I have the following integral:
$$ \frac{1}{\sqrt{2\pi\sigma^{2}}}\int_{-\infty}^{\infty} e^{-{(x - \mu)^2/(2\sigma^{2})}} x\,dx.$$
And when I use the integral calculator to solve it, some of the terms contain an erf function, in the solution of the indefinite integral. I'm interested in only the definite integral. Can it be done without using the erf function?
Hint
$$ \frac{1}{\sqrt{2\pi\sigma^{2}}}\int_{-\infty}^{\infty} e^{-{(x - \mu)^2/(2\sigma^{2})}} x\,dx= \frac{1}{\sqrt{2\pi\sigma^{2}}}\int_{-\infty}^{\infty} e^{-{x^2/(2\sigma^{2})}} (x+\mu)\,dx$$ and note that $$f(x)=-f(-x)\quad,\quad \int_\Bbb R|f(x)|\,dx<\infty\implies\int_\Bbb Rf(x)\,dx=0$$
For calculating integrals like $$I=\int_\Bbb R e^{\alpha x^2}dx$$ use $$I^2=\iint_{\Bbb R^2} e^{\alpha (x^2+y^2)}\,dx\,dy=\int_0^{2\pi} \int_0^\infty e^{\alpha r^2}r\,dr\,d\theta.$$