I'm trying to evaluate two integrals: $$f(m,c) = \int_{-\infty}^\infty \frac{x}{\sqrt{x^2 + c^2}} e^{-(x-m)^2} dx$$ and $$g(m,c) = \int_{-\infty}^\infty \frac{1}{\sqrt{x^2 + c^2}} e^{-(x-m)^2} dx$$ where $c,m\in\mathbb{R}$
For $f(m,c)$ I was trying to substitute $x$ with $t = \sqrt{x^2 + c^2}$, which eliminates the fraction, but complicates the expression under the exponent. Also, I'm not sure what the new limits of integration should be: $$f(m,c) = \int_{?}^{?} e^{-(\sqrt{t^2-c^2}-m)^2} dt$$ Maybe this integral is related, but I couldn't bring $f(m,c)$ to it.
For $g(m,c)$ I've tried to substitute $x$ with $x=c\sinh{t}$, which leads to integral: $$g(m,c) = \int_{?}^{?} e^{-(c\sinh{t}-m)^2} dt$$ with which I don't know what to do next. Maybe this integral is related to $g(m,c)$.
If $m$ is "small" you could expand the exponential term as $$e^{-(x-m)^2}=e^{-x^2}\,\sum_{n=0}^\infty P_n(x)\, m^n$$ where the first polynomials are $$\left( \begin{array}{cc} n & P_n(x) \\ 0 & 1 \\ 1 & 2 x \\ 2 & 2 x^2-1 \\ 3 & \frac{4 x^3}{3}-2 x \\ 4 & \frac{2 x^4}{3}-2 x^2+\frac{1}{2} \\ 5 & \frac{4 x^5}{15}-\frac{4 x^3}{3}+x \\ 6 & \frac{4 x^6}{45}-\frac{2 x^4}{3}+x^2-\frac{1}{6} \\ 7 & \frac{8 x^7}{315}-\frac{4 x^5}{15}+\frac{2 x^3}{3}-\frac{x}{3} \\ 8 & \frac{2 x^8}{315}-\frac{4 x^6}{45}+\frac{x^4}{3}-\frac{x^2}{3}+\frac{1}{24} \\ 9 & \frac{4 x^9}{2835}-\frac{8 x^7}{315}+\frac{2 x^5}{15}-\frac{2 x^3}{9}+\frac{x}{12} \\ 10 & \frac{4 x^{10}}{14175}-\frac{2 x^8}{315}+\frac{2 x^6}{45}-\frac{x^4}{9}+\frac{x^2}{12}-\frac{1}{120} \\ 11 & \frac{8 x^{11}}{155925}-\frac{4 x^9}{2835}+\frac{4 x^7}{315}-\frac{2 x^5}{45}+\frac{x^3}{18}-\frac{x}{60} \\ 12 & \frac{4 x^{12}}{467775}-\frac{4 x^{10}}{14175}+\frac{x^8}{315}-\frac{2 x^6}{135}+\frac{x^4}{36}-\frac{x^2}{60}+\frac{1}{720} \end{array} \right)$$
and use $$I_{2n}= \int_{-\infty}^{+\infty} \frac{x^{2n}}{\sqrt{x^2 + c^2}} e^{-x^2}\, dx$$ and face linear combinations of Bessel functions