Question

\[\int_{-inf}^{inf} e^{-x^2}e^{-x y} dx\]

Answer 1

@myininaya ,@amistre64

Answer 2

I think this is the solution\[e^{\frac{y^2}{4}}\sqrt{\pi}\]

Answer 3

show the work; I am not interested in solution

Answer 4

@zarkon

Answer 5

@Jemurray3

Answer 6

\[ \large \int_{-\infty}^\infty e^{-x^2} e^{-xy} dx = \int e^{-(x^2+xy)}dx = e^{y^2/4}\int e^{-(x^2+xy+y^2/4)}dx\]
\[ \large = e^{y^2/4}\int e^{-(x+y/2)^2}dx = \sqrt{\pi}\cdot e^{y^2 / 4}\]

Answer 7

how did you get \[\int e^{-(x^2+xy)}dx = e^{y^2/4}\]
?

Answer 8

I didn't. I multiplied the integrand by exp(-y^2/4) and to compensate multiplied the outside by exp(y^2/4).

Answer 9

thanks, man

Answer 10

Sure