Fourier Integral Representation of Delta Function and Equity of Complex Integrands In Peskin & Schroeder's QFT textbook, the following Fourier integral representation of Dirac delta function is given on page no. xxi.
$$\int d^4x \, \exp(ik\cdot x) = (2\pi)^4 \, \delta^{(4)}(k). $$
As Dirac delta function is an even function, I can write the following.
$$(2\pi)^4 \, \delta^{(4)}(-k) = \int d^4x \, \exp(-ik\cdot x) = (2\pi)^4 \, \delta^{(4)}(k). $$
Therefore, I conclude that $$\int d^4x \, \exp(ik\cdot x) = \int d^4x \, \exp(-ik\cdot x). \tag 1$$
The integrands are not equal to each other: $$\exp(ik\cdot x) \neq \exp(-ik\cdot x).$$
My Question
What does bar me from writing $\exp(ik\cdot x) = \exp(-ik\cdot x)$ from Eq. (1)? I don't think that I can use Fundamental Theorem of Calculus here to show that the integrands are equal. (If it is easier to understand in 1D, I would appreciate an answer for the 1D case as well.)
 A: You can just do a u-substitution to show that even though the integrands aren't equal, the integrals are. This is the same as saying that integral on the reals of f(x) and f(-x) are equal whenever f is any even function like 1/x^2 or e^(-x^2). Here it helps to realize that the argument of exp(ikx), ie. ikx, is imaginary, so that the integrand has absolute value equal to 1 (as opposed to being more than 1, so that you might think the integral converges in one case but diverges in the other).
A: Barring obvious analytic issues of how to interpret $\int_{\mathbb R^4} e^{ik\cdot x}\,dx$, I think you may be trying to apply something like the following fact:
Fact. If $f(x)$ and $g(x)$ are integrable functions such that $\int_Af(x)\,dx = \int_Ag(x)\,dx$ for all measurable $A$, then $f(x) = g(x)$ almost everywhere.
But it looks like you are trying to conclude that $f(x) = g(x)$ when you only know $\int f(x)\,dx = \int g(x)\,dx$ (i.e. that the integrals over the whole space are equal). This is almost never true, even if $f(x)$ and $g(x)$ were non-negative.
Or to put it in analogous terms with another more transparent example with fewer analytic subtleties, you know $\int_0^{2\pi} \sin(x)\,dx=\int_0^{2\pi}0\,dx$. What prevents you from concluding $\sin(x)=0$ for all $x\in[0,2\pi]$?

Added: A "(Lebesgue)-measurable set" is a technical term, referring to a member of the Lebesgue $\sigma$-algebra on $\mathbb R^n$. Amongst the measurable sets are all open sets, for example. Completely generally, a measurable set is one that can be well approximated by finite unions of open cubes in a certain precise sense. For this reason, I think it's useful intuition to think of a general measurable set as a finite union of open cubes, and then the fact boils down to verifying that $\int_Qf(x)\,dx = \int_Qg(x)\,dx$ for every open cube $Q$. There are many good resources to learn about measure theory. I would recommend Stein and Shakarchi Real Analysis.
In analysis and in physics (I think in E&M I have seen this, though I am blanking on a concrete example right now), the main application of the fact I know is the following. Sometimes you have a function $f(x)$ you care about and you know $\int f(x)\phi(x)\,dx = 0$ for every smooth compactly supported $\phi(x)$ (often verified with an integration by parts/Stokes' theorem). Then the fact can be verified by approximating sets that are finite unions of cubes by sums of smooth compactly supported bump functions. So we can conclude that $f(x) = 0$.
The other area I know where the fact gets applied with some frequency is in probability theory, and specifically martingale theory and its applications. Briefly, in that context if you have a $\sigma$-algebra $\mathcal F$, an $\mathcal F$-measurable random variable $Z$ you suspect is the conditional expectation of another random variable $X$ with respect to $\mathcal F$, you can prove $Z = E[X\mid\mathcal F]$ by checking that $E[Z;A] = E[X;A]$ for every $A\in\mathcal F$. Relevant in this connection is Dynkin's $\pi$-$\lambda$ theorem which is a powerful method for establishing identities such as this.
