A difficult integral of reciprocal of sum of exponentials Consider the integral: $$\int_0^\infty \frac{1}{e^{ax}+e^{bx}+e^{cx}} dx$$ where $a, b, c$ are complex numbers.
I encounter such integral when dealing with Laplace transform of co-functions. It turns out that the online integral calculator couldn’t even give any result.
I suppose the integrand is well-defined since $e^x$ has no branch point on the whole complex plane. 
Maybe the integrand has poles, but I believe the integral exists in the sense of Cauchy p.v..
I am a bit reluctant to use the substitution $y=e^x$ since $y^p$ in general has branch points for complex $p$.
Is there any method to compute such integral?
Any help will be appreciated.
EDIT: since a result in closed form is very unlikely to exist, a series as an answer will also be appreciated.
 A: I think this integral deserves to be studied more. Let's introduce a function of three variables:
$$f(a,b,c)=\int_0^\infty \frac{1}{e^{ax}+e^{bx}+e^{cx}} dx$$
The function is obviously symmetric in all the variables.
While the integral doesn't have a closed form, what's important is: how else can we define the function and determine its properties or connect it to something else.
First, let's consider partial derivatives:
$$\frac{ \partial f}{ \partial a} =-\int_0^\infty \frac{xe^{ax}}{(e^{ax}+e^{bx}+e^{cx})^2} dx$$
Same with the other two variables.
On the other hand, we have:
$$\frac{d}{dx} \frac{1}{e^{ax}+e^{bx}+e^{cx}}=-\frac{ae^{ax}+be^{bx}+ce^{cx}}{(e^{ax}+e^{bx}+e^{cx})^2}$$
Thus, using integration by parts:
$$a\frac{ \partial f}{ \partial a}+b\frac{ \partial f}{ \partial b}+c\frac{ \partial f}{ \partial c}=- \int_0^\infty \frac{x(ae^{ax}+be^{bx}+ce^{cx})}{(e^{ax}+e^{bx}+e^{cx})^2} dx=$$
$$=\frac{x}{e^{ax}+e^{bx}+e^{cx}} \bigg|_0^\infty-\int_0^\infty \frac{1}{e^{ax}+e^{bx}+e^{cx}}dx=-f$$
So our function obeys a first order linear PDE:

$$a\frac{ \partial f}{ \partial a}+b\frac{ \partial f}{ \partial b}+c\frac{ \partial f}{ \partial c}+f=0$$

The form of this equation obviously generalizes to any number of exponentials in the denominator.

Of course, this PDE doesn't tell us much more than the integral itself. Moreover, it has infinitely more solutions than just this integral.
But maybe it might be useful in some context.
As for boundary conditions, we can look back at the integral and see that 
$$\lim_{a \to +\infty} f(a,b,c)=0$$
For $a>0$ we also have:
$$\lim_{b,c \to -\infty} f(a,b,c)=\frac{1}{a}$$
To this we can add (again for $a>0$):
$$f(a,0,0)=\frac{\ln 3}{2 a}$$
Same for the other variables. The above should be more than enough as far as boundary conditions go.
If we try to solve the PDE by separation of variables we have:
$$f(a,b,c)=A(\lambda,\mu) a^{\lambda} b^{ \mu-\lambda} c^{-1-\mu}$$
For some arbitrary constants $\lambda, \mu, A(\lambda,\mu)$. Since the equation is linear we can sum any number of solutions with different parameters and different amplitudes. Not sure if we can satisfy all the initial conditions that way though, I'll have to look into this further. The general solution (assuming the separation of variables of course, there are other kinds of solutions) could look like:
$$f(a,b,c)=\int_{-\infty}^\infty \int_{-\infty}^\infty A(\lambda,\mu) a^{\lambda} b^{ \mu-\lambda} c^{-1-\mu} d \lambda d \mu$$
Where of course $A(\lambda,\mu)$ could be made zero for any range to keep the solution finite.

