Prof. Putinar's Infinite Product: Ben Orin's Addition

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Prof. Putinar,

Guillera & Sondow gave

e^{x}=\prod_{n=1}^{\infty}\left(\prod_{k=1}^{n} (1+kx)^{(-1)^{k+1}\left(\begin{array}{c}n\\k\end{array}\right)} \right) ^{\frac{1}{n}}​

for x\geq 0, to which I add

\boxed{\frac{e^{u}\Gamma (u)}{\sqrt{2\pi e}}=\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n} (k+u)^{(-1)^{k}\left(\begin{array}{c}n\\k\end{array}\right) (k+u)} \right) ^{\frac{1}{n+1}}}​

for \mbox{Re} <u>\geq 0</u>.

-Ben Orin
 
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Cool.



















:rolleyes:
 
The paper of Guillera & Sondow cited above is entitled http://arxiv.org/PS_cache/math/pdf/0506/0506319.pdf and their result (the infinite product for e^x) is equation (58) on pg. 18.
 
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An Infinite Product for e^x

Edit: Fixing my post for TeX and updating link to paper.

Prof. Putinar,

Guillera & Sondow1 gave e^{x} = \prod_{n=1}^{\infty}\left( \prod_{k=1}^{n} (kx+1) ^{(-1)^{k+1} \left( \begin{array}{c}n\\k\end{array}\right) } \right) ^{\frac{1}{n}}\mbox{ for }x\geq 0,
to it's company I add \frac{e^{u}\Gamma (u)}{\sqrt{2\pi e}}=\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n}(k+u)^{<br /> (-1)^{k}\left(\begin{array}{c}n\\k\end{array}\right)(k+u)}\right)^{\frac{1}{n+1}}\mbox{ for }\mbox{Re} <u>\geq 0.</u>



-Ben Orin

benorin@umail.ucsb.edu

1 The infinite product for e^{x} is (60) on pg. 20 of http://arxiv.org/abs/math/0506319" .)
 
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