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I am solving an integral that appears in a physics paper. $$ -\int_0^{\infty}dt\,\frac{e^{-t}}{t}\bigg[\bigg(1+\frac{3}{N}t\bigg)^N-1\bigg] $$ The paper does not give the full solution, it only gives the behavior of the solution when $N$ is big $$ -\frac{3}{2}\sqrt{\frac{2\pi}{N}}\bigg(\frac{3}{e^{2/3}}\bigg)^N $$ I have not been able to get this big $N$ behaviour but I have a better thing, I solved the integral exactly. The solution is $$ -\int_0^{\infty}dt\,\frac{e^{-t}}{t}\bigg[\bigg(1+\frac{3}{N}t\bigg)^N-1\bigg]=-\sum_{m=1}^N\frac{N!}{(N-m)!m}\bigg(\frac{3}{N}\bigg)^m $$ I want to obtain the alleged big $N$ behaviour. There is only one thing said in the paper about how this is done. The author says that the integral can be expressed in terms of confluent hypergeometric functions.

I have tried to look in the wiki page for confluent hypergeometric functions but there is nothing readily useful with my level of knowledge. Thus, does anybody here have a clue how the integral above is related to confluent hypergeometric functions?

The paper in question is http://arxiv.org/pdf/hep-ph/0505034v1.pdf. Check the footnote at page 15.

1 Answers1

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Let: $\mathrm{aa : =} \frac{x^m n!}{(- m + n) !m}$

Then Hypergeometric summing with respect to m you get:

$\mathrm{hypergeom} \left( \left\{ - n + 1 \hspace{0.17em} \mathrm{, \hspace{0.17em}} \hspace{0.17em} 1 \hspace{0.17em} \mathrm{, \hspace{0.17em}} \hspace{0.17em} 1 \right\}, \{ 2 \}, - x \right) nx$

For the proof we need
$Definition. Pochhammer: \left(z\right)_{k}=\frac{\Gamma\left(z+k\right)}{\Gamma\left(z\right)}$ for $z\in\mathbb{R}$
and when $z\in\mathbb{Z}_{+}$ use:

$\left(z\right)_{k}=\frac{\left(z+k-1\right)!}{\left(z-1\right)!}$

$\left(-z\right)_{k}\left(-x\right)^{k}=\begin{cases} x^{k}\cdot{\frac{\left(z\right)!}{\left(z-k\right)!}} & k\leq\left(z\right)\\ 0 & k>\left(z\right) \end{cases}$

The last is a polynomial. This relation can be derived by examining the poles and zeros in the Pochhammer/Gamma definition above. This particular use of the Gamma definition to extend the Pochhammer symbol seems to be lacking in both DLMF and Wikipedia. One can use a standard table like [http://functions.wolfram.com/GammaBetaErf/Pochhammer/27/01/||Wolfram-Pochhammer] .

$Theorem. \sum_{m=1}^{N}\frac{N!}{(N-m)!m}x^{m}=N\cdot x\cdot Hypergeometric(\{\text{−}N+1,1,1\},\{2\},\text{−}x)$

Proof.
Like a lot of other operations with Hypergeometric functions this can be considered a rewrite; but rewrites can embed a problem into a form that embeds calculations into domains that are well researched. Also notice that since it is a rewrite the fact that $x=\frac{3}{N}$ is irrelevant.

$ Hypergeom(\{\text{−}N+1,1,1\},\{2\},\text{−}x)={\displaystyle \sum_{k=0}^{\infty}}\frac{(-N+1)_{k}\left(1\right)_{k}\left(1\right)_{k}}{\left(2\right)_{k}}\frac{\left(-x\right)^{k}}{k!}$

Examining our case term-wise:

$\left(1\right)_{k}=k!$

$\left(2\right)_{k}=\left(k+1\right)!$

$\frac{\left(1\right)_{k}}{\left(2\right)_{k}}=\frac{1}{k+1}$

$\left(-N+1\right)_{k}\left(-x\right)^{k}=\begin{cases} x^{k}\cdot{\frac{\left(N-1\right)!}{\left(N-1-k\right)!}} & k\leq\left(N-1\right)\\ 0 & k>\left(N-1\right) \end{cases}$

Thus the Hypergeometric term is:

