Integrating exponential of second degree polynomials

Question

I'm looking to compute the value of the following integral, for small values of $|a|$. $$u_n(a,b)=\frac{1}{2}\int_{-1}^1 x^ne^{ax^2+bx}\mathrm{d}x$$

In this equation, $a,b \in \mathbb{R}$ and $n \in \mathbb{N}$.

It's easy to show that:

• $u_0(0, 0) = 1$
• $\frac{\partial}{\partial b} u_n = u_{n+1}$
• $\frac{\partial^2}{\partial b^2} u_n = \frac{\partial}{\partial a} u_n = u_{n+2}$

We can find analytic solutions for this equation using Mathematica, for example here is $u_0$: $$u_0(a,b) = e^a \frac{e^b F\left(\frac{b+2a}{2\sqrt a}\right)-e^{-b} F\left(\frac{b-2a}{2\sqrt a}\right)}{2\sqrt a}$$ where $F(x)$ is the Dawson function. We can also show that: $$\lim_{a\to0} u_0 = \frac{\sinh(b)}{b}$$ $$\lim_{b\to0} u_0 = e^a\frac{F(\sqrt a)}{\sqrt a}$$

The analytic solution works great for larger $|a|$ as there are a few libraries that can provide the value of the Dawson function. However, for small values of $|a|$, the analytic solution has a catastrophic $\frac{0}{0}$ limit. This problem only gets worse as we go to $u_1$, $u_2$, etc.

Is there a way to reformulate this problem in a way where it can be computed for small values of $|a|$? For example, perhaps the value of $u_n(a, b)$ be the limit of some iteration?

The best I could do for now is to compute a Taylor series around $a=0$: $$u_n(a,b) = \sum_k \frac{a^k}{k!} v_{2k+n}(b)$$ where $v_0 = \frac{\sinh(b)}{b}$ and we use the recurrence relation: $$b v_{n+1} = f(b) - (n+1) v_n$$ Here $f(b)$ is $\cosh(b)$ if $n$ is even and it is $\sinh(b)$ if $n$ is odd. However, to compute for $a = 0.5$, we need quite a few terms and it's slow to compute.

Edit:

Here is a small induction proof for the recurrence relation. We simply derive it by $b$ and note that $\frac{\partial}{\partial b}v_n = v_{n+1}$ as $v_n = \lim_{a\to0}u_n$ and $\frac{\partial}{\partial b}u_n = u_{n+1}$. For $n$ even, we have: $$\frac{\partial}{\partial b} bv_{n+1} = \frac{\partial}{\partial b} \left( \cosh(b) - (n+1)v_n\right)$$ $$bv_{n+2} + v_{n+1} = \sinh(b) - (n+1)v_{n+1}$$ $$bv_{n+2} = \sinh(b) - (n+2)v_{n+1}$$

We have a similar proof for the case $n$ is odd (simply exchange $\cosh$ and $\sinh$).

Answer 1

By inspecting more carefully nicoguaro's solution, I realize that in general: $$v_n(b) = (-1)^{n+1} \frac{\Gamma(n+1,b) - \Gamma(n+1,-b)}{2b^{n+1}}$$

It's straightforward to confirm that $v_0 = \frac{\sinh(b)}b$ and that $\frac{\partial}{\partial b} v_n = v_{n+1}$ for all $n \in \mathbb{N}$, so this is the same function!

So we can use the Taylor expansion (to any arbitrary order in $a$) with the recurrence relation as we can directly compute any starting $v_n$ term and then propagate to nearby values of $v_{n+2}$, $v_{n+4}$, etc. using the recurrence relation provided in the original question.

Answer 2

I think that a numerical integration might work in your case, I don't see why it should not.

Besides numerically integrating your function, you could try an asymptotic expansion for small $a$. Since you are using Mathematica, I tried to compute the integral using Wolfram Language with the function AsymptoticIntegrate, specifically I used

AsymptoticIntegrate[x^n * Exp[a*x^2 + b*x], {x, -1, 1}, a->0]

This gave me the following result that seems to be fine for the values I tested.

$$u_n(a, b) \sim \frac{1}{b^3}\left(-b^2\right)^{-n} \left[b^2 \left((-1)^n (-b)^n-b^n\right) \Gamma(1+n)+a \left((-1)^n (-b)^n-b^n\right) \Gamma(3+n)+b^{2+n} \Gamma(1+n,-b)+(-1)^{1+n} (-b)^{2+n} \Gamma(1+n,b)+a b^n \Gamma(3+n,-b)+(-1)^{1+n} a (-b)^n \Gamma(3+n,b)\right]$$