Gaussian integral with another function

Question

integraton dear friends any one can help me to solve this Gaussian integral

$$\int_{a}^{b}\frac{\exp\left(- \sum_{i=1}^n(r_i - A)^{2}\right)}{A} dA$$

Answer 1

By writing the integral as follows

$$\int_a^b \frac{1}{A}e^{-(r_1 - A)^2}\ e^{-(r_2 - A)^2} \ldots e^{-(r_n - A)^2}\ dA$$

we can easily turn it into

$$\int_a^b \prod_{k = 1}^{n} \frac{1}{A}e^{-(r_k - A)^2}$$

The productory is finite, hence we can take it out from the integral, also because Gaussian integrals are absolutely convergent.

By the change of variable

$$y = r_k - A$$ we eventually get

$$\prod_{k = 1}^{n} \int_{r_k - b}^{r_k - a} \frac{1}{r_k - y} e^{-y^2}\ dy$$

The integral is not doable yet, but due to the $e^{-y^2}$ term, which goes to zero really fast, we may attempt one more approximation, using Taylor series for that term:

$$\prod_{k = 1}^{n} \int_{r_k - b}^{r_k - a} \frac{1}{r_k - y} \sum_{p = 0}^{+\infty} \frac{(-1)^p}{p!} y^{2p}\ dy$$

$$\prod_{k = 1}^{n} \sum_{p = 0}^{+\infty} \frac{(-1)^p}{p!} \int_{r_k - b}^{r_k - a} \frac{1}{r_k - y} y^{2p}\ dy$$

That integral can be solve by the mean of Hypergeometric functions:

$$\prod_{k = 1}^{n} \sum_{p = 0}^{+\infty} \frac{(-1)^p}{p!} \left( \frac{y^{2 p+1} \, _2F_1\left(1,2 p+1;2 p+2;\frac{y}{r_k - a}\right)}{(r_k - a) (2 p+1)} - \frac{y^{2 p+1} \, _2F_1\left(1,2 p+1;2 p+2;\frac{y}{r_k - b}\right)}{(r_k - b) (2 p+1)}\right)$$

This is how far I can go. The infinite series has no known sum, so you just have to take some terms and then run the entire productory (since it's finite) to get some suitable answer. I also checked with W. Mathematica but due to the arbitrariness of $r_k, a, b$ we cannot do much.