Integral of an exponent [updated - new way of solving the problem]

Question

I have been struggling with this integral for a whole week now. The integral form is quite simple but extremely difficult for solving it. It related to the heat conduction problem, but I have simplified it so that we could see it a bit more clear. $$ I = \int_0^a \frac{\exp(x^2-b^2)}{x^2-b^2}dx $$ Where $ a\neq b >0 $. I tried using Feynman's technique by multiplying additional variable $t$ like this: $$ I(t) = \int_0^a \frac{\exp[t(x^2-b^2)]}{x^2-b^2}dx $$ Then differenciating it with regard to new variable $t$: $$ \frac{\partial I(t)}{\partial t} = \int_0^a \frac{\partial }{\partial t}{\frac{\exp[t(x^2-b^2)]}{x^2-b^2}}dx = \int_0^a \exp[t(x^2-b^2)]dx $$ $$ \frac{\partial I(t)}{\partial t} = \exp(-tb^2)\int_0^a \exp(tx^2)dx $$ The integral $\int_0^a \exp(tx^2)dx $ is well-known and is defined by $$ \int_0^a \exp(tx^2)dx = \frac{\sqrt{\pi}}{2\sqrt{t}}\operatorname{erfi}(a\sqrt{t}) $$ where $\operatorname{erfi}(z)$ is imaginary error function. The we have to take integral with regard to $t$, but at this moment I just don't know how to continue solving. Any ideas? $$ I(t) = \frac{\sqrt{\pi}}{2}\int\frac{1}{\sqrt{t}}{\operatorname{erfi}(a\sqrt{t}) \exp(-tb^2)}dt $$ We could get rid of $\sqrt{t}$ by substitution $v=\sqrt{t}$, then $t = v^2 $, $dt = 2vdv$. Then we have $$ I(v) = \sqrt{\pi}\int{\operatorname{erfi}(av) \exp(-v^2b^2)}dv $$

[Updated] I have found a new way to solve this integral by changing to a double integral, you could find the article here.

$$ I = \int_0^a \frac{\exp(x^2-b^2)}{x^2-b^2}dx = \exp(-b^2)\int_0^a \frac{\exp(x^2)}{(x-b)(x+b)}dx$$ Then we have $$ I = \frac{\exp(-b^2)}{2b}\int_0^a \frac{\exp(x^2)}{x-b} - \frac{\exp(x^2)}{x+b}dx $$ Now we could define a function $g(t) = \frac{\exp(x^2)}{x+t*b}$ with the variable $t$ changing from -1 to 1. So now we have a new form of the integral: $$ I = -\frac{\exp(-b^2)}{2b}\int_0^a \int_{-1}^{1} \frac{\partial}{\partial t} \frac{\exp(x^2)}{x+tb}dtdx $$

$$ I = \frac{\exp(-b^2)}{2}\int_0^a \int_{-1}^{1} \frac{\exp(x^2)}{(x+tb)^2}dtdx $$

$$ I = \frac{\exp(-b^2)}{2}\int_{-1}^{1} \int_0^a \frac{\exp(x^2)}{(x+tb)^2}dxdt $$

Now we have to find the inner integral with regard to $x$, then take the outer integral with regard to $t$. But it seems more difficult than the original one.

Answer 1

Hoping that it ccould be of some help.

Being stuck with the integral $$I= \sqrt{\pi}\int{\operatorname{erfi}(av)\, \exp(-v^2b^2)}\,dv$$ let $av=z$ and $k=\frac {b^2}{a^2}$ to make $$I=\frac{\sqrt{\pi} }a \int {\operatorname{erfi}(z)\, \exp(-kz^2)}\,dz$$ Now, using the infinite series expansion $$\operatorname{erfi}(z)= \frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty\frac{z^{2n+1}}{n! (2n+1)} $$ this would lead to $$I=\frac 2 a\sum_{n=0}^\infty\frac{1}{n! (2n+1)} \int z^{2n+1}\, \exp(-kz^2)\,dz$$ and $$J_n=\int z^{2n+1}\, \exp(-kz^2)\,dz=-\frac{1}{2} k^{-(n+1)} \Gamma \left(n+1,k z^2\right)$$