I'm trying to find an approximation (or exact solution if possible) for an integral of the form:
$$\int_{u=0}^b e^{-\frac{u^2 + ac}{au}}du.$$
I was thinking of somehow applying a Gauss Hermite Quadriture Expansion, but I'm not sure how I would do this. Does anyone know the best way I should go about this?
On
Note that this integral converges when $a\neq0$ and $\text{Re}(b)\text{Re}(c)\geq0$ .
Case $1$: $c=0$
Then $\int_0^be^{-\frac{u^2+ac}{au}}~du$
$=\int_0^be^{-\frac{u^2}{au}}~du$
$=\int_0^be^{-\frac{u}{a}}~du$
$=\left[-ae^{-\frac{u}{a}}\right]_0^b$
$=a(1-e^{-\frac{b}{a}})$
Case $2$: $c\neq0$
Then $\int_0^be^{-\frac{u^2+ac}{au}}~du=\int_0^be^{-\frac{u}{a}-\frac{c}{u}}~du$ , which can treat as the generalization of Evaluating $\int_{x=0}^a e^{-x}e^{-1/x}\,\mathrm dx$.
Consider $\int e^{-\frac{u}{a}-\frac{c}{u}}~du$ :
$\int e^{-\frac{u}{a}-\frac{c}{u}}~du$
$=\int\sum\limits_{n=0}^\infty\dfrac{\left(\dfrac{u}{a}+\dfrac{c}{u}\right)^{2n}}{(2n)!}du-\int\sum\limits_{n=0}^\infty\dfrac{\left(\dfrac{u}{a}+\dfrac{c}{u}\right)^{2n+1}}{(2n+1)!}du$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{C_k^{2n}\left(\dfrac{u}{a}\right)^k\left(\dfrac{c}{u}\right)^{2n-k}}{(2n)!}du-\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^{2n+1}\left(\dfrac{u}{a}\right)^k\left(\dfrac{c}{u}\right)^{2n+1-k}}{(2n+1)!}du-\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^{2n+1}\left(\dfrac{u}{a}\right)^{2n+1-k}\left(\dfrac{c}{u}\right)^k}{(2n+1)!}du$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{c^{2n-k}u^{2k-2n}}{a^kk!(2n-k)!}du-\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n-1}\dfrac{c^{2n-k+1}u^{2k-2n-1}}{a^kk!(2n-k+1)!}du-\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{c^ku^{2n-2k+1}}{a^{2n-k+1}k!(2n-k+1)!}du-\int\sum\limits_{n=0}^\infty\dfrac{c^{n+1}}{a^nn!(n+1)!u}du$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{c^{2n-k}u^{2k-2n+1}}{a^kk!(2n-k)!(2k-2n+1)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n-1}\dfrac{c^{2n-k+1}u^{2k-2n}}{a^kk!(2n-k+1)!(2k-2n)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{c^ku^{2n-2k+2}}{a^{2n-k+1}k!(2n-k+1)!(2n-2k+2)}-\sum\limits_{n=0}^\infty\dfrac{c^{n+1}\ln u}{a^nn!(n+1)!}+C$
However, the above result cannot substitute $0$ , so we can divide $\int_0^be^{-\frac{u}{a}-\frac{c}{u}}~du$ as $\int_0^me^{-\frac{u}{a}-\frac{c}{u}}~du+\int_m^be^{-\frac{u}{a}-\frac{c}{u}}~du$ , $\int_m^be^{-\frac{u}{a}-\frac{c}{u}}~du$ use the above antiderivative result, and find the suitable value of $m$ such that $\int_0^me^{-\frac{u}{a}-\frac{c}{u}}~du$ can express as known special functions.
I only able to find the suitable values of $m$ for the cases of real numbers $a$ and $c$ .
