# Tetration Forum

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I was playing around with Maple and I noticed that.

$\int_{0}^{\infty}x^{\frac{1}{x}-2}dx=\int_{0}^{\infty}x^{-x}dx$=1.995455958

I then added some parameters and came up with the following:

$\int_{0}^{\infty}x^{a/x^{b}-c}dx=\int_{0}^{\infty}x^{-ax^{b}+(c-2)}dx$

For positive a and b, and c>2.

I do not know why this is, but I find it very interesting. The self-root function is the inverse of an infinite order tetration.
(04/03/2010, 04:25 AM)Ztolk Wrote: [ -> ]I was playing around with Maple and I noticed that.

$\int_{0}^{\infty}x^{\frac{1}{x}-2}dx=\int_{0}^{\infty}x^{-x}dx$=1.995455958

I then added some parameters and came up with the following:

$\int_{0}^{\infty}x^{a/x^{b}-c}dx=\int_{0}^{\infty}x^{-ax^{b}+(c-2)}dx$

For positive a and b, and c>2.

I do not know why this is, but I find it very interesting. The self-root function is the inverse of an infinite order tetration.

have you tried substitution ?

a moebius substitution ?
What is that?
(04/07/2010, 04:04 PM)Ztolk Wrote: [ -> ]What is that?

http://en.wikipedia.org/wiki/Integration...bstitution

http://en.wikipedia.org/wiki/M%C3%B6bius_transformation

combining the above two and leaving out the restriction ad -bc = 0 if needed.

regards

tommy1729
(04/03/2010, 04:25 AM)Ztolk Wrote: [ -> ]I was playing around with Maple and I noticed that.

$\int_{0}^{\infty}x^{\frac{1}{x}-2}dx=\int_{0}^{\infty}x^{-x}dx$=1.995455958

the substitution y = 1/x proves it.

regards

tommy1729
So it does. Neat. Thanks.

I should try this with the full three parameter version.

edit: 1/x substitution checks out for the full thing.