In another question of mine at mse I had the detail of the two assumed identities $$ C_{b,p} = \int_0^\infty b^{x^p} dx $$ and $$ C_{b,p}={n! \over (- \beta)^n} $$
for some $ \small n \in \mathbb N $ and $ \small p= \frac 1n $ which I'd checked with some $0 \lt b \lt 1$, $ \small \beta=\ln(b)$ .

I played a bit with my formulae and got aware, that if we rewrite the second identity $$ C_{b,p}={ \Gamma(1+n) \over (- \beta)^n} $$

it looks as if we can extend this even to noninteger $n$ and we'll actually get by this an integral formula for the $\Gamma$ in the form of: $$ \Gamma(n) = {(-\ln(b))^n \over n }\int_0^\infty \exp \left( \ln (b) x^{1/n} \right) dx \tag 1 $$

I've looked at wikipedia and HOMF (E.Weissstein seems to be inaccessible???) but could not find a similar formula which would allow me to check more whereabouts (range of convergences in ranges of the parameters etc.), so I ask here, whether

  1. this formula is basically correct/applicable for convergent cases

  2. and what the allowed ranges for $b$ and $n$ are

  3. and (since I'm still terribly weak in the manipulation with integrals)
    how that formula (1) could be related to the standard integral definition of the $\Gamma$ if this is possible...

[update:] Example Pari/GP-code gives equality up to machine precision (200 dec digits selected)

[b=0.5,p=0.5+Pi/1000]    \\ set parameters for convergent cases
                         \\ here n =1/p shall be noninteger
 %1516 = [0.500000000000, 0.503141592654]

[w=log(b),    n=1/p ]
      %1517 = [-0.693147180560, 1.98751209322]

v1  = (-w)^n/n * intnum(x=0,9999999, b^x^p)  \\ int_0^oo 
 %1518 = 0.994784377172

v2 =  gamma(n)
 %1519 = 0.994784377172

 %1520 = -2.077003953 E-200

This is most easily handled by setting $a=-\ln b$. Assuming $n>0$, you must have $a>0$ (i.e., $b<1$), else the integral in (1) won't converge. You can now substitute $x=(u/a)^n$ in the integral, to get $$\int_0^\infty \exp(-ax^{1/n})\,dx=\frac{n}{a^n}\int_0^\infty u^{n-1}e^{-u}\,du=\frac{n}{a^n}\Gamma(n).$$ This should work for any $n>0$ and $a>0$. The integral in (1) will also diverge if $n<0$, since then the integrand goes to $1$ as $x\to\infty$.

  • $\begingroup$ @Did Oh, duh. Yeah, I should have spotted that, but I got interrupted and had to run just before posting. Thanks for pointing it out, I have edited the answer accordingly. $\endgroup$ – Harald Hanche-Olsen Jan 20 '13 at 19:54

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