I am reading through my textbook, and there is a derivation of the $$\sin(2\pi x)\leftrightarrow \frac{1}{2i} \delta(\nu -1)-\frac{1}{2i} \delta(\nu+1)$$ transform pair using generalized functions, where $\varphi(x)$ is a test function:
\begin{align*} \int \mathcal{F} \{\sin(2\pi x)\} \varphi (\nu)\, d\nu &= \int_{-\infty}^{\infty} \sin(2\pi x) \Phi(x)\, dx\\ &=\int_{-\infty}^{\infty} \Phi(x)\frac{1}{2i}e^{i2\pi x}\, dx - \int_{-\infty}^{\infty} \Phi(x)\frac{1}{2i}e^{-i2\pi x}\, dx\\ &= \frac{1}{2i}\varphi(1) - \frac{1}{2i} \varphi(-1). \end{align*}
Which is then shown to be the Fourier Transform Pair. My question, however, is how they went from: \begin{equation} \int_{-\infty}^{\infty} \Phi(x)\frac{1}{2i}e^{i2\pi x}\, dx - \int_{-\infty}^{\infty} \Phi(x)\frac{1}{2i}e^{-i2\pi x}\, dx \end{equation}
to \begin{equation} \frac{1}{2i}\varphi(1) - \frac{1}{2i} \varphi(-1) \end{equation}
As this is not obvious to me.