The inverse gamma function.

[tommy1729]

Because of recent events here, I felt the need to talk about the inverse gamma function.

Im talking about the entire approximation of exp^[1/2] and the Taylor coefficients being O ( 1/ Gamma( n (arc2sinh(n) - 1) ) ).

Cleary from the viewpoint of numerical analysis it seems usefull to have a good approximation of the inverse gamma function.
The functional inverse that is.

Now my honesty forces me to quote the source of an intresting approximation of the inverse gamma function :

David W Cantrell

http://mathforum.org/kb/message.jspa?messageID=342552

In case that page gets removed I post the main formula here :

A is the positive zero of the digamma function ( 1.4616... )
B = sqrt(2 pi) / e - Gamma(A). ( 0.0365... )
L(x) = ln( (x+c) / sqrt(2 pi) )
W(x) is the Lambert W function. ( the functional inverse of x exp(x))
Gamma^[-1](x) = approx = 1/2 + L(x)/W(L(x)/e)
The error term goes to 0 as x goes to +oo.

Together with the approximation for the Lambert W :

LambertW(x) = ln(x) - ln(ln(x)) + ln(ln(x)) / ln(x)

this gives a practical way to compute the inverse gamma function.


I think there is also a page by wolfram about the inverse gamma but I cant find it ? ( And no, I dont mean the inverse regularized gamma or the statistical inverse gamma distribution )


I would like to see an integral representation of the inverse gamma function too.


regards

tommy1729

I always like posting my own results ^_^, for \( \Re(z) > 0 \)

\( \phi(z) = \frac{1}{\G(z)}\int_0^\infty \frac{w^{z-1}}{w!}\,dw \)

then by some of my fractional calculus theorems. For \( \Re(w) > 0 \) and \( \sigma > 0 \)

\( \frac{1}{w!} = \frac{1}{2\pi i}\int_{\sigma - i\infty}^{\sigma+i \infty} \G(\xi)\phi(\xi)w^{-\xi}\,d\xi \)


Is that a good integral expression?

Dear James.

For starters if you are trying to find the integral I asked for :

1) I asked for the functional inverse of the Gamma function.
Not the reciprocal.
The whole OP was about the functional inverse of the Gamma function.
Although I could have stated that more clearly when I asked about the integral representation ...
2) ... Also defining f(x) By M^[-1] M^[1] f(x) seems a bit lame.
That looks similar to saying x = sqrt(x)^2 or x = exp(ln(x)).
3) despite 1) and 2) why do you wonder if that is OK ? You know the mellin inversion theorem.

Thanks anyway.

Maybe a second attempt.

Im not sure such an integral representation exists btw.


regards

tommy1729