5
$\begingroup$

The last piece I am left with in my proof is to compute the following two series:

$$\sum_{i=1}^{n-1}\dfrac{\Gamma(i-d)\Gamma(n-i+d)}{\Gamma(i+1)\Gamma(n-i)(n-i-d)}, \sum_{i=1}^{n-1} \dfrac{\Gamma(n-d-i)\Gamma(i+d)}{(i-d)\Gamma(i+1)\Gamma(n-i)},$$

here $n\geq 2$, $d\in(-0.5, 0.5)$. According to Maple that would be

$$ \begin{align} & \sum_{i=1}^{n-1}\dfrac{\Gamma(i-d)\Gamma(n-i+d)}{\Gamma(i+1)\Gamma(n-i)(n-i-d)}= \\ &=\dfrac{\Gamma(-d)}{2}\left[\dfrac{d\Gamma(-d)\Gamma(d)\Gamma(n-2d)\Gamma(n)-2\Gamma(n+d)\Gamma(n-d)\Gamma(-2d)}{(n-d)\Gamma(n)\Gamma(n-d)\Gamma(-2d)}\right] \\ &= \dfrac{d\Gamma(-d)^2\Gamma(d)\Gamma(n-2d)}{2\Gamma(n-d+1)\Gamma(-2d)}-\dfrac{\Gamma(-d)\Gamma(n+d)}{(n-d)\Gamma(n)} \end{align} $$ and $$ \begin{align} &\sum_{i=1}^{n-1} \dfrac{\Gamma(n-d-i)\Gamma(i+d)}{(i-d)\Gamma(i+1)\Gamma(n-i)}= \\ &=\dfrac{\Gamma(d)}{2}\dfrac{d\Gamma(-d)^2\Gamma(n-2d)\Gamma(n)+2\Gamma(n-d)^2\Gamma(-2d)}{d\Gamma(-2d)\Gamma(n-d)\Gamma(n)} \\ &= \dfrac{\Gamma(d)\Gamma(-d)^2\Gamma(n-2d)}{2\Gamma(-2d)\Gamma(n-d)} + \dfrac{\Gamma(n-d)\Gamma(d)}{d\Gamma(n)}. \end{align} $$

I have already simplified these series as much as possible and could not find any more elementary properties of the $\Gamma$ function that would make these expressions more convenient. Therefore, I thought I should use one of the definitions of the $\Gamma$ function. Also I believe it should be something that involves integrals or sums (rather than products) in order to change the order of summation and apply $\sum_{i=1}^{n-1}$ first.

However, writing the first series as $$ \sum_{i=1}^{n-1}\dfrac{\int_0^\infty e^{-u}u^{i-d-1}du\int_0^\infty e^{-v}v^{n-i+d-1}dv}{(n-i-d)\int_0^\infty e^{-z}z^idz\int_0^\infty e^{-w}w^{n-i-1}dw\int_0^\infty e^{-r}r^{n-i-d-1}dr} = $$ $$ \sum_{i=1}^{n-1}\dfrac{\int^\infty_0 \int^\infty_0 e^{-u-v}u^{i-d-1}v^{n-i+d-1}dudv}{(n-i-d) \int^\infty_0 \int^\infty_0 \int^\infty_0 e^{-z-w-r}z^{i}w^{n-i-1}r^{n-i-d-1}dzdwdr} $$

does not help at all because of denominator.

I also tried to rewrite it (both the series and the desired result) using the Beta function: $$ \begin{align} \sum_{i=1}^{n-1}\dfrac{\Gamma(i-d)\Gamma(n-i+d)}{\Gamma(i+1)\Gamma(n-i)(n-i-d)} &= \sum_{i=1}^{n-1}\dfrac{B(d+1,i-d)B(n-i+d,1-d)B(n-i-d,1) (i-n)}{\Gamma(1+d)\Gamma(-d)d}\\ &= \dfrac{B(-d,-d)B(n-2d,d)B(n-i-d,1) + 2B(-d,n+d)}{2(n-d)} \end{align} $$ and to use a couple of definitions: $$ B(x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,d\theta, \qquad \mathrm{Re}(x)>0,\ \mathrm{Re}(y)>0, $$ $$ B(x,y) = \int_0^\infty\dfrac{t^{x-1}}{(1+t)^{x+y}}\,dt, \qquad \mathrm{Re}(x)>0,\ \mathrm{Re}(y)>0. $$ However, even though they allowed me to use geometric progression, resulting expressions were too cumbersome (or should I try harder?). Even if this is a correct way, I cannot see how to go from a triple integral (geometric progression, cumbersome expression) to a triple + single integral, i.e. $$ \dfrac{B(-d,-d)B(n-2d,d)B(n-i-d,1) + 2B(-d,n+d)}{2(n-d)} $$ (here I ignore requirements $\mathrm{Re}(x)>0$, $\mathrm{Re}(y)>0$, which can be easily satisfied).

