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I could not find a tag for incomplete gamma function, so I've just used the gamma function one.

A rather curious fact about the normalised incomplete gamma function$${1\over n!}\operatorname{\Gamma}(n,\mu)=\int_\mu^\infty x^{n-1}e^{-x}dx$$ (or is that the complementary normalised incomplete gamma function?) is that it is equal to the hypervolume cut out by the hypersurface (a hyperhyperbola, maybe!) $$\prod_{k=1}^n x_k=e^{-\mu}$$ (the $x_k$ being coordinates in $n$-dimensional space) of the positive unit hyperbox with a corner at the origin, bounden by all the planes $x_j=0$ & $x_j=1$. (So it doesn't matter very much whether it's the IGF or the complementary IGF that is under consideration, as the other function is just the hypervolume in the other part of the box.

I know a rather gross proof of this that consists in integrating the whole of the present function's representation as such a hypervolume along the new axis to get the next function up in the series of increasing $n$: the general case of this can without tremendous difficulty be 'abstracted' from the first two or three by broaching yet another curious theorem of binomial coefficients; but I am not much enamoured of the idea of setting it out, and I suppose few of you are strongly desirous of seeing it.

But the question is - does anyone know a slick proof of the theorem that is the chief subject of this post? - by which I mean a proof that proceeds along somekind of shortcut or 'wormhole' through the space of general function-theory, without recourse to just taking the thing apart & reconstructing it cog-by-cog, as in my method of proof ... but I think the meaning of slick proof is well-enough known amongst us! I feel that there ought to be and probably is such a proof - my intuition just cannot rest in the idea that there isn't one.

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  • $\begingroup$ Change of variable $y_k = \log(x_k)$ ? $\endgroup$ – reuns Dec 4 '18 at 0:31
  • $\begingroup$ So what would by the volume enclosed in that space? It would be just a hyperplane across the hypercorner, I think, wouldn't it. The region would have a simpler shape ... but then to recover the original hypervolume there would have to be a scaling that varies from point-to-point within the area, that would have to be built-into the integral. Unfortunately, I don't think there's any elementary relation between exponential of integral & integral of exponential (of function): I was wondering about that when I failed to find out what integral of gamma function is, but got one for lnΓ. $\endgroup$ – AmbretteOrrisey Dec 4 '18 at 0:41
  • $\begingroup$ @reuns -- Maybe that scaling I mentioned wouldn't be too bad: coz if x_k=e^-y_k, then Π{k=1 to n}dx_j=-Π{k=1 to n}e^y_k.dy_k, so the function to be integrated over the hypervolume is Π{k=1 to n}e^y_k ... but I'm having difficulty transforming the boundaries, as in the "x" formulation it's the part of the hyperbox made by the planes x_k=0 & x_k=l (need to fix this in the question) cut off by the hyperhyperbola Πx=e^-μ: but in the "y" formulation the hyperhyperbola translates to the plane Σy=μ; but because y is from 0 to ∞, how does the box transform!? $\endgroup$ – AmbretteOrrisey Dec 4 '18 at 1:49
  • $\begingroup$ Oh right! I think I've got it now! I see what I was missing before! I think you're right actually. ¶ I'll see what these other comments say. I'll bet they're saying what you said ... & I'll bet you won't believe that I hadn't seen them when I wrote this! $\endgroup$ – AmbretteOrrisey Dec 4 '18 at 3:20
  • $\begingroup$ @reuns I've made another comment above this one, but I forgot to put an "@" @ the beginning of it (hahaha! see what I did there!?). I think the fog has cleared, now! ... or is clearing. Those comments though, that I'd gotten notification of, & expected to be admonishments apprising me of how slow I am - they were actually about something completely different. ¶ Thankyou for your contribution! $\endgroup$ – AmbretteOrrisey Dec 4 '18 at 3:39

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