Domain simply connected for a differential form $\omega$

Question

I have the domain $\Omega=\{(x,y): 2y+x>0\}$ and the differential form

$$\omega=\frac{y}{2y+x}dx+\left(\log (2y+x)+\frac{2y}{2y+x}\right)dy.$$

I would like to evaluate $\int_{\gamma} \omega$ with $\gamma :[0,\pi]\to R^2$ : $\gamma(t)=\left(\sin t,\dfrac{e^t}{t+1}\right)$.

$\omega$ is closed but $\Omega$ is a simply connected domain (so can I say $\omega$ is exact and deduce that $\int_{\gamma} \omega$=0)?

Answer 1

Let write the $1$-form as $$ \omega=A\:dx+B\:dy $$ with $$ A=\frac{y}{2y+x},\quad B=\log (2y+x)+\frac{2y}{2y+x}. $$ One has $$ \frac{\partial A}{\partial y}=-\frac{2 y}{(x+2 y)^2}+\frac1{x+2 y},\quad \frac{\partial B}{\partial x}=-\frac{2 y}{(x+2 y)^2}+\frac1{x+2 y} $$ giving

$$ \begin{align} d\omega=&\left(\frac{\partial A}{\partial x}dx+ \frac{\partial A}{\partial y}dy\right)\wedge dx+\left(\frac{\partial B}{\partial x}dx+ \frac{\partial B}{\partial y}dy\right)\wedge dy \\\\&=\frac{\partial A}{\partial y}dy\wedge dx+\frac{\partial B}{\partial x}dx\wedge dy \\\\&=\left(-\frac{\partial A}{\partial y}+\frac{\partial B}{\partial x}\right)dx\wedge dy \\\\&=0. \end{align} $$

Thus $\omega$ is a closed $1$-form.

Observe that

$$ \begin{align} \int_{\gamma} \omega=&\int_0^\pi\frac{\frac{e^t}{t+1}}{\frac{2e^t}{t+1}+\sin t}\:\cos t \: dt+\int_0^\pi\left(\log \left(\frac{2e^t}{t+1}+\sin t\right)+\frac{2\frac{e^t}{t+1}}{2\frac{e^t}{t+1}+\sin t}\right)\frac{te^t}{(1+t)^2}\:dt \\\\&=\int_0^\pi\frac{d}{dt}\frac{e^t \log\left(\frac{2e^t}{1+t}+\sin t\right)}{1+t}\:dt \\\\&=\frac{\pi e^{\pi} +e^{\pi}\ln\left(\frac{2}{1+\pi }\right)}{1+\pi }-\ln 2. \end{align} $$