When $a>0$ and $c>0$ ,
$\int_0^me^{-\frac{u}{a}-\frac{c}{u}}~du=\int_0^\frac{m}{\sqrt{ac}}e^{-\frac{\sqrt{ac}u}{a}-\frac{c}{\sqrt{ac}u}}~d(\sqrt{ac}u)=\sqrt{ac}\int_0^\frac{m}{\sqrt{ac}}e^{-\sqrt\frac{c}{a}\left(u+\frac{1}{u}\right)}~du$
$\because\int_0^\infty e^{-\sqrt\frac{c}{a}\left(u+\frac{1}{u}\right)}~du$
$=\int_{-\infty}^\infty e^{-\sqrt\frac{c}{a}(e^u+e^{-u})}~d(e^u)$
$=\int_{-\infty}^\infty e^{u-2\sqrt\frac{c}{a}\cosh u}~du$
$=\int_{-\infty}^0e^{u-2\sqrt\frac{c}{a}\cosh u}~du+\int_0^\infty e^{u-2\sqrt\frac{c}{a}\cosh u}~du$
$=\int_\infty^0e^{-u-2\sqrt\frac{c}{a}\cosh(-u)}~d(-u)+\int_0^\infty e^{u-2\sqrt\frac{c}{a}\cosh u}~du$
$=\int_0^\infty e^{-u-2\sqrt\frac{c}{a}\cosh u}~du+\int_0^\infty e^{u-2\sqrt\frac{c}{a}\cosh u}~du$
$=2\int_0^\infty e^{-2\sqrt\frac{c}{a}\cosh u}\cosh u~du$
$=2K_1\left(2\sqrt{\dfrac{c}{a}}\right)$ (according to http://people.math.sfu.ca/~cbm/aands/page_376.htm)
But $\int_0^\infty e^{-\sqrt\frac{c}{a}\left(u+\frac{1}{u}\right)}~du$
$=\int_0^1e^{-\sqrt\frac{c}{a}\left(u+\frac{1}{u}\right)}~du+\int_1^\infty e^{-\sqrt\frac{c}{a}\left(u+\frac{1}{u}\right)}~du$
$=\int_0^1e^{-\sqrt\frac{c}{a}\left(u+\frac{1}{u}\right)}~du+\int_1^0e^{-\sqrt\frac{c}{a}\left(\frac{1}{u}+u\right)}~d\left(\dfrac{1}{u}\right)$
$=\int_0^1e^{-\sqrt\frac{c}{a}\left(u+\frac{1}{u}\right)}~du+\int_0^1\dfrac{1}{u^2}e^{-\sqrt\frac{c}{a}\left(u+\frac{1}{u}\right)}~du$
$=\int_0^1\left(1+\dfrac{1}{u^2}\right)e^{-\sqrt\frac{c}{a}\left(u+\frac{1}{u}\right)}~du$
$=2\int_0^1e^{-\sqrt\frac{c}{a}\left(u+\frac{1}{u}\right)}~du-\int_0^1\left(1-\dfrac{1}{u^2}\right)e^{-\sqrt\frac{c}{a}\left(u+\frac{1}{u}\right)}~du$
$\therefore\int_0^1e^{-\sqrt\frac{c}{a}\left(u+\frac{1}{u}\right)}~du=\dfrac{1}{2}\int_0^\infty e^{-\sqrt\frac{c}{a}\left(u+\frac{1}{u}\right)}~du+\dfrac{1}{2}\int_0^1\left(1-\dfrac{1}{u^2}\right)e^{-\sqrt\frac{c}{a}\left(u+\frac{1}{u}\right)}~du=K_1\left(2\sqrt{\dfrac{c}{a}}\right)+\dfrac{1}{2}\int_0^1e^{-\sqrt\frac{c}{a}\left(u+\frac{1}{u}\right)}~d\left(u+\dfrac{1}{u}\right)=K_1\left(2\sqrt{\dfrac{c}{a}}\right)-\dfrac{1}{2}\left[\sqrt{\dfrac{a}{c}}e^{-\sqrt\frac{c}{a}\left(u+\frac{1}{u}\right)}\right]_0^1=K_1\left(2\sqrt{\dfrac{c}{a}}\right)-\dfrac{1}{2}\sqrt{\dfrac{a}{c}}e^{-2\sqrt\frac{c}{a}}$
$\therefore m$ should take $\sqrt{ac}$ and $\int_0^\sqrt{ac}e^{-\frac{u}{a}-\frac{c}{u}}~du=\sqrt{ac}K_1\left(2\sqrt{\dfrac{c}{a}}\right)-\dfrac{a}{2}e^{-2\sqrt\frac{c}{a}}$
