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I need an asymptotic expansion/closed form for $$\sum_{k=1}^{\infty}\int_{0}^{\infty}(B(x+n+k,n+1))^2\ dx$$ where $B(m,n)$ is the Beta function and $n\in\mathbb{N}$.

Denote $$I_n=\sum_{k=1}^{\infty}\int_{0}^{\infty}(B(x+n+k,n+1))^2\ dx$$ By definition of Beta function $$I_n=\sum_{k=1}^{\infty}\int_{0}^{\infty}\left(\frac{\Gamma(x+n+k)\Gamma(n+1)}{\Gamma(x+2n+k+1)}\right)^2\ dx$$ $$I_n=(n!)^2\sum_{k=1}^{\infty}\int_{0}^{\infty}\left(\frac{\Gamma(x+n+k)}{\Gamma(x+2n+k+1)}\right)^2\ dx$$ Now interchanging the summation and integral (which needs to be justified) we get $$I_n=(n!)^2\int_{0}^{\infty}\sum_{k=1}^{\infty}\left(\frac{\Gamma(x+n+k)}{\Gamma(x+2n+k+1)}\right)^2\ dx$$ Now the infinite sum above is see here

$$I_n=(n!)^2\int_{0}^{\infty} \left(\frac{\Gamma(x+n+1)}{\Gamma(x+2n+2)}\right)^2{}_3F_2\left ( 1,x+n+1,x+n+1;x+2n+2,x+2n+2;1 \right)dx$$

where ${}_3F_2$ represents the hypergeometric function.

Thank you!

Edit: @Gary and @TymaGaidash have in their elegant answers found that $$\sum_{k=1}^{\infty}\int_{0}^{\infty}\operatorname B^2(x+n+k,n+1)\ dx =\binom{2n}n\gamma-\sum_{j=0}^n\binom nj^2(2(H_{n-j}-H_j)\ln((j+n)!)+H_{n+j})$$ So if we define$$a_n:= \sum_{j=0}^n\binom nj^2(2(H_{n-j}-H_j)\ln((j+n)!) $$ I need an asymptotic expansion for $a_n$ as $n\to \infty$. I believe that we can apply Laplace's method of asymptotic expansion of integrals (see here p.$322$)

Question: I need to prove that the limit $$ \lim_{n\to \infty} \frac{n\ 4^{2n}}{e^{2n}}\{-d_{2n} a_n\}\leq \frac{3}{4}$$ where $d_{2n}=\text{LCM}(1,2,...,2n)$, $\{x\}$ is the fractional part of $x$ and $a_n$ is defined as above.

Edit Can we at least simplify the above fractional part $\{-d_{2n} a_n\}$? Can we use the (anti) symmetry in $a_n$ to get a simpler form of $\{-d_{2n} a_n\}$?

I would really appreciate an answer which I can accept. Thank you!

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  • $\begingroup$ The integrand is $\frac1{(x+n+k)_{n+1}}$, a Pochhammer symbol which has a finite sum representation after partial fractions, which allows us to take the integral as a finite sum of logarithms. However, it is hard to see a pattern for partial fraction coefficients. $\endgroup$ Jul 29, 2023 at 14:58
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    $\begingroup$ @TymaGaidash Thanks a lot for your inputs $\endgroup$
    – Max
    Jul 29, 2023 at 15:02
  • $\begingroup$ @Max hi see mathworld.wolfram.com/DixonsTheorem.html . $\endgroup$ Aug 1, 2023 at 9:30
  • $\begingroup$ @Max Sorry I haven't the skill to do that . Ask for Tyma Gaidash or Gary . $\endgroup$ Aug 1, 2023 at 12:27
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    $\begingroup$ @Max Your post starts "I need an asymptotic expansion/closed form for...". There are answers for that below. Later you write "I need an asymptotic expansion for $a_n$ as $n→∞$". There are answers for that too. If you find them incomplete, please indicate why you do so. The last question should be moved to a new question separate from this one. $\endgroup$
    – Gary
    Aug 6, 2023 at 1:29

