Reading the Schwartz quantum field theory p.235 I came across the following integration (Calculating Born approximation) :
$$ \frac{m_e^2}{4\pi^2}(\int d^{3}x e^{-i \vec{k}\cdot \vec{x}} \frac{e^2}{4\pi |\vec{x}|})^2=\frac{m_e^2}{4\pi^{2}} (\frac{e^2}{|\vec{k}|^2})^2 $$
My question is, literally, $$\int d^3x e^{ -i\vec{k}\cdot \vec{x}}\frac{1}{|\vec{x}|} = 4\pi \frac{1}{|\vec{k}|^2}$$ ?
If this is true, we maybe prove the above integration. And is it true? Can anyone help?
My first attempt : I will try to show that $$\int d^3k e^{ -i\vec{k}\cdot \vec{x}}\frac{1}{|\vec{k}|} = 4\pi \frac{1}{|\vec{x}|^2}.$$ Then by changing $\vec{k}$ and $\vec{x}$, we get the desired result.
By introducing Spherical coordinate and letting $r:=|\vec{x}|$, we have
$$\int d^3k e^{ -i\vec{k}\cdot \vec{x}}\frac{1}{|\vec{k}|} = \int^{\infty}_{0}d|\vec{k}| \int^{\pi}_{0} d\theta \int^{2\pi}_{0}d\phi \frac{1}{|\vec{k}|}e^{-i|\vec{k}||\vec{x}|cos\theta}|\vec{k}|^2 sin\theta$$
$$ = \int^{2\pi}_{0}d\phi \int^{\infty}_{0}d|\vec{k}|\int^{\pi}_{0}d\theta e^{-i|\vec{k}|rcos\theta}|\vec{k}|sin\theta \overset{\mathrm{u=cos\theta}}{=} 2\pi \int^{\infty}_{0}|\vec{k}|d|\vec{k}| \int^{-1}_{1} -due^{-i|\vec{k}|ru}$$
$$= 2\pi \int^{\infty}_{0}|\vec{k}|d|\vec{k}|\int^{1}_{-1}du e^{-i|\vec{k}|r u} = 2\pi \int^{\infty}_{0}|\vec{k}|d|\vec{k}|[\frac{1}{-i|\vec{k}|r}e^{-i|\vec{k}|ru}]^{1}_{-1} $$ $$ = 2\pi \int^{\infty}_{0}|\vec{k}|d|\vec{k}| \frac{1}{-i|\vec{k}|r}(e^{-i|\vec{k}|r} - e^{i|\vec{k}|r}) = \frac{2\pi i}{r} \int^{\infty}_{0} d|\vec{k}|(e^{-i|\vec{k}|r} - e^{i |\vec{k}|r})$$
And note that
$$ \int^{\infty}_{0} d|\vec{k}|e^{-i|\vec{k}|r} = \frac{1}{-ir}e^{-ir|\vec{k}|} |^{\infty}_{0} = \frac{1}{-ir}\cdot 0 - (\frac{1}{-ir}\cdot 1) = \frac{1}{ir} $$
and
$$ \int^{\infty}_{0} d|\vec{k}|(-e^{i|\vec{k}|r}) = -\frac{1}{ir}e^{ir|\vec{k}|}|^{\infty}_{0} = -\infty + \frac{1}{ir} $$ .
So,
$$\frac{2\pi i}{r} \int^{\infty}_{0} d|\vec{k}|(e^{-i|\vec{k}|r} - e^{i |\vec{k}|r}) = \frac{2\pi i}{r} [\frac{1}{ir}-\infty+\frac{1}{ir}] = \frac{2\pi i}{r} [\frac{2}{ir} -\infty] =4\pi \frac{1}{r^2} -\infty = 4 \pi \frac{1}{|\vec{x}|^2}-\infty $$.
An issue that makes me annoying is the appearence of $\infty$. My calculation is correct? Where did I made mistake? Can any one point out?