Hence when $a>0$ and $c>0$ , $\int_0^be^{-\frac{u}{a}-\frac{c}{u}}~du$
$=\int_0^\sqrt{ac}e^{-\frac{u}{a}-\frac{c}{u}}~du+\int_\sqrt{ac}^be^{-\frac{u}{a}-\frac{c}{u}}~du$
$=K_1\left(2\sqrt{\dfrac{c}{a}}\right)-\dfrac{1}{2}\sqrt{\dfrac{a}{c}}e^{-2\sqrt\frac{c}{a}}+\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{c^{2n-k}u^{2k-2n+1}}{a^kk!(2n-k)!(2k-2n+1)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n-1}\dfrac{c^{2n-k+1}u^{2k-2n}}{a^kk!(2n-k+1)!(2k-2n)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{c^ku^{2n-2k+2}}{a^{2n-k+1}k!(2n-k+1)!(2n-2k+2)}-\sum\limits_{n=0}^\infty\dfrac{c^{n+1}\ln u}{a^nn!(n+1)!}\right]_\sqrt{ac}^b$
$=K_1\left(2\sqrt{\dfrac{c}{a}}\right)-\dfrac{1}{2}\sqrt{\dfrac{a}{c}}e^{-2\sqrt\frac{c}{a}}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{c^{2n-k}(b^{2k-2n+1}-a^{k-n+\frac{1}{2}}c^{k-n+\frac{1}{2}})}{a^kk!(2n-k)!(2k-2n+1)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n-1}\dfrac{c^{2n-k+1}(b^{2k-2n}-a^{k-n}c^{k-n})}{2a^kk!(2n-k+1)!(k-n)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{c^k(b^{2k-2n+2}-a^{k-n+1}c^{k-n+1})}{2a^{2n-k+1}k!(2n-k+1)!(n-k+1)}-\sum\limits_{n=0}^\infty\dfrac{c^{n+1}\left(\ln b-\dfrac{\ln a+\ln c}{2}\right)}{a^nn!(n+1)!}$
$=K_1\left(2\sqrt{\dfrac{c}{a}}\right)-\dfrac{1}{2}\sqrt{\dfrac{a}{c}}e^{-2\sqrt\frac{c}{a}}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{a^{k-n+\frac{1}{2}}c^{n+\frac{1}{2}}-b^{2k-2n+1}c^{2n-k}}{a^kk!(2n-k)!(2n-2k-1)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n-1}\dfrac{a^{k-n}c^{n+1}-b^{2k-2n}c^{2n-k+1}}{2a^kk!(2n-k+1)!(n-k)}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{a^{k-n+1}c^{2k-n+1}-b^{2k-2n+2}c^k}{2a^{2n-k+1}k!(2n-k+1)!(n-k+1)}+\sum\limits_{n=0}^\infty\dfrac{c^{n+1}(\ln a-2\ln b+\ln c)}{2a^nn!(n+1)!}$
When $a<0$ and $c>0$ ,
$\int_0^me^{-\frac{u}{a}-\frac{c}{u}}~du=\int_0^\frac{m}{\sqrt{-ac}}e^{-\frac{\sqrt{-ac}u}{a}-\frac{c}{\sqrt{-ac}u}}~d(\sqrt{-ac}u)=\sqrt{-ac}\int_0^\frac{m}{\sqrt{-ac}}e^{\sqrt{-\frac{c}{a}}\left(u-\frac{1}{u}\right)}~du$
$\because\int_0^1e^{\sqrt{-\frac{c}{a}}\left(u-\frac{1}{u}\right)}~du$
$=\int_\infty^0e^{\sqrt{-\frac{c}{a}}(e^{-u}-e^u)}~d(e^{-u})$
$=\int_0^\infty e^{-u-2\sqrt{-\frac{c}{a}}\sinh u}~du$
$=-\int_0^\infty e^{-2\sqrt{-\frac{c}{a}}\sinh u}\sinh u~du+\int_0^\infty e^{-2\sqrt{-\frac{c}{a}}\sinh u}\cosh u~du$
