What is the Series Expansion of the following function?

Question

I was wondering what the expansion series of the function

$$ f(x) = -\frac{1}{x^3} \cdot \frac{1}{\Gamma(x) \cdot \Gamma(-(\exp(\frac{2}{3}\pi\cdot i))x) \cdot \Gamma(-(\exp(\frac{4}{3}\pi \cdot i))x)} $$ is, at $x = 0$. I'm also interested in the method behind computing the series expansion.

You might think: "why don't you just paste this equation in wolframalpha and then see what the expansion series is?" Well, I did exactly that, but wolframalpha couldn't compute it! Does that mean the expansion series doesn't exist or is it more likely that it's just too 'hard' for the computational knowledge engine to find the series?

Thanks,

Max

EDIT For more information: this function is function (27) at this page, when n=3. I offer my apologies for not stating the question well at first.

Answer 1

Mathematica can handle (rather easily) that function. For example, the expansion at x=0 to order 5.

$f(x) = -1-2 \gamma x-2 \gamma ^2 x^2+\displaystyle \frac{1}{6} x^3 \left(-8 \gamma ^3-\psi ^{(2)}(1)\right)-$

$-\displaystyle\frac{1}{3} x^4 \left(\gamma \left(2 \gamma ^3+\psi ^{(2)}(1)\right)\right)+\frac{1}{60} x^5 \left(-16 \gamma ^5-20 \gamma ^2 \psi ^{(2)}(1)+\psi ^{(4)}(1)\right)+O\left(x^6\right)$

(same notation than the one used in Mariano's post).

Answer 2

Up to a constant and a power of $x$, your function is $\frac1{x\Gamma(x)}$. Mathematica tells me that the Taylor series at zero is $$ 1+\gamma x+\left(\frac{\gamma ^2}{2}-\frac{\pi ^2}{12}\right) x^2+\frac{1}{12} x^3 \left(2 \gamma ^3-\gamma \pi ^2-2 \psi ^{(2)}(1)\right)+\frac{x^4 \left(60 \gamma ^4-60 \gamma ^2 \pi ^2+\pi ^4-240 \gamma \psi ^{(2)}(1)\right)}{1440}+\dots$$

($\gamma$ is Euler's constant, and the $\psi$'s are poly-gamma functions.)