Let $\Gamma=ABCD$, where:
$$A(0,0),B(a,0),C(a,\alpha),D(0,\alpha)$$
$$\omega=Pdx+Qdy=e^{-x^2+y^2}\cos(2xy)dx+e^{-x^2+y^2}\sin(2xy)$$
with $\alpha\in\mathbb{R}$, and $a$ they don't say anything about it. However I have to use this to calculate:
$$\int_0^{\infty}e^{-t^2}cos(2\alpha t)dt$$
But since $\Gamma$ is a closed curve then $\int_{\Gamma}\omega= 0$. However how do I parametrize this and what is the intuition behind it?
My attempt:
when trying to parametrize I said: $\Gamma = AB\cup BC\cup CD\cup DA$
and so:
$AB:$ $$x(t) = t$$ $$y(t) = 0$$ $t\in [0,a]$
$BC:$ $$x(t)=a$$ $$y(t) = t$$ $t\in [0,\alpha ]$
$CD:$ $$x(t)=a-t$$ $$y(t)=\alpha$$ $t\in [0,a]$
$DA:$ $$x(t)=0$$ $$y(t) = \alpha -t$$
$t\in [0,\alpha]$
Now how do I construct the integral?
I solved it myself actually:
Using that parametrization you get that:
$$0=\int_{\Gamma}\omega=\int_{0}^{a}e^{-t^2}dt+\int_0^{\alpha}e^{-a^2+t^2}cos(2at)dt-\int_0^ae^{-t^2+\alpha^2}cos(2\alpha t)dt$$
If we note that desired integral with $I$. We can see that from the last expression we get:
$$Ie^{\alpha^2}=\int_{0}^{a}e^{-t^2}dt+\int_0^{\alpha}e^{-a^2+t^2}cos(2at)dt$$
Letting $a\to\infty$
We get: $$\boxed{I=\frac {\sqrt{\pi}}{2}e^{-\alpha^2}}$$