$=\dfrac{\pi}{2}Y_1\left(2\sqrt{-\dfrac{c}{a}}\right)-\dfrac{1}{2}\int_0^\pi\sin\left(2\sqrt{-\dfrac{c}{a}}\sin u-u\right)du-\left[\dfrac{1}{2}\sqrt{-\dfrac{a}{c}}e^{-2\sqrt{-\frac{c}{a}}\sinh u}\right]_0^\infty$ (according to http://people.math.sfu.ca/~cbm/aands/page_360.htm)
$=\dfrac{\pi}{2}Y_1\left(2\sqrt{-\dfrac{c}{a}}\right)+\dfrac{\pi}{2}\mathbf{H}_1\left(2\sqrt{-\dfrac{c}{a}}\right)+\dfrac{1}{2}\sqrt{-\dfrac{a}{c}}$ (according to http://mathworld.wolfram.com/WeberFunctions.html)
$\therefore m$ should take $\sqrt{-ac}$ and $\int_0^\sqrt{-ac}e^{-\frac{u}{a}-\frac{c}{u}}~du=\dfrac{\pi\sqrt{-ac}}{2}Y_1\left(2\sqrt{-\dfrac{c}{a}}\right)+\dfrac{\pi\sqrt{-ac}}{2}\mathbf{H}_1\left(2\sqrt{-\dfrac{c}{a}}\right)-\dfrac{a}{2}$
Hence when $a<0$ and $c>0$ , $\int_0^be^{-\frac{u}{a}-\frac{c}{u}}~du$
$=\int_0^\sqrt{-ac}e^{-\frac{u}{a}-\frac{c}{u}}~du+\int_\sqrt{-ac}^be^{-\frac{u}{a}-\frac{c}{u}}~du$
$=\dfrac{\pi\sqrt{-ac}}{2}Y_1\left(2\sqrt{-\dfrac{c}{a}}\right)+\dfrac{\pi\sqrt{-ac}}{2}\mathbf{H}_1\left(2\sqrt{-\dfrac{c}{a}}\right)-\dfrac{a}{2}+\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{c^{2n-k}u^{2k-2n+1}}{a^kk!(2n-k)!(2k-2n+1)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n-1}\dfrac{c^{2n-k+1}u^{2k-2n}}{a^kk!(2n-k+1)!(2k-2n)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{c^ku^{2n-2k+2}}{a^{2n-k+1}k!(2n-k+1)!(2n-2k+2)}-\sum\limits_{n=0}^\infty\dfrac{c^{n+1}\ln u}{a^nn!(n+1)!}\right]_\sqrt{-ac}^b$
$=\dfrac{\pi\sqrt{-ac}}{2}Y_1\left(2\sqrt{-\dfrac{c}{a}}\right)+\dfrac{\pi\sqrt{-ac}}{2}\mathbf{H}_1\left(2\sqrt{-\dfrac{c}{a}}\right)-\dfrac{a}{2}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{c^{2n-k}(b^{2k-2n+1}-(-a)^{k-n+\frac{1}{2}}c^{k-n+\frac{1}{2}})}{a^kk!(2n-k)!(2k-2n+1)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n-1}\dfrac{c^{2n-k+1}(b^{2k-2n}-(-1)^{k-n}a^{k-n}c^{k-n})}{2a^kk!(2n-k+1)!(k-n)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{c^k(b^{2k-2n+2}+(-1)^{k-n}a^{k-n+1}c^{k-n+1})}{2a^{2n-k+1}k!(2n-k+1)!(n-k+1)}-\sum\limits_{n=0}^\infty\dfrac{c^{n+1}\left(\ln b-\dfrac{\ln(-a)+\ln c}{2}\right)}{a^nn!(n+1)!}$