In case all this is not terribly difficult I would like only a hint first.


This does not really help, but since there seems to be some particular way to evaluate this kind of series I tried to take some simpler one first and just got annoyed by such an elegant answer (the one I got from Maple, of course):

$$\sum_{i=1}^{n-1}\dfrac{\Gamma(i-d)\Gamma(n-i+d)i}{\Gamma(i+1)\Gamma(n-i)}=\Gamma(1-d)\Gamma(1+d)(n-1)=\dfrac{d(1-n)\pi}{\sin(\pi(1+d))}$$

(comparing with the very first series above $\frac{1}{n-d-i}$ here is just replaced with $i$).


Edit: the desired results also can be rewritten to

$$ \begin{align} & \sum_{i=1}^{n-1}\dfrac{\Gamma(i-d)\Gamma(n-i+d)}{\Gamma(i+1)\Gamma(n-i)(n-i-d)} \\ &= \dfrac{\Gamma(-d)\Gamma(n+d)}{(n-d)\Gamma(n)} \left(\dfrac{d\Gamma(2d)\Gamma(n)\Gamma(d-n+1)}{\Gamma(n+d)\Gamma(2d-n+1)} - 1\right), \\ &\sum_{i=1}^{n-1} \dfrac{\Gamma(n-d-i)\Gamma(i+d)}{(i-d)\Gamma(i+1)\Gamma(n-i)} \\ &=-\dfrac{\Gamma(d)\Gamma(n-d)}{d\Gamma(n)} \left(\ \dfrac{\Gamma(n)\Gamma(n-2d)\Gamma(1-d)^2}{\Gamma(1-2d)\Gamma(n-d)^2} -1\right). \end{align} $$

$\endgroup$
  • $\begingroup$ The numerator has $\Gamma(i-d)\Gamma(n-[i-d])$, which might perhaps be rephrased using a combination between the reflection formula and the multiplication formula. $\endgroup$ – Lucian Dec 24 '13 at 16:07
  • $\begingroup$ @Lucian, thank you for an idea. I might be missing something, but I cannot see how to use the multiplication formula in this case. It seems that after applying $\Gamma(s+1)=s\Gamma(s)$ multiple times we are able to use the reflection formula, but then we are also left with a product $(n-i+d-1)\dots(1-i+d)$ which does not help since there is a single term in the denominator that contains $d$. $\endgroup$ – Julius Dec 24 '13 at 22:20
6
$\begingroup$