6 Answers 6

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This is not a complete answer. Using the limit representation in Tyma Gaidash's answer, and the known results $$ H_{n + j + r} = \log (n + j + r) + \gamma + o(1), \quad r\to+\infty, $$ (with $\gamma$ being the Euler–Mascheroni constant) and $$ \sum\limits_{j = 0}^n {\binom{n}{j}^2 } = \binom{2n}{n}, $$ we can derive the explicit formula $$ \boxed{I_n = \binom{2n}{n}\gamma - \sum\limits_{j = 0}^n {\binom{n}{j}^2 (2(H_{n - j} - H_j )\log (( n+j)!) + H_{n + j} )} ,} $$ provided that $$ \mathop {\lim }\limits_{r \to + \infty } \sum\limits_{j = 0}^n {\binom{n}{j}^2 \left( {2(H_{n - j} - H_j )\log ((n + j + r)!) + \log (n + j + r)} \right)} = 0. $$ Denote the expression under the limit by $S_n(r)$. Performing the change of summation index $j\to n-j$ and taking the average with the original expression, we find $$ S_n (r) = \sum\limits_{j = 0}^n {\binom{n}{j}^2 \left( {(H_{n - j} - H_j )\log \frac{{(n + j + r)!}}{{(2n - j + r)!}} + \frac{1}{2}\log ((2n - j + r)(n + j + r))} \right)} . $$ Stirling's formula and the Maclaurin series of the logarithm then yields $$ S_n (r) = (\log r)\sum\limits_{j = 0}^n {\binom{n}{j}^2 \left( {(H_{n - j} - H_j )(2j - n) + 1} \right)} + o(1) $$ as $r\to+\infty$. Therefore, it remains to show that $$ \sum\limits_{j = 0}^n {\binom{n}{j}^2 ((H_{n - j} - H_j )(2j - n) + 1)} =0 $$ for all $n\ge 1$. Because of the symmetry in $H_{n - j} - H_j$, this may be further simplifed to the claim that $$ \sum\limits_{j = 0}^n {\binom{n}{j}^2 (2j(H_{n - j} - H_j )+ 1)} =0 $$ for all $n\ge 1$. Computer algebra software confirms this for $n=1,2,3\ldots,200$.

Addendum. The limit expression in the original version of Tyma Gaidash's answer was $$ I_n=\lim_{r\to\infty}\sum_{j=0}^n\binom n j^2\left(2 (H_{n-j}-H_j)\ln\left(\frac{(n+j+r)!}{(n+j)!}\right)+H_{n+j+r}-H_{n+j}\right). $$

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  • $\begingroup$ +1. Beautiful answer and well written. Can we show that $$ \sum\limits_{j = 0}^n {\binom{n}{j}^2 (2j(H_{n - j} - H_j )+ 1)} =0 $$ It seems to be true as you said for $n=1,2,3,...,200$. But how to show that it is $0$ for all $n$? $\endgroup$
    – Max
    Jul 31, 2023 at 4:18
  • $\begingroup$ @Max I have not succeeded in showing it for all $n\ge 1$ yet. That is why I say at the beginning that this is not a complete answer. $\endgroup$
    – Gary
    Jul 31, 2023 at 4:19
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    $\begingroup$ @Max You again totally modified your question by adding a new problem that does not follow from the previous ones in a straightforward manner. Post it as a separate question with explanation why you consider that particular limit. You should provide context. $\endgroup$
    – Gary
    Aug 5, 2023 at 8:51
  • $\begingroup$ Thanks for answering. I have accepted your answer. Its a request that please answer my question here or in the new question which I have asked here: math.stackexchange.com/questions/4746850/… $\endgroup$
    – Max
    Aug 7, 2023 at 16:04
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Here are the asked details after applying the Thomae formula $$I_n=n!^2\int_0^{\infty}\Gamma(x+n+1)\frac{\Gamma(2n+2)}{\Gamma(2n+3)\Gamma(3n+3+x)}\, _3\,F_2(n+1,n+1,2n+2;3n+3+x,2n+3;1)dx$$

$$=\frac{n!^2}{2n+2}\int_0^{\infty}\frac{\Gamma(x+n+1)}{\Gamma(3n+3+x)}\sum_{k=0}^{\infty}\frac{1}{k!}\frac{((n+1)_k)^2(2n+2)_k}{(2n+3)_k(3n+3+x)_k}$$$$=n!^2\int_0^{\infty}\frac{\Gamma(x+n+1)}{\Gamma(3n+3+x)}\sum_{k=0}^{\infty}\frac{1}{k!}\frac{((n+1)_k)^2}{(2n+2+k)(3n+3+x)_k}dx$$ Up to the appearance of a $2n+2+k$ in the denominator, we get a $_2\, F_1(n+1,n+1; 3n+3+x;1)$ function. Therefore another presentation is $$I_n=n!^2\int_0^{\infty}\frac{\Gamma(x+n+1)}{\Gamma(3n+3+x)}\left(\int_0^1z^{2n+2}\, _2\,F_1(n+1,n+1; 3n+3+x;z)dz\right)dz$$ I may have done some errors of computation...