$\frac{\left(N-1\right)!\cdot\left(k\right)!\cdot\left(k\right)!}{\left(N-1-k\right)!\cdot\left(k+1\right)!}\frac{x^{k}}{k!}=\frac{\left(N-1\right)!}{\left(N-1-k\right)!\cdot\left(k+1\right)}x^{k}$

Now we assign $m=k+1$ and sum

${\displaystyle \sum_{m=1}^{\infty}}\frac{\left(N-1\right)!}{\left(N-m\right)!\cdot\left(m\right)}x^{m-1}$

and add some indexing corrections to get

$N\cdot x\cdot Hypergeometric(\{\text{−}N+1,1,1\},\{2\},\text{−}x)$ $=N\cdot x\cdot{\displaystyle \sum_{m=1}^{\infty}}\frac{\left(N-1\right)!}{\left(N-m\right)!\cdot\left(m\right)}x^{m-1}=\sum_{m=1}^{N}\frac{N!}{(N-m)!m}x^{m}$

rrogers
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  • How does this provide an asymptotics when $N\to\infty$? – Did May 25 '16 at 15:53
  • @rrogers I appreciate the effort rrogers but I don't quite understand your answer. Any further clarification would be appreciated! – PhoenixPerson May 25 '16 at 19:22
  • @Did Your right I only answered 1/2. I will fix it. – rrogers May 26 '16 at 22:20
  • @AnarchistBirdsWorshipFungus I did the explanation but haven't quite found a asymptotic approach. For another linkage to Confluent Hypergeometric functions look at DLMF 13.4.3: http://dlmf.nist.gov/13.4.E4 . Although this particular case a=0 is explicitly forbidden; it can be avoided by some tricks. But they are not illuminating. This sort of reasoning should be amenable to a Turing machine type of basis/operation; like the "Method of Coefficients" written on top of CoQ. I asked on math.stackexchange but got no responses. "tumbleweed" which is essentially a "raspberry" ; a slang term. – rrogers May 30 '16 at 14:13
  • @AnarchistBirdsWorshipFungus Incidentally the {1},{2} and multiplication by x combination can be replaced by integration and be removed if your interested. A neat trick I haven't found a use for. – rrogers May 30 '16 at 14:24
  • @rrogers thaks for the clarification. One thing though, is there no way by takingthe $N$ big limit to write the $3F1$ we have as a $2F1$? – PhoenixPerson May 31 '16 at 09:26
  • @rrogers about the neat trick you mention. Anything that could be useful will be welcome – PhoenixPerson May 31 '16 at 12:36
  • @AnarchistBirdsWorshipFungus Sure: how do you suggest the posting? An addition to the above, a new answer, or... I set up a chat.stackexchange room https://chat.stackexchange.com/rooms/40512/generalized-hypergeometric-functions If you want to discuss this; although I am not full knowledgeable about this editor. I basically cut and paste Latex from Lyx to do the math. – rrogers May 31 '16 at 14:16
  • @rrogers for conciseness I would add an edit. Thanks! – PhoenixPerson May 31 '16 at 14:18
  • @rrogers what is the dot in $(z)!$ that you write in the formula for the pochhammer function you write in the 7th line? – PhoenixPerson May 31 '16 at 14:43
  • @AnarchistBirdsWorshipFungus That should have been a \cdot and is fixed. Incidently all of the \cdot are eye candy to replace implied multiplication anywhere I think that jamming symbols together might be misleading. For instance I might put one in if I wanted to multiply 5 times 4. I do have the integration as an exercise; it's simple and eliminates tha {1},{2} but haven't been managed to make it usefull since eliminating them leaves a 1 for the bottom part i.e. 2F0 ; and straight-normal generalized hypergeometric integration gives a zero divisor that I have to dodge or cancel. – rrogers May 31 '16 at 22:28
  • There are two (?) ways to avoid that but controlling the expressions to be usefull has taken time. – rrogers May 31 '16 at 22:29
  • @AnarchistBirdsWorshipFungus Would you like to continue this discussion (not results) in the chat room? The system apparently thinks it's getting too extended to be usefull. – rrogers May 31 '16 at 22:39
  • maybe you are interested in my follow-up question http://math.stackexchange.com/questions/1807126/can-you-prove-the-following-formula-for-hypergeometric-functions where I state a conjecture for the asymtotic expansion – PhoenixPerson Jun 01 '16 at 12:02