We can use the Generalized Hypergeometric function \begin{equation*} \,{}_pF_q(a_1,\ldots,a_p;b_1,\ldots,b_q;z) = \sum_{n=0}^\infty \frac{(a_1)_n\dots(a_p)_n}{(b_1)_n\dots(b_q)_n} \, \frac {z^n} {n!}, \end{equation*} with Pochhammer symbol defined by \begin{align*} (a)_0 &= 1, \\ (a)_n &= a(a+1)(a+2)...(a+n-1)=\dfrac{\Gamma(a+n)}{\Gamma(a)}, && n \geq 1, \end{align*} and evaluate the first series in the following way: \begin{align*} &\sum_{i=1}^{n-1}\dfrac{\Gamma(i-d)\Gamma(n-i+d)}{\Gamma(i+1)\Gamma(n-i)(n-i-d)} \\ &= \dfrac{\Gamma(1-d)\Gamma(n-1+d)}{\Gamma(n-1)(n-1-d)} {}_4F_3(1, 1 - d, 2 - n, 1 + d - n; 2, 2 + d - n, 2 - d - n; 1)) \\ &= \dfrac{\Gamma(1-d)\Gamma(n-1+d)}{\Gamma(n-1)(n-1-d)} \sum_{m=0}^\infty \dfrac{(1)_m(1-d)_m(2-n)_m(1+d-n)_m}{(2)_m(2+d-n)_m(2-d-n)_m}\dfrac{1}{m!} \\ &= \dfrac{\Gamma(1-d)\Gamma(n-1+d)}{\Gamma(n-1)(n-1-d)} \sum_{m=0}^\infty \dfrac{(1-d)_m(2-n)_m(1+d-n)_m}{(m+1)(2+d-n)_m(2-d-n)_m}\dfrac{1}{m!} \\ &= \dfrac{\Gamma(1-d)\Gamma(n-1+d)}{\Gamma(n-1)(n-1-d)} \sum_{m=0}^\infty \dfrac{(1-d)_m(2-n)_m(1+d-n)_m}{(2+d-n)_m(2-d-n)_m}\dfrac{1}{(m+1)!} \\ &= \dfrac{\Gamma(1-d)\Gamma(n-1+d)}{\Gamma(n-1)(n-1-d)} \dfrac{(1+d-n)(1-n-d)}{-d(1-n)(d-n)}\sum_{m=0}^\infty \dfrac{(-d)_{m+1}(1-n)_{m+1}(d-n)_{m+1}}{(1+d-n)_{m+1}(1-d-n)_{m+1}}\dfrac{1}{(m+1)!} \\ &= \dfrac{\Gamma(-d)\Gamma(n+d)}{(n-d)\Gamma(n)} ({}_3F_2(-d,1-n,d-n;-d-n+1,d-n+1;1)-1) \\ &= \dfrac{\Gamma(-d)\Gamma(n+d)}{(n-d)\Gamma(n)} \left(\dfrac{(2d-n+1)_{n-1}(1)_{n-1}}{(d-n+1)_{n-1}(d+1)_{n-1}}-1\right) \\ &= \dfrac{\Gamma(-d)\Gamma(n+d)}{(n-d)\Gamma(n)} \left(\dfrac{\Gamma(2d)\Gamma(n)\Gamma(d-n+1)\Gamma(d+1)}{\Gamma(d)\Gamma(n+d)\Gamma(2d-n+1)} - 1\right)\\ &= \dfrac{\Gamma(-d)\Gamma(n+d)}{(n-d)\Gamma(n)} \left(\dfrac{d\Gamma(2d)\Gamma(n)\Gamma(d-n+1)}{\Gamma(n+d)\Gamma(2d-n+1)} - 1\right),\\ \end{align*} here we used Saalschütz's theorem: \begin{equation*} {}_3F_2 (a,b, -n;c, 1+a+b-c-n;1)= \frac{(c-a)_n(c-b)_n}{(c)_n(c-a-b)_n}. \end{equation*}

In the same way we evaluate the second series: \begin{align*} &\sum_{i=1}^{n-1} \dfrac{\Gamma(n-d-i)\Gamma(i+d)}{(i-d)\Gamma(i+1)\Gamma(n-i)} \\ &= \dfrac{\Gamma(1+d)\Gamma(n-d-1)}{\Gamma(n-1)(1-d)} {}_4F_3(1, 1-d, 2-n, 1+d; 2, 2-d, d-n+2; 1)) \\ &= \dfrac{\Gamma(1+d)\Gamma(n-d-1)}{\Gamma(n-1)(1-d)} \dfrac{(1-d)(d-n+1)}{d^2(n-1)}({}_3F_2(-d,d,1-n;1-d,d-n+1,1)-1) \\ &= -\dfrac{\Gamma(d)\Gamma(n-d)}{d\Gamma(n)} \left(\ \dfrac{(1)_{n-1}(1-2d)_{n-1}}{((1-d)_{n-1})^2} -1\right) \\ &= -\dfrac{\Gamma(d)\Gamma(n-d)}{d\Gamma(n)} \left(\ \dfrac{\Gamma(n)\Gamma(n-2d)\Gamma(1-d)^2}{\Gamma(1-2d)\Gamma(n-d)^2} -1\right). \\ \end{align*}

$\endgroup$
0
$\begingroup$

I'm not sure how useful this hint will be, but here it is:

$$\text{Since we know that}\;\;\Gamma(s+1)=s\Gamma (s)\;,\;\;\text{we get that}$$

$$\dfrac{\Gamma(i-d)\Gamma(n-i+d)}{\Gamma(i+1)\Gamma(n-i)(n-i-d)}=\frac{\color{red}{\Gamma(i-d)}\cdot\prod_{k=1}^d(n-i+d-k)\color{green}{\Gamma(n-i)}}{\prod_{k=0}^d(k-d)\cdot\color{red}{\Gamma(i-d)}\color{green}{\Gamma(n-i)}\Gamma(n-i-d)}$$

$\endgroup$
  • $\begingroup$ Just added to my question that $d\in(-0.5, 0.5)$ so, unfortunately, this does not apply. $\endgroup$ – Julius Dec 24 '13 at 16:07
  • $\begingroup$ Well, it still applies but it doesn't come out as pretty as expected...:) $\endgroup$ – DonAntonio Dec 24 '13 at 16:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.