Wikipedia:Reference desk/Archives/Mathematics/2013 February 27

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February 27

The Gamma function and Euler's constant (e)

We know that e is defined as the (discrete) sum of the multiplicative inverses of the factorials (of natural numbers): $e=\sum _{n=0}^{\infty }{\frac {1}{n!}}$ . Now, since the integral is the continuous equivalent of a discrete sum, and the gamma function the generalization of the factorial function, it would follow that $\int _{0}^{\infty }{\frac {dx}{\Gamma (x)}}$ would be the continuous equivalent of the mathematical definition of the number e. When I tried to compute this value with Mathematica, it yielded a value close, though not identical, to that of the number e: 2.807770... instead of 2.71828..., the error being about 0.089... I guess my question would be two-fold: (1) is this correct ?, and (2) does it have any meaning ? (Like, for instance, the constant $\gamma$, which is defined as the difference between the [discrete] sum of the inverses of natural numbers, a.k.a the harmonic series, and the [continuous] integral of the same: $\gamma =\sum _{n=1}^{\infty }{\frac {1}{n}}-\int _{1}^{\infty }{\frac {dx}{x}}$). — 79.113.224.159 (talk) 15:56, 27 February 2013 (UTC)

Shouldn't the analogous integration run from 1 to infinity? Because $\Gamma (x)=(x-1)!$. Icek (talk) 23:27, 27 February 2013 (UTC)

No. That yields 2.2665..., which lies even further away from e. (Basically, starting off at 1, or any point other than 0, would be a random choice, and a simple look at the function's graph suffices to understand why). — 79.113.224.159 (talk) 05:57, 28 February 2013 (UTC)

0 is just as an arbitrary choice as any other number. Besides the point was that 0! = Γ(1), which is in fact corroborated and not refuted by your link. Lastly there is no reason why taking the continuous limit of an summation must yield the integral which is the closest approximation to the sum. — Preceding unsigned comment added by 202.65.245.7 (talk) 13:56, 28 February 2013 (UTC)

...and Gamma(1) is basically a vertical segment, whose surface (which is what the integral function ultimately calculates) is 0. In order for us to have a surface, we must also have a width, not just a height (which is the geometrical reflection of the function's value). We gain that width by starting a step before, at 0, instead of 1. Obviously, a discrete sum is the sum of segment-lengths, so we don't need to create any surfaces there. — 79.113.197.109 (talk) 17:37, 28 February 2013 (UTC)

Arguing this way you should start integrating at 0 for the harmonic series... which would make the Euler-Mascheroni constant infinite. Icek (talk) 20:51, 28 February 2013 (UTC)

You're right, I even wrote the definition of $\gamma$ wrong the first time (now it's corrected). I guess the explanation is that this is the domain on which the two functions are defined, or that they both have in common: 0 to $\infty$. In the case of $\gamma$, if we'd extend the domain to include the interval [0, 1], we'd get:

$\gamma _{new}=\gamma +\lim _{x\to 0}({\frac {1}{x}}-\int _{0}^{1}{\frac {dx}{x}})=\gamma +\lim _{x\to 0}({\frac {1}{x}}+\ln x)=\gamma +\lim _{y\to \infty }(y-\ln y)=\gamma +\lim _{t\to \infty }(e^{t}-t)=\infty$.

This is known as the Fransén–Robinson constant. DTLHS (talk) 05:59, 28 February 2013 (UTC)

Thanks ! :-) — 79.113.224.159 (talk) 06:08, 28 February 2013 (UTC)

Resolved