Can I evaluate this integral in closed-form $\int\limits_{0}^{\infty} x \exp \left(- \sum\limits_{k=0}^{N}a_k x^{k}\right)dx$? I have the following integral and I'm trying to represent it in a closed-form even if it is in terms of some special functions.
$$\int_0^\infty x \exp \left(- \sum_{k=0}^N a_k x^k \right)\,dx$$
For the case of $N=1$ and $N=2$, it is straightforward. But I failed to evaluate it for $N>2$. I looked at the extended incomplete gamma function in [1], but it is really not useful in this case; it can be used if I only have $x^{N_1}$ and $x^{N_2}$ in the exponent.
Any idea about a possible way to represent it?

Clarification:
$a_k>0 \  \forall k \in \{0,1, ... ,N\}$.
 A: First, here it is in a very simple case, where the polynomial has just one term, with coefficient $1$:
$$
\int_0^\infty x \exp(-x^k) \, dx
$$
\begin{align}
u & = x^k \\
x & = u^{1/k} \\
dx & = (1/k)u^{(1/k)-1} \, du
\end{align}
$$
\int_0^\infty x \exp(-x^k) \, dx = \int_0^\infty u^{1/k} \exp(-u) \frac 1 k u^{(1/k) -1 } \, du = \frac 1 k \int_0^\infty u^{(2/k)-1} e^{-u} \,du = \frac{\Gamma(2/k)} k. 
$$
However, consider this:
\begin{align}
u & = \text{some polynomial in }x \\
x & =\text{what???} \\
dx & = \text{what?????}
\end{align}
Polynomials in general don't have solutions by radicals. And polynomials generally are not one-to-one. And if you allow one of these functions, not expressible by radicals, how do you differentiate it and thereby fill in this blank?: $dx = \text{what?}$
So it appears to me you probably won't find a closed form except in narrow special cases.
A: One thing that's rather obvious is that it diverges if $a_N < 0$.  Other than that and some special cases, there's basically no hope of a closed-form solution in general.
One thing you can try is this.  After scaling to make $a_N = 1$, write your integral as
$$ G(t) = \int_0^\infty x \exp(-t p(x) - x^N)\; dx $$
where $p(x)$ is a polynomial of degree $< N$. Expand out $\exp(-t p(x))$ in a power series:
$$ G(t) = \sum_{n=0}^\infty \dfrac{(-t)^n}{n!} \int_0^\infty x p(x)^n \exp(-x^N)\; dx $$
and expand out the polynomial $xp(x)^n$, then use
$$ \int_0^\infty x^m \exp(-x^N)\; dx = \dfrac{1}{N} \Gamma\left(\frac{m+1}{N}\right) $$
EDIT: 
It might be useful to mention a theorem of Liouville: the only cases where  $q(x) \exp(-p(x))$   (with $q(x)$ and $p(x)$ rational functions, $p(x)$ non-constant) has an antiderivative that is an
elementary function are those where that antiderivative is of the form $A(x) \exp(-p(x))$ with $A(x)$ rational. 
In particular, if $q(x)$ and $p(x)$ are polynomials, so is $A(x)$.  We can 
bound the degree of $A(x)$ because $A'(x) - p'(x) A(x) = q(x)$,
and then determine whether such a polynomial $A(x)$ exists.  In particular, this can't happen if $p'$ has higher degree than $q$.
Of course, having an elementary antiderivative is not a prerequisite for having a closed-form integral from $0$ to $\infty$, but it helps.
