Question on Multiplicative Factor for a Standard Fourier Transform I am reading a physics text and not sure how one goes from 1.4 to 1.5

Is the standard Fourier transform for $1/x$ not -$i / \pi$, so where does the factor of $4 \pi/k^2 $ come from?
 A: The equation has solution in the context of distributions though classical solution is possible.I think the use of Fourier Transform cannot be justified as the potential in question is not necessarily an $ L^1$ function nor is it an $ L^2$ function.
But here is what I know :
Let a continuous function $f(x,y,z)$ have first order partial derivatives $f_x,f_y,f_z$ exist at every point and the function and all its first order partial derivatives are absolutely integrable. 
If for $x',y',z' \in R$ there is a non negative integrable function $h(x',y',z')$ such that for some measurable set $A \subset R^3$ with finite Lebesegue outer measure:     
$|f_x(x+x',y',z')|< h(x',y',z')$ almost everywhere in $A$
$|f_y(x',y+y',z')|< h(x',y',z')$ almost everywhere in $A$
$|f_z(x',y',z+z') |< h(x',y',z')$ almost everywhere in $A$
$ r'=\sqrt{|x-x'|^2+|y-y'|^2+|z-z'|^2}$
let:
$$g(x,y,z)=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac {f(x',y',z')}{4\pi r'}dx'dy'dz',\ \ \text{s.t.}\ \ \nabla^2g= \frac {\partial^2g}{\partial x^2}+\frac {\partial^2g}{\partial y^2}+\frac {\partial^2g}{\partial z^2}$$
Then $\nabla^2g=-f$ 
Proof :
Theorem 1. If a sequence of  absolutely continuous functions {$f_n$} converges pointwise to some $f$  and if the sequence of  derivatives {$f_n’$}  converges almost everywhere to some  $g$ and if {$f_n’$} is uniformly integrable then $\lim\limits_{n\mapsto \infty} f_n’ = g= f’$ almost everywhere. Where the derivative of  $f$ is $f’$. If the convergence is pointwise and $ g $ is continuous then $ f'$ = $ g $ everywhere.
Proof : by FTC  $f_n(x) – f_n(a) = \int_a^x f_n’ dx$
By Vitali convergence theorem : $\lim\limits_{n\mapsto \infty}\int_a^x f_n’ dx = \int_a^x g dx$
Therefore $\lim\limits_{n\mapsto \infty}( f_n(x) – f_n(a))= \int_a^x g dx$
$f(x)-f(a) = \int_a^x g dx$
$f(x)’=g$ almost everywhere
If the convergence is pointwise and $ g $ is continuous then $ f'$ = $ g $ everywhere.
Theorem 2. Divergence Theorem 
Theorem 3. Leibniz Integral Rule : Measure theoretic version 
$ r'=\sqrt{|x-x'|^2+|y-y'|^2+|z-z'|^2}$
$ r=\sqrt{|x'|^2+|y'|^2+|z'|^2}$
Define $f_N=\int_{-N}^{N} \int_{-N}^{N} \int_{-N}^{N}f(x’,y’,z’)\frac {1}{4\pi r’}erf(\frac {r’N}{\sqrt 2})dx'dy'dz'$
$erf(\frac {r’N}{\sqrt 2})=\frac {2}{\sqrt \pi}\int_0^{\frac {r’N}{\sqrt 2}}e^{-t^2}dt$ 
$\lim\limits_{N\mapsto \infty}erf(\frac {r’N}{\sqrt 2})=1$
$\lim\limits_{N\mapsto \infty} f_{N} = g$
$\frac {1}{4\pi r’}erf(\frac {r’N}{\sqrt 2})$ can be developed into a power series of $r'$ by simply plugging Taylor's expansion of $erf(\frac {r’N}{\sqrt 2})$
By theorem 3 $\frac {\partial f_{N}}{\partial x}=\int_{-N}^{N} \int_{-N}^{N} \int_{-N}^{N} \frac {f(x',y',z')(x'-x)erf(\frac {r’ N}{\sqrt 2})}{4\pi r'^3}dx'dy'dz'-$
$\int_{-N}^{N} \int_{-N}^{N} \int_{-N}^{N}\frac {N(x' - x)f(x',y',z') e^{-(\frac {r’^2 N^2}{ 2})}}{\sqrt 2 4\pi r'^2}dx'dy'dz'$
The existence of$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac {f(x',y',z')}{4\pi r'}dx'dy'dz' $ and  $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac {f(x',y',z')(x'-x)}{4\pi r'^3}dx'dy'dz' $ which is also continuous can be shown by converting to polar coordinates.