In the other hand, there are completely different ways of presentation of $I_n$ by replacing $\sum_{k=1}^{\infty}B(n+k+x,n+1)^2$ by $$\int_0^1\int_0^1(t^xt^n(1-t)^n))(s^xs^n(1-s)^n))\sum_{k=1}^{\infty}(ts)^{k-1}dtds$$$$=\int_0^1\int_0^1(t^xt^n(1-t)^n))(s^xs^n(1-s)^n))\frac{1}{1-ts}dtds.$$ This leads to $$I_n=\int_0^1\int_0^1\frac{(st(1-t)(1-s))^n}{(1-ts)(-\log st)}dtds.$$

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  • $\begingroup$ +1 for your elegant answer. $\endgroup$
    – Max
    Jul 30, 2023 at 18:29
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$\def\H{\operatorname H}$ Applying partial fractions on $\frac{1}{(x(x+1)(x+2)\dots(x+n))^2}$ uses the harmonic numbers:

$$I_n=\sum_{k=1}^\infty\int_0^\infty\sum_{j=0}^n\binom nj^2\left(2\frac{\H_j-\H_{n-j}}{x+j+k+n}+\frac1{(x+j+k+n)^2}\right)dx$$

Integrating and evaluating the negative antiderivative at $x=0$:

$$I_n=\sum_{j=0}^n\binom n j^2\sum_{k=1}^\infty\left(2(\H_{n-j}-\H_j)\ln(j+k+n)+\frac1{j+k+n}\right)$$

@Gary sums over $k$ with the Euler-Mascheroni constant, to get:

$$\boxed{\sum_{k=1}^{\infty}\int_{0}^{\infty}\operatorname B^2(x+n+k,n+1)\ dx =\binom{2n}n\gamma-\sum_{j=0}^n\binom nj^2(2(\H_{n-j}-\H_j)\ln((j+n)!)+\H_{n+j})}$$

Results: $$I_1=2\gamma-\frac52+2\ln(2)\\I_2=6\gamma-\frac{131}{12}+6\ln(2)+3\ln(3)\\I_3=20\gamma-\frac{523}{12}+\frac{11}3\ln(4)+\frac{38}3\ln(5)+\frac{11}3\ln(6)\\I_4=70\gamma-\frac{143657}{840}+\frac{25}6\ln(5)+\frac{185}6\ln(6)+\frac{185}6\ln(7)+\frac{25}6\ln(8)\\I_5=252\gamma-\frac{168401}{252}+\frac{137}6\ln(6)+\frac{881}{15}\ln(7)+\frac{627}5\ln(8)+\frac{881}{15}\ln(9)+\frac{137}{30}\ln(10)\\I_6=924\gamma-\frac{14468369}{5544}+\frac{49}{10}\ln(7)+\frac{973}{10}\ln(8)+\frac{1799}5\ln(9)+\frac{1799}5\ln(10)+\frac{973}{10}\ln(11)+\frac{49}{10}\ln(12)\\\vdots$$

