Let $f$ be smooth with compact support. Consider the double layer potential (up to a constant) $ u(x_1,x_2)=-2\pi\int_{-\infty}^{\infty}\frac{\partial}{\partial x_2}\Gamma(x_1-y_1,x_2)f(y_1)\,dy_1= $ $ \int_{-\infty}^{\infty} \frac{x_2f(y_1)}{x^2 + x_2^2} dx, $ where $\Gamma(x)=-\frac1{2\pi}\log|x|$ is a fundamental solution for the Laplace equation. As is known $u$ is smooth up to the boundary for smooth $f$. We have $ \frac{\partial u(0,0)}{\partial x_2}= \lim_{\epsilon \to 0+} \frac{u(0,\epsilon)-u(0,0)}\epsilon= \lim_{\epsilon \to 0+} \frac{u(0,\epsilon)-f(0)}\epsilon= \lim_{\epsilon \to 0+}\int_{-\infty}^{\infty} \frac{ f(y_1)}{x^2 + \epsilon^2} dx, $ which is the required value. To calculate it note that $ \frac{\partial u(x_1,x_2)}{\partial x_2} = -2\pi\int_{-\infty}^{\infty}\frac{\partial^2}{\partial x_2^2}\Gamma(x_1-y_1,x_2)f(y_1)\,dy_1= $ $ 2\pi\int_{-\infty}^{\infty}\frac{\partial^2}{\partial y_1^2}\Gamma(x_1-y_1,x_2)f(y_1)\,dy_1= 2\pi\int_{-\infty}^{\infty}\Gamma(x_1-y_1,x_2)f''(y_1)\,dy_1 $ because $\Gamma$ satisfies the Laplace equation. The last integral converges uniformly for $|x|\le 1$ so taking the limit $x\to0$ gives $ \frac{\partial u(0,0)}{\partial x_2}=2\pi\int_{-\infty}^{\infty}\Gamma(0-y,0)f''(y)\,dy_1=-\int_{-\infty}^{\infty}\log|y|f''(y)\,dy. $