Solving or approximating integral of $e^{-u^2}$ divided by quadratic function

Question

I am trying to solve or approximate the following type of integrals: $$\int_a^b \frac{\beta^2}{\beta^2 - c} e^{-u^2} du$$ or $$\int_a^b \frac{c}{\beta^2 - c} e^{-u^2} du,$$ with $\beta = u \sqrt{2} \sigma + \mu$.

The limits are $a = -8.5$ and $b$ varies from -0.62 to -2.2. Or $a$ varies from -0.5 to 3 and $b\to\infty$.

The other values are constant:

$c = 0.31$

$\sigma = 0.05$

$\mu = 1.7$

This integral is for finding a probability distribution from a product of distributions (https://en.wikipedia.org/wiki/Product_distribution)

I would like an approximate answer to see more clearly the effect of different parameters instead of having to compute the integral numerically each time.

Answer 1

For the values that you are giving us, it seems that $\sigma \ll 1$. So you could try to use the approximation (expansion in $\mu\sigma \ll \mu^2 ,\mu^2 -c$) $$ \int_a^b \frac{\beta^2}{\beta^2 -c} e^{-u^2}\,du \approx \frac{\sqrt{\pi} \mu^2}{2 (\mu^2 -c)} (\operatorname{erf}(b) -\operatorname{erf}(a)) +\frac{\sqrt 2 c\mu \sigma}{(\mu^2 - c)^2} (e^{-b^2} - e^{-a^2}) \,. $$ Similarly, $$ \int_a^b \frac{c}{\beta^2 -c} e^{-u^2}\,du \approx \frac{\sqrt{\pi} c}{2 (\mu^2 -c)} (\operatorname{erf}(b) -\operatorname{erf}(a)) +\frac{\sqrt 2 c \mu \sigma}{(\mu^2 - c)^2} (e^{-b^2} - e^{-a^2}) \,. $$

By the way, if course you have to limit the boundaries of integration such that $\beta^2 \neq c$ nowhere in the region of integration.

Answer 2

I do not think that you could obtained a closed form expression for the antiderivative.

Using your notations, $$I=\int \frac{e^{-u^2}}{\beta^2 - c} \, du=\frac{1}{2 \sigma ^2}\int \frac{e^{-u^2}}{u^2+\alpha u +\gamma }\,du$$ whith $$\alpha=\frac{\sqrt{2} \mu }{\sigma }\qquad \text{and} \qquad \gamma=\frac{\mu ^2-c}{2 \sigma ^2}$$ Using partial fraction $$\frac 1{u^2+\alpha u +\gamma }=\frac 1{(u-a)(u-b)}=\frac 1{a-b}\left(\frac 1{u-a} -\frac 1{u-b}\right)$$ where $a$ and $b$ are the roots of the quadratic equation in $u$. So, we face two integrals $$I_k=\int \frac{e^{-u^2}}{u-k} \, du$$ which do not seem to be known if $k\neq 0$.

If $k\ll 1$ (which is not the case at all with your data), we could approximate $I_k$ axpanding as a series around $k=0$ and write $$I_k=\sum_{n=0}^\infty \int \frac{e^{-u^2}}{u^{n+1}} k^n\, du=-\frac 12\sum_{n=0}^\infty \Gamma \left(-\frac{n}{2},u^2\right)\, k^n$$