I need to determine functions $P,Q\in C^2(\mathbb R^2)$ provided that the line integral \begin{equation} I=\int_LP(x+\alpha,y+\beta)dx+Q(x+\alpha,y+\beta)dy \end{equation} over closed curve $L$ doesn't depend on $\alpha$ and $\beta$.
I am confused because it seems to me that for any two functions $P,Q\in C^2(\mathbb R^2)$ such that $\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$ integral $I$ doesn't depend on $\alpha,\beta$ because of Green theorem.
What am I missing?
On
We will prove that $\int_LP(x+\alpha,y+\beta)dx+Q(x+\alpha,y+\beta)dy$ doesn't depend on $\alpha$ and $\beta$ if and only if $\iint_D\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}dxdy=0$, where $D$ is a region enclosed by $L$.
If $\iint_D\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}dxdy=0$, then $\int_LP(x+\alpha,y+\beta)dx+Q(x+\alpha,y+\beta)dy$ for every $\alpha$ and $\beta$ by Green theorem.
Lets prove the converse. We have
\begin{equation} 0=\frac{1}{\sqrt{\alpha^2+\beta^2}}\left(\int_LP(x+\alpha,y+\beta)dx+Q(x+\alpha,y+\beta)dy-\int_LP(x,y)dx+Q(x,y)dy\right)\\ =\int_L\frac{P(x+\alpha,y+\beta)-P(x,y)}{\sqrt{\alpha^2+\beta^2}}dx+\frac{Q(x+\alpha,y+\beta)-Q(x,y)}{\sqrt{\alpha^2+\beta^2}}dy. \end{equation} Thus\begin{equation} 0=\lim_{(\alpha,\beta)\to (0,0)}\int_L\frac{\frac{\partial P}{\partial x}\alpha+\frac{\partial P}{\partial y}\beta}{\sqrt{\alpha^2+\beta^2}}dx+\frac{\frac{\partial Q}{\partial x}\alpha+\frac{\partial Q}{\partial y}\beta}{\sqrt{\alpha^2+\beta^2}}dy =\iint_D\left(\frac{\frac{\partial^2 Q}{\partial x^2}\alpha+\frac{\partial^2 Q}{\partial y\partial x}\beta}{\sqrt{\alpha^2+\beta^2}}-\frac{\frac{\partial^2 P}{\partial x\partial y}\alpha-\frac{\partial^2 P}{\partial y^2}\beta}{\sqrt{\alpha^2+\beta^2}}\right)dxdy=\\ \lim_{(\alpha,\beta)\to (0,0)}\iint_D\frac{ \frac{\partial{\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)} } {\partial x} \alpha + \frac{\partial{\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)} } {\partial y} \beta }{\sqrt{\alpha^2+\beta^2}}dxdy=\\ \lim_{(\alpha,\beta)\to (0,0)}\iint_D\left(\frac{\partial Q}{\partial x}(x+\alpha,y+\beta)-\frac{\partial P}{\partial y}(x+\alpha,y+\beta)+\frac{\partial Q}{\partial x}(x,y)-\frac{\partial P}{\partial y}(x,y)\right)dxdy=\\ 2\iint_D\frac{\partial Q}{\partial x}(x,y)-\frac{\partial P}{\partial y}(x,y)dxdy \end{equation}
Differentiating under the integral sign tells us that (omitting the arguments $(x+alpha,y+\beta)$ for convenience) \begin{align*} 0 &= \frac{\partial}{\partial\alpha}\int_L P(x+\alpha,y+\beta)\,dx + Q(x+\alpha,y+\beta)\,dy =\int_L \frac{\partial P}{\partial x}dx + \frac{\partial Q}{\partial x}dy \quad\text{and} \\ 0 &= \frac{\partial}{\partial\beta}\int_L P(x+\alpha,y+\beta)\,dx + Q(x+\alpha,y+\beta)\,dy =\int_L \frac{\partial P}{\partial y}dx + \frac{\partial Q}{\partial y}dy \end{align*} Now, letting $\Omega$ be the plane region bounded by $L$, we deduce from Green's Theorem that $$\int_\Omega \Big(\frac{\partial^2 Q}{\partial x^2} - \frac{\partial^2 P}{\partial y\partial x}\Big)\,dx\,dy = 0 = \int_\Omega \Big(\frac{\partial^2 Q}{\partial x\partial y} - \frac{\partial^2 P}{\partial y^2}\Big)\,dx\,dy = 0.$$ Certainly any functions $P$ and $Q$ making both these integrands identically $0$ will satisfy the given hypotheses. Without knowing $L$ specifically, that's all I can say.