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note that $$e^{a\partial/\partial x}f(x)=f(x+a)$$ is the translation operator, so your exponent of the delta function gives $2\pi \delta(x-in)$, which is indeed consistent with

$$\int_{-\infty}^\infty e^{izy}dy=2\pi\delta(z)$$

for $z=x-in$.