$\int_{a}^{b}\frac{1}{x}e^{-\left(x+c\right)^{2}}dx$ I know how to solve $\int_{a}^{b}\frac{1}{x}e^{-x^{2}}dx$ using upper incomplete gamma function (we can assum that
$\left|b\right|>\left|a\right|$) $$\int_{a}^{b}\frac{1}{x}e^{-x^{2}}dx=\int_{a}^{b}\frac{2x}{2x^{2}}e^{-x^{2}}dx\stackrel{\begin{array}{c}
t=x^{2}\\
dt=2xdx
\end{array}}{=}\int_{a^{2}}^{b^{2}}\frac{1}{2t}e^{-t}dt=\frac{1}{2}\left[\Gamma\left(0,a^{2}\right)-\Gamma\left(0,b^{2}\right)\right]$$
there is something that I can do when I have some phase in the exponent:
$$\int_{a}^{b}\frac{1}{x}e^{-\left(x+c\right)^{2}}dx=?$$
Even an evaluation will help,
thanks!
 A: Here will be an evaluation with the Lerch Transcendent function, but we will use the Incomplete Beta function instead,   using a series expansion. The indefinite integral will be easier. We need to do the substitution or the integral will not converge with the same series expansion:
$$\int \frac{e^{-(x+c)^2}}{x}dx\mathop=^{x+c=t}=\int \frac{e^{-t^2}}{t-c}dt=\int\frac1{t-c}\sum_{n=0}^\infty \frac{(-1)^n t^{2n}}{n!}dt$$
Remember that the Incomplete Beta function equals the following integral with a domain extension if needed:
$$\text B_z(a,b)=\int_0^z t^{a-1} (1-t)^{b-1}dt$$
Therefore for all variables positive after term integration:
$$\int\frac1{t-c}\sum_{n=0}^\infty \frac{(-1)^n t^{2n}}{n!}dt=C-\sum_{n=0}^\infty \frac{(-1)^nc^{2n}\text B_\frac tc(2n+1,0) }{n!}$$
The main series expansion for the function is:
$$\text B_z(a,0)=z^a\sum_{n=0}^\infty \frac{z^n}{a+n} =Φ(z,1,a)$$
Therefore:
$$C-\sum_{n=0}^\infty \frac{(-1)^nc^{2n}\text B_\frac tc(2n+1,0) }{n!}=C-\sum_{m=0}^\infty \frac{(-1)^mc^{2m}}{m!} \left(\frac tc\right)^{2m+1}\sum_{n=0}^\infty\frac{\left(\frac tc\right)^n}{2m+n+1}$$
Which has no Horn function closed form seemingly.
Let me finish. Please correct me and give me feedback!
Update:
Using this Approach0 search, there were two non-closed form solutions in the question for:

Supposedly Simple Integral $$\int_0^x\frac{e^{-t^2}}{a^2+t^2}dt,$$

I also tried a few other series expansions, but they also had no closed form because of the coefficients of the gamma function arguments which could not use the Gauss Multiplication formula.
So please be our guest and help with a closed form.
