Recently, I encountered a difficulty in proving the equation,
$$\int \mathrm d^4x\, \frac{e^{-ipx}}{x^4} =\pi^2 \ln(p^2+i\epsilon)\quad .$$
Here, $x$ is the coordinate, $p$ is the momentum in Minkowski space, $px=p^{\mu}x_{\mu}$, $x^2=x^{\mu}x_{\mu}$ and $x^4 =(x^2)^2$. The Minkowski sign convention is $g_{00}=-g_{11}=-g_{22}=-g_{33}=1$.
If anyone knows how to prove such an equation, could you please show me the hint?
First, I'll prove a formula that I always like to keep in hand. What is the $d$-dimensional Fourier transform of a rotationally-symmetric function $f(x)$? That is, $$ \tilde{f}(p) = \int d^d x \, e^{i \mathbf{p} \cdot \mathbf{x}} f(x). $$ Here, the magnitudes are $x = |\mathbf{x}|$, $p = |\mathbf{p}|$. I'm going to work with the Euclidean norm, $\mathbf{p} \cdot \mathbf{x} = p_1 x_1 + p_2 x_2 + \cdots + p_d x_d$, and we'll analytiically continue later.
We can set this up in hyperspherical coordinates. Further details can be found on Wikipedia. The coordinates are $$ \begin{align} x_1 &= x \cos \varphi_1 \\ x_2 &= x \sin \varphi_1 \cos \varphi_2 \\ x_3 &= x \sin \varphi_1 \sin \varphi_2 \cos \varphi_3 \\ & \vdots \\ x_{d-1} &= \sin \varphi_1 \cdots \sin \varphi_{d-2} \cos \varphi_{d-1} \\ x_{d} &= \sin \varphi_1 \cdots \sin \varphi_{d-1} \sin \varphi_{d-1} \end{align} $$ Here, $\varphi_{1}$, $\varphi_{2}$, ..., $\varphi_{d-2}$ range over $[0,\pi]$, and $\varphi_{d-1} \in [0,2 \pi)$. The spherical volume element in these coordinates is $$ d^d x = x^{d-1} \sin^{d-2} \varphi_1 \sin^{d-3} \varphi_2 \cdots \sin \varphi_{d-2} \, dx \, d\varphi_1 \, \cdots \, d\varphi_{d-1}. $$ We choose our coordinates so that the $x_1$ coordinate is in the same direction as $\mathbf{p}$, that is, $\mathbf{p} = (p,0,\cdots,0)$. So the integral is $$ \tilde{f}(p) = \int d^d x \, e^{i x p \cos \varphi_1} f(x). $$ We first perform the $\varphi_1$ integral. After some transformations, we recognize an integral representation of a Bessel function (equation 10.9.4 on the link): $$ \int_0^{\pi} d \varphi_1 \, e^{i p x \cos \varphi_1} \sin^{d-2} \varphi_1 = 2 \int_0^1 dz \, \cos(xpz) (1 - z^2)^{(d-3)/2} = \frac{ \pi^{1/2} \Gamma\left( \frac{d-1}{2} \right)}{\left(\frac{1}{2}xp\right)^{(d-2)/2}} J_{\frac{d-2}{2}}(xp). $$ The rest of the integral is the surface area of the $(n-2)$-sphere, which is the familiar expression $2\pi^{(d-1)/2}/\Gamma[(d-1)/2]$. So, $$ \tilde{f}(p) = \frac{(2 \pi)^{d/2}}{p^{(d-2)/2}} \int_0^{\infty} dx \, x^{d/2} \, J_{\frac{d-2}{2}}(xp) f(x). $$ This is a bit of a digression, but if you work with dimensional regularization a lot (or you like to work in arbitrary dimensions), this is a very useful formula to have around.
With this, the rest of your calculation becomes very straight-forward. You have $f(x) = 1/x^4$, so the integral becomes $$ \tilde{f}(p) = \frac{(2 \pi)^{d/2}}{p^{d - 4}} \int_0^{\infty} dx \, x^{d/2-4} J_{\frac{d-2}{2}}(x). $$ This is another point where one looks up properties of special functions (eq 10.22.43 in the link). We see that the integral only converges when $4<d<9$, but in this range of $d$, we have $$ \tilde{f}(p) = \frac{2^{d-4} \pi^{d/2}}{p^{d-4}} \Gamma\left( \frac{d-4}{2} \right). $$ If we expand this near $d=4$, we get $$ \tilde{f}(p) = -\frac{2 \pi^2}{4 - d} + C - \pi^2 \log(p^2), $$ where $C \approx 19.28$ is a constant. One can Wick rotate back to your preferred metric.
The addition of an $i \epsilon$ on one side of your equation but not the other involves the addition of some physical (e.g. "Feynman") prescription for your problem.
I may find another answer for this question, and I am not sure whether it is right or not?
We know the propagator for the Klein–Gordon field can be expressed by
$\lim\limits_{m->0} \int \frac{d^4p}{(2\pi)^4} \frac{e^{-ipx}}{p^2-m^2+i \epsilon}=\frac{i}{4 \pi^2} \frac{1}{x^2-i \epsilon}$. (1)
You may find the above equation in Eq.(27) of the book "Quantum Electrodynamics"(Fourth Edition, Walter Greiner and Joachim Reinhardt), and we can have
$ \int d^4x \frac{e^{ipx}}{x^2-i \epsilon}=-i4 \pi^2 \frac{1}{p^2+i \epsilon}$. (2)
Then we consider
$ \int d^4x \frac{e^{ipx}}{ (x^2-i \epsilon)^2}=A*ln(B*p^2+ C)$, (3)
where A, B and C are constants to be determined. Since the right side of Eq.(3) is dimensionless and only depends on scalar $p^2$, it can only described by the right side.
We apply the operator $-\frac{\partial}{\partial p^{\mu}}\frac{\partial}{\partial p_{\mu}}$ on the both sides of Eq.(3), in principle we will get Eq.(2). Then we obtain $A=i \pi^2$, $B=1$ and $C=i \epsilon$.
In the question I ask, I find that the denominator should be $(x^2-i \epsilon)^2$ instead of $(x^2)^2$after I checked other references, since the integral $ \int d^4x \frac{e^{ipx}}{ (x^2-i \epsilon)^2}$ is related to operator expansion of two electromagnetic currents.