# Path Integral on a circle: extension of integral range to the real line.

In "Path integrals in physics, vol.1. Stochastic processes and quantum mechanics, Chaichian M., Demichev A", eq-(2.4.43) they have: $$\langle \phi,t|\phi_0,t_0\rangle_{\text{circle}}\approx\prod_{n=1}^N\int_0^{2\pi}d\phi_n\prod_{n=1}^{N+1}\int_{\mathbb{R}}\frac{dk_n}{2\pi}\sum_{l_n\in\mathbb{Z}}\\\times\exp\left(i\sum_{n=1}^{N+1}[k_n(\phi_n-\phi_{n-1}+2\pi l_n)]-\frac{\epsilon}{\hbar}H(\hbar k_n,\phi_n)\right).$$ They then say that the sum over $$l_n$$ along with the integration over $$\phi_n$$ in the range $$[0,2\pi)$$ is equivalent to integrating $$\phi_n$$ over the whole real line.

I don't quite understand why this happens. Looking at just the $$\phi_1$$ terms we have

$$\int_0^{2\pi}d\phi_1\int_{\mathbb{R}}\frac{dk_1}{2\pi}\sum_{l_1\in\mathbb{Z}}\int_0^{2\pi}d\phi_2\int_{\mathbb{R}}\frac{dk_2}{2\pi}\times\sum_{l_2\in\mathbb{Z}}\mathrm{e}^{i2\pi k_1l_1}\mathrm{e}^{i[k_1(\phi_1-\phi_0)-\frac{\epsilon}{\hbar}H(\hbar k_1,\phi_1)]}\mathrm{e}^{i2\pi k_2l_2}\mathrm{e}^{i[k_2(\phi_2-\phi_1)-\frac{\epsilon}{\hbar}H(\hbar k_2,\phi_2)]}.$$ Now the summation for $$l_1$$ has a $$k_1$$ in the exponent, so just extending the integration range doesn't really make sense to me.

The end result they acquire is $$\langle \phi,t|\phi_0,t_0\rangle_{\text{circle}}\approx\sum_{l_{N+1}\in\mathbb{Z}}\prod_{n=1}^N\int_{\mathbb{R}}d\phi_n\prod_{n=1}^{N+1}\int_{\mathbb{R}}\frac{dk_n}{2\pi}\\\times\exp\left(i\sum_{n=1}^{N+1}[k_n(\phi_n-\phi_{n-1}+2\pi l_{N+1}\delta_{n,N+1})]-\frac{\epsilon}{\hbar}H(\hbar k_n,\phi_n)\right).$$

The seemed to have replaced $$\int_0^{2\pi}d\phi_n\sum_{l_n\in\mathbb{Z}}\mathrm{e}^{2\pi i k_nl_n}$$ with $$\int_{\mathbb{R}}d\phi_n$$. Any insights on how this is done would be greatly appreciated.

I'm trying to lead on to the path integral on the torus, so I want to understand the circle first.