In an Introduction to Quantum Field Theory by M. E. Peskin & D. V. Schroeder (eq. 2.56 on page 30) the following relation for the retarded Green's function was established: $$(\partial^2 + m^2) D_R(x - y) = -i\delta^{(4)}(x - y).\tag{2.56}$$ After that, I assume they use the Fourier representation of a delta function: $$\int \frac{d^4p}{(2\pi)^4}e^{-ip \cdot (x-y)}\,$$ and the Fourier transformation of $ D_R(x - y)$: $$D_R(x - y)=\int \frac{d^4p}{(2\pi)^4}e^{-ip \cdot (x-y)} \tilde{D}_R(p)\,\tag{2.57}$$ to arrive at the following relation: $$\int \frac{d^4p}{(2\pi)^4} ((-p^2 + m^2)\tilde{D}_R(p) + i)e^{-ip \cdot (x-y)} = 0\,$$ THE CONFUSING PART IS THAT THEY CLAIM THAT THE EXPRESSION UNDER THE INTEGRAL MUST BE EQUAL TO ZERO: $$(-p^2 + m^2)\tilde{D}_R(p) + i = 0.\tag{p.30}$$ For me this does not make any sense because the fact that the integral is zero DOES NOT mean that the function under the integral is identically zero. Simple example would be the integral of $\cos(x)$ on the interval from 0 to $2\pi$.
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7$\begingroup$ Your penultimate equation is a Fourier transform, and if the Fourier transform of a function vanishes, then the function must vanish, roughly speaking. $\endgroup$– Tobias FünkeCommented Jul 6 at 16:54
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$\begingroup$ Please give a more detailed reference (book title, authors, edition, chapter, page and equation numbers). $\endgroup$– Tobias FünkeCommented Jul 6 at 16:58
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1$\begingroup$ @TobiasFünke That seems to be an answer, not a comment. $\endgroup$– ACuriousMind ♦Commented Jul 7 at 9:21
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$\begingroup$ "For me this does not make any sense because the fact that the integral is zero DOES NOT mean that the function under the integral is identically zero." Sure, but this particular integral you are interested in is zero for every value of $z = (x-y)$. You may have learned something about complete sets of functions. The plane waves are a complete set in this context and so the vanishing of the integral for all $z$ implies the vanishing of the integrand. (This is a basic result of Fourier analysis, which might account for all the downvotes the question is receiving...) $\endgroup$– hftCommented Jul 7 at 23:02
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$\begingroup$ @hft To be fair, P&S (or indeed most physicists on the subject, myself included) never specify the conditions imposed on the solution that make plane waves a complete basis. Show a pure mathematician that particular passage of P&S, and they are sure to raise the same question. I, for one, consider the down votes too harsh. $\endgroup$– T.P. HoCommented Jul 8 at 3:21
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