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  • $\begingroup$ +1 for your answer. Please edit your answer for general $n$ $\endgroup$
    – Max
    Jul 30, 2023 at 18:30
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    $\begingroup$ $$ I_n = \binom{2n}{n}\gamma - \sum\limits_{j = 0}^n {\binom{n}{j}^2 (2(H_{n - j} - H_j )\log ((j + n)!) + H_{n + j} )} $$ $\endgroup$
    – Gary
    Jul 31, 2023 at 0:10
  • $\begingroup$ If you replace the upper limit of integration by $r$ and put a $\lim_{r\to+\infty}$ in the front, you will end up with the formula at the bottom of my incomplete answer. $\endgroup$
    – Gary
    Jul 31, 2023 at 4:51
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This is just to add to the nice posted solutions the heuristic approach for finding the asymptotics of $I_n$ at $n\to\infty$. We can use the approach by @Gérard Letac and present $I_n$ as $$I_n=\sum_{k=1}^{\infty}\int_{0}^{\infty}(B(x+n+k,n+1))^2\ dx=\int_0^1\int_0^1\frac{\big(st(1-t)(1-s)\big)^n}{(1-ts)(-\log st)}dtds$$ Making the substitutions $s=x+\frac12;\,t=y+\frac12$ $$I_n=\int_{-1/2}^{1/2}\int_{-1/2}^{1/2}\frac{\big(\frac14-x^2\big)^n\big(\frac14-y^2\big)^n}{\big(\frac34-\frac x2-\frac y2-xy\big)\left(-\ln\big(\frac14+\frac x2+\frac y2+xy\big)\right)}dxdy$$ $$=\frac1{4^{2n}}\int_{-1}^1\int_{-1}^1\frac{(1-x^2)^n(1-y^2)^n}{(3-x-y-xy)\big(\ln 4-\ln(1+x+y+xy)\big)}dxdy\tag{1}$$ At $n\to\infty$ the integral gets the main contribution at small $\,x,\,y\,$; for the main asymptotic term we get $$I_n\sim\frac1{4^{2n}6\ln2}\int_{-1}^1\int_{-1}^1(1-x^2)^n(1-y^2)^ndxdy\sim\frac\pi{6\ln2}\frac{4^{-2n}}n$$ We can also get higher asymptotic terms - just decomposing the integrand in (1) near $\,x,\,y =0$ , and to make sure that $$\boxed{\,\,I_n=\frac\pi{6\ln2}\frac{4^{-2n}}n\left(1+\frac1n\Big(\frac1{4\ln^22}-\frac1{12\ln2}-\frac{23}{36}\Big)+O\left(\frac1{n^2}\right)\right)\,\,}$$ $\bf{Addendum}$

Asymptotics of $a_n:= \sum_{j=0}^n\binom nj^2(2(H_{n-j}-H_j)\ln((j+n)!)$

The answer was obtained by @Gary; I just want to add a bit different way of calculation by means of using Stirling's formula $$I_n=\binom{2n}n\gamma-\sum_{j=0}^n\binom nj^2(2(H_{n-j}-H_j)\ln((j+n)!)+H_{n+j})=\binom{2n}n\gamma-a_n-b_n$$ where we denoted $b_n=\sum_{j=0}^n\binom nj^2H_{n+j}$

As $\binom nj^2$ reaches a sharp maximum near $j=\frac n2$, we can choose $n$ even for a while, $j=\frac n2+k $ and present $b_n$ as $$b_n=\sum_{k=-n/2}^{n/2}\left(\frac{n!}{\big(\frac n2-k\big)!\big(\frac n2+k\big)!}\right)^2H_{\frac{3n}2+k}$$ Using the Stirling's formula for $p!$ ( for $p\gg1$) $$b_n\sim\sum_{k=-n/2}^{n/2}\left(\frac{\sqrt{2\pi n}\left(\frac ne\right)^n}{\sqrt{2\pi(\frac n2+k)}\sqrt{2\pi(\frac n2-k)}\Big(\frac{\frac n2+k}e\Big)^{\frac n2+k}\Big(\frac{\frac n2-k}e\Big)^{\frac n2-k}}\right)^2H_{\frac{3n}2+k}$$ The terms decline sharply as soon as $k$ excides $\sqrt n\,$ , so we can switch from summation to integration. Given that $H_{\frac{3n}2+k}$ is slowly changing function, we are allowed just to take its value at $k=0$ and use the asymptotics $H_\frac{3n}2=\gamma+\ln\frac{3n}2+O\big(\frac1n\big)$

After straightforward manipulations we get $$b_n\sim\frac{2\cdot4^n\big(\gamma+\ln\frac32+\ln n\big)}{\pi\sqrt n}\int_{-\infty}^\infty e^{-4t^2}dt=\frac{4^n}{\sqrt{\pi n}}\left(\gamma+\ln\frac32+\ln n\right)$$ Due to the fact that $$\binom{2n}n\sim \frac{4^n}{\sqrt{\pi n}}\to\infty\,\,\text{at}\,\,n\to\infty$$ and $I_n\to0$ at $n\to\infty$, we conclude that $$\boxed{\,\,a_n\sim-\frac{ 4^n}{\sqrt{\pi n}}\left(\ln\frac32+\ln n\right)\,\,}$$

