Integral of Exponential raised by exponential: $\int_{-\infty}^{\infty}\mathrm{d}x(1-\exp[\frac{-b}{\sqrt{2\pi}a}\exp(-\frac{x^2}{2a^2})]).$

Question

I am interested in the following integral

$$\int_{-\infty}^{\infty}\mathrm{d}x\left(1-\exp\left[\frac{-b}{\sqrt{2\pi}a}\exp\left(-\frac{x^2}{2a^2}\right)\right]\right).$$

Does anyone know how to solve this integral?

Answer 1

Using the identity $1-e^{-z}=\sum_{n\geq 1}\left(-1\right)^{n-1}\frac{z^n}{n!}$, the desired integral is $\sum_{n\geq 1}\frac{I_n}{n!}$ with $$I_n:=\int_{-\infty}^{\infty}\text{d}x\exp\left[\frac{b^n}{a^n\sqrt{2\pi}^n}\exp\left(-\frac{nx^2}{2a^2}\right)\right]=\frac{b^n}{a^n\sqrt{2\pi}^n}\sqrt{\frac{2\pi a^2}{n}}.$$Define $C:=\sqrt{2\pi a^2},\,q:=\frac{b}{a\sqrt{2\pi}}$, so the desired integral is $$C\sum_{n\geq 1}\left(-1\right)^{n-1}\frac{q^n}{n!\sqrt{n}}.$$I'm not aware of any closed-form expression for this, but it clearly converges.