I am having a difficult time evaluating an integral unlike any integral I have seen before. To get right into things here is the integral:
$$\frac{A}{\sigma_o\sqrt{2\pi}}\int_{-\infty}^\infty [\sin(kx)-A\delta\sin^2(kx)]e^{-\left(\frac{(x-x')^2}{2\sigma^2}\right)[1-2A\delta\sin(kx)]}dx$$
One very important thing to know about this is that $\delta$ is very small, so any term with $\delta^2$ can be immediately thrown out.
Previously I had solved this integral by distributing the exponent, then splitting that up into two exponents:
$$e^{-\left(\frac{(x-x')^2}{2\sigma^2}\right)}e^{A\delta\sin(kx)\left(\frac{(x-x')^2}{\sigma^2}\right)}$$
Then I used a Taylor series on the second exponent to write it as:
$$1+A\delta\sin(kx)\left(\frac{(x-x')^2}{\sigma^2}\right)$$ You can ignore the remaining terms because they involve $\delta^2$ . From here I was able to solve the integral. However the solution did not give me the correct answer so I concluded that the Taylor series was not an appropriate approximation.
So from here I've gotten stuck. I don't know how to integrate a gaussian with $\sin(kx)$ in the exponent, and I don't know of any other way to get it out of there. Any help on this would be greatly appreciated and I'll do my best to answer any questions.