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    $\begingroup$ (+1) Numerically confirmed. $\endgroup$
    – Gary
    Aug 1, 2023 at 23:24
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    $\begingroup$ Instead of editing the question over and over again and adding new problems, please ask a new question on the site and refer back to this one. $\endgroup$
    – Gary
    Aug 2, 2023 at 21:51
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This answers the question about the asymptotics of $a_n$ provided that the conjectured formula for $I_n$ in my other answer is correct. Note that since $H_k=\psi(k+1)+\gamma$, we can write $$ a_n = - I_n - \sum\limits_{j = 0}^n {\binom{n}{j}^2 \psi (n + j + 1)} = - \sum\limits_{j = 0}^n {\binom{n}{j}^2 \psi (n + j + 1)} + o(1). $$ The $o$-term follows from Svyatoslav's answer. Now by the asymptotic result $\psi(k+1)=\log k +\mathcal{O}(k^{-1})$, we have \begin{align*} \sum\limits_{j = 0}^n {\binom{n}{j}^2 \psi (n + j + 1)} & = \sum\limits_{j = 0}^n {\binom{n}{j}^2 \log (n + j)} + \mathcal{O}(1)\sum\limits_{j = 0}^n {\binom{n}{j}^2 \frac{1}{{n + j + 1}}} \\ & = \sum\limits_{j = 0}^n {\binom{n}{j}^2 \log (n + j)} + \mathcal{O}\!\left( {\frac{1}{n}} \right)\binom{2n}{n}. \end{align*} Since $\log n \le \log(n+j)\le \log n+\log 2$, it follows that $$ a_n \sim -\binom{2n}{n}\log n $$ as $n\to+\infty$.

Addendum. Performing the change of summation index $j\to n−j$ and taking the average with the original expression, we find $$ \sum\limits_{j = 0}^n {\binom{n}{j}^2 \log (n + j)} = \sum\limits_{j = 0}^n {\binom{n}{j}^2 \log \sqrt {(n + j)(2n - j)} } . $$ Now \begin{align*} \log \sqrt {(n + j)(2n - j)} & = \log \left( {\frac{{3n}}{2}} \right) + \log \sqrt {1 - \frac{4}{9}\left( {\frac{1}{2} - \frac{j}{n}} \right)^2 } \\ & = \log \left( {\frac{{3n}}{2}} \right) + \mathcal{O}(1)\left( {\frac{1}{2} - \frac{j}{n}} \right)^2 . \end{align*} Numerics suggest that $$ \sum\limits_{j = 0}^n {\binom{n}{j}^2 \left( {\frac{1}{2} - \frac{j}{n}} \right)^2 } \sim \frac{1}{{8n}}\binom{2n}{n}, $$ as $n\to+\infty$. This would lead to the more precise result $$ a_n = -\binom{2n}{n}\left(\log \left( {\frac{{3n}}{2}} \right) +\mathcal{O}\!\left( {\frac{1}{n}} \right)\right) $$ as $n\to+\infty$.

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    $\begingroup$ (+1) Yes, the approach is clear. I got the same result in a bit different way $\endgroup$
    – Svyatoslav
    Aug 2, 2023 at 2:43
  • $\begingroup$ ($+1$) for your elegant answer. Thank you! $\endgroup$
    – Max
    Aug 2, 2023 at 3:08
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    $\begingroup$ Instead of editing the question over and over again and adding new problems, please ask a new question on the site and refer back to this one. $\endgroup$
    – Gary
    Aug 2, 2023 at 21:51
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A way to simplify $_3\,F_2$ is to use the Thomae formula which says

$$s(A,B,C,D,E)=\frac{\Gamma(C)}{\Gamma(D)\Gamma(E)}_3\,F_2(A,B,C;D,E;1)= $$ $$s(D-C,E-C,D+E-A-B-C, D+E-A-C,D+D-B-C)$$ In your integral formula using $A=1, \ B=x+n+1, $ etc, the new $_3\,F_2$ seems to contain $x$ only in $D+E-A-C.$ I became aware of this formula while writing

Letac, G. and Piccioni, M. (2012) 'Random continued fractions with beta hypergeometric distribution. Ann. Probab. vol 40, number 3, 1105-1134

containing all references and our proof of the Thomae formula.

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    $\begingroup$ Thank you. Sorry, I cannot see how to get l a closed form from this. Please help me regarding this $\endgroup$
    – Max
    Jul 29, 2023 at 10:39
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    $\begingroup$ +1. Please help me to write a closed form for $I_n$ $\endgroup$
    – Max
    Jul 29, 2023 at 10:41
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    $\begingroup$ May be it does not exist. My suggestion only reduces the number of products of $x\mapsto \Gamma(x+c)$ in the sum of the series. $\endgroup$ Jul 29, 2023 at 10:54
  • $\begingroup$ Please answer so that I can with utmost respect accept your answer. $\endgroup$
    – Max
    Aug 5, 2023 at 7:56

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