Integrating a Complex Exponential Function

Suppose $w=\exp(2i\pi/3)$. How would I go about integrating

$$\int\frac{3dx}{e^x+e^{wx}+e^{w^2x}}$$

Is there a transformation i can use? This is an entire function; there is no $x$ that will produce poles. If I said that

$$\frac{3}{e^x+e^{wx}+e^{w^2x}}=\sum_{n=0}^\infty A_{3n}\frac{x^{3n}}{(3n)!}$$

Then can’t I say that

$$\int\frac{3dx}{e^x+e^{wx}+e^{w^2x}}=\int \sum_{n=0}^\infty A_{3n}\frac{x^{3n}}{(3n)!}=\sum_{n=0}^\infty A_{3n}\int\frac{x^{3n}}{(3n)!}=\sum_{n=0}^\infty A_{3n}\frac{x^{3n+1}}{(3n+1)!}$$

But this wont give me information regarding the generating function I need. Any suggestions?

Solutions Collecting From Web of "Integrating a Complex Exponential Function"

Well, I bet you are aware of the identity:
$$\frac{e^x+e^{\omega x}+e^{\omega^2 x}}{3}=\sum_{n\geq 0}\frac{x^{3n}}{(3n)!}\tag{1}$$
hence by assuming
$$ \frac{3}{e^{x}+e^{\omega x}+e^{\omega^2 x}}=\sum_{n\geq 0}A_{3n} \frac{x^{3n}}{(3n)!} \tag{2}$$
and considering the Cauchy product of $(1)$ and $(2)$ we get $A_0=1$ and
$$ \sum_{n=0}^{m}\binom{3m}{3n}A_{3n} = 0,\tag{3}$$
so our coefficients can be computed by recursion and they give a generalization of Bernoulli numbers. Continuing the analogy, since integrals like $\int_{0}^{+\infty}\frac{dx}{\cosh x}\,dx$ and $\int_{0}^{+\infty}\frac{x^k\,dx}{\sinh x}\,dx$ are related with the Riemann $\zeta$ function, my bet is that your integral is related with another $\zeta$-function, the Epstein zeta function.