Who’s That Pokémon?

Before you scream “IT’S PIKACHU”, hear me out. So today, while bored, I randomly came up with the following troll function:

\displaystyle \mathsf{mlbaker}(z) := \int_0^\infty \frac{z^t}{\Gamma(t+1)} \, dt

The idea is as follows. You note that the exponential function is given by

\displaystyle e^z = \sum_{k=0}^\infty \frac{z^k}{k!}

and change the infinite series into an improper integral, substituting \Gamma(t+1) for the factorial function in the original sum, because \Gamma(n) = (n-1)! when n is an integer (it “extends” the factorial function to the real line — in fact, it is meromorphic in the complex plane). Although Wolfram|Alpha ignores me whenever I enter this function as input (I think it’s in denial), I calculated that \mathsf{mlbaker}(1) \approx 2.266 — remember that e \approx 2.718. The function seems to behave in a pseudo-exponential fashion, as one might expect: the usual “exponential rules” seem to be satisfied, with some degree of error. Anyways, I’m wondering if we can define something like an inverse of this thing, or find its power series expansion. Heh, maybe once we find the power series expansion, we can apply the same trick once again, and get a big sequence of \mathsf{mlbaker}s… it’ll be just like those times as a kid, I’d angle two mirrors towards each other and peer down an infinite hall of copies of myself. How horrifying. I think one copy of myself is enough for this world.

To be honest I’m not even sure for what values it’s well-defined. An interesting thing is that when we use the power series to evaluate e^x for negative x, we get an alternating infinite series. However, (-1)^t for non-integer t is a bit trickier, and it causes \mathsf{mlbaker}(z) to start taking complex values… :O

Another question: what’s the asymptotic behaviour of \mathsf{mlbaker}(z)?

EDIT: Math.SE thread.


About mlbaker

just another guy trying to make the diagrams commute.
This entry was posted in analysis, random and tagged , , . Bookmark the permalink.

One Response to Who’s That Pokémon?

  1. Tony says:

    I always find it so fascinating when you begin to integrate functions on a complex basis and you end up with this convoluted function that seems to make absolutely no sense at all (until we google it). There are so many depths that it’s hard to explore every level…

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