Definite integral of a combination of exponents

Question

Is there a closed-form formula (e.g. in terms of some known special functions) of

$$\int_{-\infty}^{\infty}\frac{\exp(-x^2)}{1+b\exp(-\alpha x)}d x$$

Answer 1

We can try to find a solution in terms of series, but the fact is that we cannot go too much further due to the two parameter $a$ and $b$, since we don't know a priori if they are both positive, both negative or else.

Let's start by splitting the integral into two pieces, and we will then apply the "same" method of the geometric series, changing some things with respect to what we have.

$$\int_{-\infty}^{+\infty} = \int_{-\infty}^0 + \int_0^{+\infty}$$

Let's start with the second one.

Integral 2

$$\int_0^{+\infty} \frac{e^{-x^2}}{1 + be^{-ax}}\ \text{d}x$$

Since the range extends to $[0, +\infty)$, the denominator $1 + be^{-ax}$ si surely small, hence we can use the geometric series:

$$\frac{1}{1 + be^{-ax}} = \sum_{k = 0}^{+\infty} \left(-be^{-ax}\right)^k$$

hence the integral easily becomes

$$\sum_{k = 0}^{+\infty} (-b)^k \int_0^{+\infty} e^{-x^2 - akx}\ \text{d}x$$

The integral is a well known integral, and the result is expressed in terms of the Complementary Error Function (you can find the proof almost in every textbook of special function, or simply on Google):

$$\int_0^{+\infty} e^{-x^2 - akx}\ \text{d}x = \frac{\sqrt{\pi}}{2} \large e^{\frac{a^2k^2}{4}}\small \text{erfc}\left(\frac{ak}{2}\right)$$

Hence the second piece will give you:

$$(2) ~~~~~~~ \frac{\sqrt{\pi}}{2}\sum_{k = 0}^{+\infty} (-b)^k \large e^{\frac{a^2k^2}{4}}\small \text{erfc}\left(\frac{ak}{2}\right)$$

Let's now focus on the first.

Integral One

The procedure is quite the same, except that the range changes, so we cannot use the same term for the geometric series. Instead let's do a trick by collecting $be^{-ax}$ at the denominator like

$$\frac{e^{-x^2}}{1 + be^{-ax}} = \frac{e^{-x^2}}{be^{-ax}(1 + b^{-1}e^{ax})} = \frac{1}{b}e^{-x^2 + ax} \frac{1}{1 + b^{-1} e^{ax}}$$

According to the range $(-\infty, 0]$ now the term $b^{-1}e^{ax}$ is small, hence again the geometric series comes in handy:

$$\frac{1}{1 + b^{-1} e^{ax}} = \sum_{k = 0}^{+\infty} \left(-\frac{1}{b}e^{ax}\right)^k$$

Hence the integral becomes

$$\sum_{k = 0}^{+\infty} \left(-\frac{1}{b}\right)^k \int_{-\infty}^0 \large e^{-x^2 + ax(1+k)}\ \text{d}x$$

Like before, the integral is really straightforward and you get:

$$\small \frac{1}{b}\int_{-\infty}^0 e^{-x^2 + ax(1+k)}\ \text{d}x = \frac{\sqrt{\pi}}{2} \large e^{\frac{a^2(1+k)^2}{4}}\small\ \text{erfc}\left(\frac{a(1+k)}{2}\right)$$

That is finally

$$(1) ~~~~~~~ \frac{\sqrt{\pi}}{2b}\sum_{k = 0}^{+\infty}\left(-\frac{1}{b}\right)^k\large e^{\frac{a^2(1+k)^2}{4}}\small\text{erfc}\left(\frac{a(1+k)}{2}\right)$$

The whole integration then is $(2) + (1)$ hence:

$$\int_{-\infty}^{+\infty} \frac{e^{-x^2}}{1 + be^{-ax}}\ \text{d}x = \frac{\sqrt{\pi}}{2}\sum_{k = 0}^{+\infty} \left\{\left(\frac{1}{b}\cdot\left(-\frac{1}{b}\right)^{k}\right)\large e^{\frac{a^2(1+k)^2}{4}}\small\text{erfc}\left(\frac{a(1+k)}{2}\right) + (-b)^k \large e^{\frac{a^2k^2}{4}}\small \text{erfc}\left(\frac{ak}{2}\right)\right\}$$