Delta Function Identity in Peskin & Schroeder In Notation and Conventions of their QFT textbook (page no. xxi), Peskin and Schroeder mentions the following identity:
$$\int d^4x \, e^{ik\cdot x} = (2\pi)^4 \delta^{(4)}(k).$$
They define the Fourier transforms in four dimensions as follows.
$$f(x) = \int \frac{d^4k}{(2\pi)^4} e^{-ik\cdot x} \, \tilde{f}(k), \quad \tilde{f}(k) = \int d^4 x  \, e^{ik\cdot x} f(x).$$
I am having trouble to see how the identity follows from these definitions. Could you please clarify?
 A: Substituting the expression for $ f(x)$ into the expression for $\tilde f(k)$ yields
$$\tilde f(k^\prime) = \int \mathrm{d}^4x\,e^{ik^\prime x} \int\frac{\mathrm{d}^4\,k}{(2\pi)^4} e^{-ik x} \, \tilde f(k) =\int\frac{\mathrm{d}^4\,k}{(2\pi)^4}\,  \underbrace{\int \mathrm{d}^4x \,e^{ix(k^\prime-k)}}_{\equiv g(k^\prime-k)} \tilde f(k) \quad. $$
We therefore require that
$$ \tilde f(k^\prime)  = \int\frac{\mathrm{d}^4\,k}{(2\pi)^4}\, g(k^\prime-k) \,\tilde f(k) \quad . $$
But this is just the defining property of the delta distribution and thus:
$$g(k^\prime- k)  = (2\pi)^4\, \delta^{(4)}(k^\prime-k) \quad ,$$
which eventually shows
$$(2\pi)^4\, \delta^{(4)}(k) = \int \mathrm{d}^4x \,e^{ikx} \quad .$$
A: Notice that $f(x) = 1$ so you have
$$
\hat{f}(k) = \int \mathrm{d}^4x\, e^{ik\cdot x} 
$$
which is supposedly equal to $(2\pi)^4\delta^{(4)}(k)$. If you now consider the backwards transformation and plug that in you get a consistent result
$$
f(x) = \int\frac{\mathrm{d}^4\,k}{(2\pi)^4} e^{-ik\cdot x} (2\pi)^4 \delta^{(4)}(k) = e^{0} = 1 
$$
A: Given that $$f(x) = \int \frac{d^4k}{(2\pi)^4} e^{-ik\cdot x} \, \tilde{f}(k) \tag 1$$
and,
$$\tilde{f}(k) = \int d^4 x  \, e^{ik\cdot x} f(x). \tag 2$$
Let $$\tilde{f}(k) = \delta(k) = \left\{ \begin{array}{rl}
                                         \infty &\mbox{ if $x=0$} \\
                                          0 &\mbox{otherwise.}
                                         \end{array} \right.$$
Then the inverse Fourier transformation of $\tilde{f}(k) = \delta(k)$ is $f(x)$ given by
$$f(x) = \int \frac{d^4k}{(2\pi)^4} e^{-ik\cdot x} \, \delta(k) = \frac{1}{(2\pi)^4}.$$
Then it follows that the Fourier transformation of $\frac{1}{(2\pi)^4}$ is $\tilde{f}(k) = \delta(k)$:
$$ \int d^4 x  \, e^{ik\cdot x} \frac{1}{(2\pi)^4} = \tilde{f}(k) = \delta(k) \\
\Rightarrow \boxed{(2\pi)^4 \delta(k) = \int d^4 x  \, e^{ik\cdot x}}.$$
As the Dirac delta $\delta(k)$ is an even function, we get
$$(2\pi)^4 \delta(k) = \int d^4 x  \, e^{-ik\cdot x}.$$
Therefore, $$(2\pi)^4 \delta(k) = \int d^4 x  \, e^{\pm ik\cdot x}.$$
For more discussion, please check here.
