If we want to solve the Klein-Gordon equation $(\Box+m^2)\phi=0$ we can use the Fourier transform: $$\phi(x)=\int_\mathbb{R^4} \mathrm{e}^{\mathrm{i}kx}c_k\:\mathrm{d^4}k$$ Which gives: $$(\Box+m^2)\phi(x)= \int_ \mathbb{R^4} (m^2-k^2)\mathrm{e}^{\mathrm{i}kx}c_k\:\mathrm{d^4}k=0 $$ My question is now why it’s allowed to commute differentiation and integration? I know from the measure theory version of the Leibniz rule that it‘s possible when $(m^2-k^2)c_k\in L(\mathbb{R}^n)$ but that’s useless without knowing $c_k$.
2026-03-27 13:46:04.1774619164
Fourier transform: Why is it allowed to differentiate under the integral?
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