Integral representation of Kronecker delta

Question

I have seen this expression:

$$ \frac{1}{V}\int_Ve^{\pm i(\vec{n}-\vec{m})\cdot\vec{x}}d^3r=\delta_{\vec{n},\vec{m}} $$

How can I prove it? Intuitively it seems right, but I would like to do it rigorously.

Answer 1

If $\vec n=\vec m$, the integrand is $1$, so $$\frac 1V\int d^3r=\frac1V V=1$$ Let's do the integral in spherical coordinates for $\vec n\ne\vec m$, such that the $\theta$ angle is measured with respect to $\vec n -\vec m$. Then $$I=\frac1V\int_0^Rr^2dr\int_0^\pi\sin\theta d\theta\int_0^{2\pi}d\phi e^{i(\vec n-\vec m)\vec r}$$ In the exponent, let's use $|\vec n-\vec m|=a$. Then the exponent is $iar\cos\theta$. Note that $V=\frac 43 \pi R^3$, and the integral over $\phi$ is $2\pi$. Then $$I=\frac{3}{4\pi R^3}2\pi\int_0^R r^2dr\int_0^\pi\sin\theta d\theta e^{iar\cos\theta}$$ Change of variable $x=iar\cos\theta$. Then $dx=-iar\sin\theta d\theta$. When $\theta=0$ you have $x=iar$, and for $\theta=\pi$ you have $x=-iar$. Then $$I=\frac{3}{2R^3}\int_0^Rr^2dr\int_{iar}^{-iar}e^xdx\frac{1}{-iar}\\=\frac{3}{2R^3}\int_0^Rr^2dr\frac 1{iar}(e^{iar}-e^{-iar})\\=\frac{3}{2R^3}\int_0^Rr^2dr\frac 1{ar}2\sin(ar)$$ You can still see that in the limit $a=0$, which is $\vec n=\vec m$, $\frac{\sin(ar)}{ar}=1$, so $I=1$. If $a\ne 1$, change variable of integration $y=ar$, $dy=adr$, and the upper limit is now $Ra$:$$I=\frac3{R^3a^3}\int_0^{Ra}y\sin y dy$$ You can see that if $R$ is finite, the value of the integral might not be $0$. But you always have $-1\le\sin 1\le y$, so $$|I|\le \frac{3}{R^3a^3}\int_0^{Ra}y dy=\frac{3}{R^3a^3}\frac{R^2a^2}2$$ In the limit $R\to \infty$, with $\vec n\ne \vec m$, you get $I=0$