$=\dfrac{\pi\sqrt{-ac}}{2}Y_1\left(2\sqrt{-\dfrac{c}{a}}\right)+\dfrac{\pi\sqrt{-ac}}{2}\mathbf{H}_1\left(2\sqrt{-\dfrac{c}{a}}\right)-\dfrac{a}{2}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{(-a)^{k-n+\frac{1}{2}}c^{n+\frac{1}{2}}-b^{2k-2n+1}c^{2n-k}}{a^kk!(2n-k)!(2n-2k-1)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n-1}\dfrac{(-1)^{k-n}a^{k-n}c^{n+1}-b^{2k-2n}c^{2n-k+1}}{2a^kk!(2n-k+1)!(n-k)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{k-n}a^{k-n+1}c^{2k-n+1}+b^{2k-2n+2}c^k}{2a^{2n-k+1}k!(2n-k+1)!(n-k+1)}+\sum\limits_{n=0}^\infty\dfrac{c^{n+1}(\ln(-a)-2\ln b+\ln c)}{2a^nn!(n+1)!}$
When $a<0$ and $c<0$ ,
$\int_0^be^{-\frac{u}{a}-\frac{c}{u}}~du=\int_0^{-b}e^{-\frac{-u}{a}-\frac{c}{-u}}~d(-u)=\int_{-b}^0e^{\frac{u}{a}+\frac{c}{u}}~du$
Please follow the case of $a>0$ and $c>0$ .
When $a>0$ and $c<0$ ,
$\int_0^be^{-\frac{u}{a}-\frac{c}{u}}~du=\int_0^{-b}e^{-\frac{-u}{a}-\frac{c}{-u}}~d(-u)=\int_{-b}^0e^{\frac{u}{a}+\frac{c}{u}}~du$
Please follow the case of $a<0$ and $c>0$ .
I do not know much about numerical integration, but the following trick may help you avoid exponents diverge ad infinitum.
We are dealing with the integral
$$ I = \int_{0}^{b} \exp\left( - \frac{u}{a} - \frac{c}{u} \right) \, du. $$
With the substitution $u = \sqrt{a c} x^{2}$, for $\beta = \left( b / \sqrt{ac} \right)^{1/2}$ we have
\begin{align*} I &= 2\sqrt{ac} \int_{0}^{\beta} x \exp\left\{ - \sqrt{\frac{c}{a}} \left( x^{2} + x^{-2} \right) \right\} \, dx \\ &= 2\sqrt{ac} e^{-2\sqrt{c/a}} \int_{0}^{\beta} x \exp\left\{ - \sqrt{\frac{c}{a}} \left( x - x^{-1} \right)^{2} \right\} \, dx. \end{align*}
With the substitution $y = x - x^{-1}$ and
$$ \alpha := \frac{\beta + \sqrt{\beta^{2} + 4}}{2} = \frac{\sqrt{b} + \sqrt{b + 4\sqrt{ac}}}{2\sqrt[4]{ac}}, \qquad \gamma := \sqrt{\frac{c}{a}}, $$
we find that
\begin{align*} I &= 2a\gamma e^{-2\gamma} \int_{-\infty}^{\alpha} \frac{\left(\sqrt{y^{2}+4}+y\right)^{2}}{4 \sqrt{y^{2}+4}} e^{-\gamma\,y^{2}} \, dy. \end{align*}
Now with aid of Mathematica, we can check that
$$\int_{-\infty}^{0} \frac{\left(\sqrt{y^{2}+4}+y\right)^{2}}{4 \sqrt{y^{2}+4}} e^{-\gamma y^{2}} \, dy = \frac{1}{2} e^{2 \gamma} K_1(2 \gamma)-\frac{1}{4 \gamma}, $$
where $K_{n}(z)$ is the modified Bessel function of the second kind. This allows us to write
\begin{align*} I &= a\gamma e^{-2\gamma} \left[ e^{2\gamma} K_1\left(2\gamma \right)-\frac{1}{2\gamma} + 2 \int_{0}^{\alpha} \frac{\left(\sqrt{y^{2}+4}+y\right)^{2}}{4 \sqrt{y^{2}+4}} e^{-\gamma\,y^{2}} \, dy \right], \end{align*}
with each term being much tractable.