much for your elegant solution! $\endgroup$
    – Svyatoslav
    Feb 15, 2023 at 1:44
  • 1
    $\begingroup$ The second approach, especially isolating the pole from the rest, is quite interesting and informative! I guess this kind of technique is not uncommon in the field of analytic combinatorics, which I didn't study much. $\endgroup$ Feb 15, 2023 at 2:23
  • 3
    $\begingroup$ @SangchulLee Yes, it is called Darboux's method. $\endgroup$
    – Gary
    Feb 15, 2023 at 2:28
  • 1
    $\begingroup$ I would also like to mention that Laplace' method formally allows to find some kind of "approximation" at $n\to\infty$ of $$b_n(\lambda)=e^{-n}\int_0^\infty\frac{x^\lambda \,e^{-x}}{x\,B(n;x)}dx$$ The straightforward evaluation gives $$b_n(\lambda)\sim n^\lambda (e-1)^{-n-\lambda}$$ The approximation is rather good for small $\lambda \,(\sim 1)$ even at small $n\sim 10$, but quickly becomes inaccurate at bigger $\lambda$. At $\lambda=1$ approximation is valid even at $n=3: \,0.3485 \,(exact) \,vs. 0.3441 \,(approx)$, and can also be obtained by Darboux' method $\endgroup$
    – Svyatoslav
    Feb 15, 2023 at 23:02
1
$\begingroup$

This is an attempt to generalize the asymptotic. Namely, to consider a bit more general case: $$b_n(\lambda)=e^{-n}\int_0^\infty\frac{x^\lambda \,e^{-x}}{x\,B(n;x)}dx\,,\,\,\lambda\geqslant0\tag{1}$$ Using the second approach proposed by @Gary, we get $$\sum\limits_{n = 1}^\infty {b_n z^n } = z\int_0^{ + \infty } {\frac{{x^\lambda}}{{(e - z)^{x + 1} }}} dx = \frac{\Gamma(1+\lambda)\,z}{{(e - z)\log^{\lambda+1} (e - z)}}\tag{2}$$ For $\lambda=0, 1, 2,...$ we can obtain the asymptotic in a closed form.

For example, for $\lambda=1$, denoting $\epsilon=e-1-z$ and keeping only the singular part $$\frac{z}{{(e - z)\log^2 (e - z)}}=\frac{e-1}{\epsilon^2}-\frac{1}{\epsilon}+H_1(\epsilon)\tag{3}$$ where $H_1(\epsilon)$ is analytical in the disk $\,|\epsilon|<e-1$.

Expanding the singular part of (3) into the series, $$[z^n]\left(\frac{e-1}{(e-1-z)^2}-\frac{1}{e-1-z}\right)=\frac{1}{(e-1)^{n+1}n!}\big((n+1)!-n!\big)=\frac{n}{(e-1)^{n+1}}\tag{4}$$ It is not clear whether we can find a good approximation for non-integer $\lambda$, though the main asymptotic term for arbitrary $\lambda$ is still $\displaystyle\frac{n^\lambda}{(e-1)^{n+\lambda}}$.

But for $\lambda=k=0, 1, 2, ...$ such closed form does exist. To get it, it is convenient to use the first approach proposed by @Gary. $$b_n (k)= \frac{{e^{ - n} }}{\Gamma (n)}\int_0^{ + \infty } {\frac{\Gamma (x + n)\,x^ke^{ - x}}{\Gamma (x + 1)} dx}=\frac{\partial^k}{\partial\alpha^k}\,\bigg|_{\alpha=0}\frac{{e^{ - n} }}{\Gamma (n)}\int_0^{ + \infty } {\frac{\Gamma (x + n)\,x^ke^{ - x(1-\alpha)}}{\Gamma (x + 1)} dx}\tag{5}$$ Changing the order of integration and using Ramanujan's formula $$\frac{{e^{ - n} }}{\Gamma (n)}\int_0^{ + \infty } {\frac{\Gamma (x + n)\,x^k}{\Gamma (x + 1)}e^{ - x(1-\alpha)} dx}=\frac{e^{-n}}{\Gamma(n)}\int_0^\infty e^{-s}s^{n-1}ds\bigg(\int_0^\infty\frac{\Big(se^{-(1-\alpha)}\Big)^x}{\Gamma(1+x)}dx\bigg)$$ $$=\big(e-e^\alpha\big)^{-n}-\int_{-\infty}^\infty\frac{1}{(e+e^{\alpha+y})^n}\frac{dy}{y^2+\pi^2}\tag{6}$$ The second term in (6) is exponentially small vs. the first one.

For example, for $k=1$ $$\frac{\partial}{\partial\alpha}\,\bigg|_{\alpha=0}\int_{-\infty}^\infty\frac{1}{(e+e^{\alpha+y})^n}\frac{dy}{y^2+\pi^2}$$ $$=-n\int_{-\infty}^\infty\frac{1}{(e+e^y)^n}\frac{dy}{y^2+\pi^2}+ne\int_{-\infty}^\infty\frac{1}{(e+e^y)^{n+1}}\frac{dy}{y^2+\pi^2}\sim O\left(\frac{n}{e^n}\right)$$ It mean that the full asymptotics (if we drop exponentially small terms) is given by $$\boxed{\,\,b_n(k)=e^{-n}\int_0^\infty\frac{x^k \,e^{-x}}{x\,B(n;x)}dx\sim\frac{\partial^k}{\partial\alpha^k}\,\bigg|_{\alpha=0}\big(e-e^\alpha\big)^{-n}\,,\,\,k=0, 1, 2, ...\,\,}$$ where all terms are valid and should be kept.

For several first $k$ we get $$\qquad\qquad b_n(1)=\frac{n}{(e-1)^{n+1}}\,\,\big(in \,agreement \,with \,(4)\,\big)$$ $$b_n(2)=\frac{n(n+e)}{(e-1)^{n+2}}$$ $$b_n(3)=\frac{n\Big(n^2+3e(n+1)+e^2\Big)}{(e-1)^{n+3}}$$ etc.

The numeric check is in perfect agreement with the approximation. Exactly the same answer can also be obtained by double application of Laplace' method, though in this case we cannot evaluate the error.

$\endgroup$
1
  • 1
    $\begingroup$ The generating function method will work. Note that you have an algebraic singularity at $\mathrm{e}-1$. $\endgroup$
    – Gary
    Feb 18, 2023 at 3:52

You must log in to answer this question.

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