Can I evaluate this integral in closed-form $\int\limits_{0}^{\infty} x \exp \left(- \sum\limits_{k=0}^{N}a_k x^{k}\right)dx$?

Question

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?

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Clarification:

$a_k>0 \ \forall k \in \{0,1, ... ,N\}$.

Answer 1

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.

Answer 2

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.