Now, the integral also allows us to derive a number of simple functional equations, by changing the variable. For example:
$$f(a,b,c)=\frac{1}{a} f \left(1,\frac{b}{a},\frac{c}{a}  \right)$$
As well as a more symmetric one:
$$f(a,b,c)=\frac{1}{abc} f \left(\frac{1}{ab},\frac{1}{bc},\frac{1}{ca}  \right)$$
In general:
$$f(\alpha a,\alpha b,\alpha c)=\frac{1}{\alpha} f \left(a,b,c  \right)$$

Additionally, if we change the variable $$e^{-x}=t$$ we obtain: $$f(a,b,c)=\int_0^1 \frac{dt}{t^{1-a}+t^{1-b}+t^{1-c}}$$
As Robert Israel points our in the comments, if the parameters are rational multiplies of each other, we can usually transform this to an integral of a rational function, which always has a closed form.

All in all, I would consider this integral to be as interesting as Carlson elliptic integrals, as it too can be interpreted as the generalized several numbers mean. Might be worth looking into its properties more.
The fact that it's related to means can be trivially seen. Really, let $0<a \leq b \leq c$:
$$\frac{1}{3c}= f(c,c,c) \leq f(a,b,c) \leq f(a,a,a)= \frac{1}{3a}$$
$$a \leq \frac{1}{3 f(a,b,c)} \leq c$$
Then there exists a number $a \leq \mu \leq c$ such that:
$$f(a,b,c)=f( \mu, \mu, \mu) = \frac{1}{3 \mu}$$
As $ \mu$ represents a generalized mean of $a,b,c$, there may be some iterative algorithm which converges to it (just like arithmetic-geometric mean and Carlson three number algorithm for his integrals).
The above arguments work for any function of the form:
$$f_n(a_1,a_2,\dots,a_n)=\int_0^\infty \frac{dx}{e^{a_1x}+e^{a_2 x}+ \cdots + e^{a_n x}}$$
A: Here's a very sloppily derived series representation. I just went through the algebra without looking out for convergence and divergence problems, so I have no idea how useful this will be (in fact, it appears that the final result hardly ever converges).
$$\begin{align}
\int_0^\infty \frac{dx}{e^{ax}+e^{bx}+e^{cx}}
&= \int_0^\infty \frac{e^{-cx}}{e^{(a-c)x}+e^{(b-c)x}+1}dx\\
&= \int_0^\infty \frac{e^{-cx}}{1-(-e^{(a-c)x}-e^{(b-c)x})}dx\\
&= \int_0^\infty e^{-cx}\sum_{n=0}^\infty (-1)^n(e^{(a-c)x}+e^{(b-c)x})^n dx\\
&= \sum_{n=0}^\infty (-1)^n\int_0^\infty e^{-cx}(e^{(a-c)x}+e^{(b-c)x})^n dx\\
&= \sum_{n=0}^\infty (-1)^n\int_0^\infty e^{-cx}\sum_{k=0}^n\binom{n}{k}e^{(a-c)xk}e^{(b-c)x(n-k)} dx\\
&= \sum_{n=0}^\infty (-1)^n \sum_{k=0}^n\binom{n}{k}\int_0^\infty e^{-cx}e^{(a-c)xk}e^{(b-c)x(n-k)} dx\\
&= \sum_{n=0}^\infty (-1)^n \sum_{k=0}^n\binom{n}{k}\int_0^\infty e^{(a-c)xk+(b-c)x(n-k)-cx} dx\\
&= \sum_{n=0}^\infty (-1)^n \sum_{k=0}^n\binom{n}{k}\int_0^\infty e^{((a-c)k+(b-c)(n-k)-c)x} dx\\
&= \sum_{n=0}^\infty (-1)^n \sum_{k=0}^n\binom{n}{k}\int_0^\infty e^{(ak+bn-bk-cn-c)x} dx\\
&= \sum_{n=0}^\infty (-1)^n \sum_{k=0}^n\binom{n}{k} \bigg[ \frac{e^{(ak+bn-bk-cn-c)x}}{ak+bn-bk-cn-c} \bigg]_{0}^{\infty} \\
&= \sum_{n=0}^\infty (-1)^n \sum_{k=0}^n\binom{n}{k}\frac{1}{ak+bn-bk-cn-c}  \\
&= \sum_{n=0}^\infty (-1)^n \sum_{k=0}^n\binom{n}{k}\frac{1}{(a-b)k+(b-c)n-c}  \\
\end{align}$$
