$$ \int_{-\infty}^{\infty} dk e^{-ak^2+bk} = \sqrt{\frac{\pi}{a}}e^{\frac{b^2}{4a}} $$
As always any help appreciated - thank you!
Asked
Active
Viewed 40 times
0
-
I don’t think you need contour integration if you’re willing to use basic Gaussian identities. Just complete the square and change variables slightly and use the normal distribution pdf. – J.G Jun 06 '20 at 23:43
-
If you already know $$\int_{-\infty}^\infty e^{-x^2};dx$$ then it's easy. But if not, see https://math.stackexchange.com/a/34857/442 to do it by integrating around a parallelogram. – GEdgar Jun 06 '20 at 23:59
1 Answers
0
Use the integral of the exponential function, the function of $\exp$ is $$ e^{-ak^2 - bk} = e^{-(ak^2 - 2bk\frac{1}{2}\frac{\sqrt{a}}{\sqrt{a}}+b^2\frac{1}{4}\frac{1}{a})} \cdot e^{b^2\frac{1}{4}\frac{1}{a}} = e^{-(\sqrt{a}k -\frac{b}{2 \sqrt{a}})^2} \cdot e^{b^2\frac{1}{4}\frac{1}{a}} $$
Alex
- 19,262