Now we can apply theorem $1$  to conclude  $\lim\limits_{N\mapsto \infty} \nabla f_{N} =\nabla g$
$\nabla^2 f_{N} =-\int_{-N}^N \int_{-N}^N \int_{-N}^N \frac {N^3f(x',y',z')e^{-(\frac {r’^2 N^2}{ 2})}}{(\sqrt{2\pi})^3}dx'dy'dz'$
$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac {f(x',y',z')(x'-x)}{4\pi r'^3}dx'dy'dz' = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac {f(x'+x,y'+y,z'+z)(x')}{4\pi r^3}dx'dy'dz'$ 
it follows from theorem 3 : 
$\nabla^2  g=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac {f_x(x'+x,y'+y,z'+z)(x')}{4\pi r^3}dx'dy'dz'+\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac {f_y(x'+x,y'+y,z'+z)(y')}{4\pi r^3}dx'dy'dz'+\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac {f_z(x'+x,y'+y,z'+z)(z')}{4\pi r^3}dx'dy'dz'$
Now application of  dominated convergence theorem ,the fact that $\lim\limits_{N\mapsto \infty} \nabla f_{N} =\nabla f$ and using theorem 2 : $\int_a^b \int_a^b \int_a^b(\nabla .\nabla  f_N )dxdydz = \int_{R^2} \nabla f_N .dA$
$ \int_{R^2} \nabla g .dA=\lim\limits_{N\mapsto \infty}\int_{R^2} \nabla f_{N} .dA$
$\int_a^b  \int_a^b \int_a^b \lim\limits_{N\mapsto \infty} (\nabla^2  f_{N} )=\lim\limits_{N\mapsto \infty} \int_a^b  \int_a^b \int_a^b  (\nabla^2  f_{N} )$
And $\int_a^b \int_a^b \int_a^b(\nabla^2  g )dxdydz =\int_{R^2} \nabla g .dA=\lim\limits_{N\mapsto \infty}\int_{R^2} \nabla f_{N} .dA=\lim\limits_{N\mapsto \infty} \int_a^b  \int_a^b \int_a^b  (\nabla^2  f_{N} )=\int_a^b  \int_a^b \int_a^b \lim\limits_{N\mapsto \infty} (\nabla^2  f_{N} )$
$\int_a^b \int_a^b \int_a^b((\nabla^2  g )-\lim\limits_{N\mapsto \infty} (\nabla^2  f_{N} ))dxdydz=0$
Since this is true for any $a$ and $b$ ,we conclude $\nabla^2g=\lim\limits_{N\mapsto \infty} (\nabla^2  f_{N} )=-f$ .
A: For $ k\ne 0$
$$F(k) = \iiint \frac{1}{|x|} e^{-i <k, x>}d^3 x$$
$\frac1{|x|}$ is radial thus $F(k)= F(|k|,0,0)$. We obtain $$F(k) = \iiint \frac{1}{|x|} e^{-i |k| x_1}d^3 x\\= \iiint \frac{1}{|y/|k||} e^{-i y_1}d^3 (y/|k|)=\iiint \frac{|k|}{|y|} e^{-i y_1} \frac{d^3y}{|k|^3} = \frac{F(1,0,0)}{|k|^2}$$
From there it suffices to show $F(1,0,0) = 4\pi$ so that $$\phi = \rho \ast \frac{ -G}{|x|} \implies \hat{\phi}(k)=\iiint \phi(x) e^{-i <k, x>}d^3 x =-G F(k)\hat{\rho}(k) =-G\frac{4\pi}{|k|^2}  \hat{\rho}(k)$$
