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Changed to a similar notation as the answer's. Also removed some unnecessary text. Added some other relevant text and a footnote.
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It seems I am stuck with a (at a first sight) trivial problem.

It's from the "Quarks and Leptons" (Halzen, Martin) book page $141$, where one considers the following integral:

$$\tag{1} T_{fi} = -i\int \!d^4x \, A(t,\vec{x})\,B(t,\vec{x})\frac{1}{|\vec{q}|^2}. $$$$\tag{1} T_{fi} = -i\int \!d^4x \, J_0^A(t_A,\vec{x}_A)\,J_0^B(t_A,\vec{x}_A)\frac{1}{|\vec{q}|^2}. $$ In equation $(1)$, $A$$J_0^A$ and $B$$J_0^B$ are the zeroth component of two electron currents.: $$J_\mu(x) = j_\mu\mathrm{exp}[(p_f-p_i)\cdot x].$$

Now, according to the authors, one can rewrite $(1)$ by making use of the Fourier transform

$$\tag{2} \frac{1}{|q|^2} = \int\! d^3x\, e^{i\vec{q}\cdot\vec{x}}\frac{1}{4\pi|\vec{x}|}, $$ to the following $$ \tag{3} T_{fi} = -i\int \!dt\int d^3x_1\int d^3x_2 \, \frac{A(t,\vec{x_1})\,B(t,\vec{x_2})}{4\pi|\vec{x_2}-\vec{x_1}|}. $$$$ \tag{3} T_{fi}^{Coul} = -i\int \!dt_A\int d^3x_A\int d^3x_B \, \frac{J_0^A(t,\vec{x}_B)\,J_0^B(t,\vec{x}_B)}{4\pi|\vec{x}_B-\vec{x}_A|}. $$

Plugging $(2)$ intoEquation $(1)$$(3)$ is just not enough, there seems to be a delta function missing somewhere. Maybe it'sthen interpreted as the notation or I have misunderstood their statementsinstantaneous$^1$ Coulomb interaction between the charges of the particles, i.e. perhaps $(1)$$J_0^A$ and $(2)$ should not be equal to each other, rather just proportional or something?$J_0^B$.

Anyone got any idea? This is of course not any homeworkThe derivation of any kind, just for funthis is given in the answer below.

I'm sure I've seen this integral$^1$I.e. interaction without retardation at time (similar problem) in classical electrodynamics, it looks very familiar$t_A$.

It seems I am stuck with a (at a first sight) trivial problem.

It's from the "Quarks and Leptons" (Halzen, Martin) book page $141$, where one considers the following integral:

$$\tag{1} T_{fi} = -i\int \!d^4x \, A(t,\vec{x})\,B(t,\vec{x})\frac{1}{|\vec{q}|^2}. $$ In equation $(1)$, $A$ and $B$ are two electron currents.

Now according to the authors, one can rewrite $(1)$ by making use of the Fourier transform

$$\tag{2} \frac{1}{|q|^2} = \int\! d^3x\, e^{i\vec{q}\cdot\vec{x}}\frac{1}{4\pi|\vec{x}|}, $$ to the following $$ \tag{3} T_{fi} = -i\int \!dt\int d^3x_1\int d^3x_2 \, \frac{A(t,\vec{x_1})\,B(t,\vec{x_2})}{4\pi|\vec{x_2}-\vec{x_1}|}. $$

Plugging $(2)$ into $(1)$ is just not enough, there seems to be a delta function missing somewhere. Maybe it's the notation or I have misunderstood their statements, i.e. perhaps $(1)$ and $(2)$ should not be equal to each other, rather just proportional or something?

Anyone got any idea? This is of course not any homework of any kind, just for fun.

I'm sure I've seen this integral (similar problem) in classical electrodynamics, it looks very familiar.

It seems I am stuck with a (at a first sight) trivial problem.

It's from the "Quarks and Leptons" (Halzen, Martin) book page $141$, where one considers the following integral:

$$\tag{1} T_{fi} = -i\int \!d^4x \, J_0^A(t_A,\vec{x}_A)\,J_0^B(t_A,\vec{x}_A)\frac{1}{|\vec{q}|^2}. $$ In equation $(1)$, $J_0^A$ and $J_0^B$ are the zeroth component of two electron currents: $$J_\mu(x) = j_\mu\mathrm{exp}[(p_f-p_i)\cdot x].$$

Now, according to the authors, one can rewrite $(1)$ by making use of the Fourier transform

$$\tag{2} \frac{1}{|q|^2} = \int\! d^3x\, e^{i\vec{q}\cdot\vec{x}}\frac{1}{4\pi|\vec{x}|}, $$ to the following $$ \tag{3} T_{fi}^{Coul} = -i\int \!dt_A\int d^3x_A\int d^3x_B \, \frac{J_0^A(t,\vec{x}_B)\,J_0^B(t,\vec{x}_B)}{4\pi|\vec{x}_B-\vec{x}_A|}. $$

Equation $(3)$ is then interpreted as the instantaneous$^1$ Coulomb interaction between the charges of the particles, $J_0^A$ and $J_0^B$.

The derivation of this is given in the answer below.

$^1$I.e. interaction without retardation at time $t_A$.

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EDIT: I think I've got it, I will post the answer later.

It seems I am stuck with a (at a first sight) trivial problem.

It's from the "Quarks and Leptons" (Halzen, Martin) book page $141$, where one considers the following integral:

$$\tag{1} T_{fi} = -i\int \!d^4x \, A(t,\vec{x})\,B(t,\vec{x})\frac{1}{|\vec{q}|^2}. $$ In equation $(1)$, $A$ and $B$ are two electron currents.

Now according to the authors, one can rewrite $(1)$ by making use of the Fourier transform

$$\tag{2} \frac{1}{|q|^2} = \int\! d^3x\, e^{i\vec{q}\cdot\vec{x}}\frac{1}{4\pi|\vec{x}|}, $$ to the following $$ \tag{3} T_{fi} = -i\int \!dt\int d^3x_1\int d^3x_2 \, \frac{A(t,\vec{x_1})\,B(t,\vec{x_2})}{4\pi|\vec{x_2}-\vec{x_1}|}. $$

Plugging $(2)$ into $(1)$ is just not enough, there seems to be a delta function missing somewhere. Maybe it's the notation or I have misunderstood their statements, i.e. perhaps $(1)$ and $(2)$ should not be equal to each other, rather just proportional or something?

Anyone got any idea? This is of course not any homework of any kind, just for fun.

I'm sure I've seen this integral (similar problem) in classical electrodynamics, it looks very familiar.

EDIT: I think I've got it, I will post the answer later.

It seems I am stuck with a (at a first sight) trivial problem.

It's from the "Quarks and Leptons" (Halzen, Martin) book page $141$, where one considers the following integral:

$$\tag{1} T_{fi} = -i\int \!d^4x \, A(t,\vec{x})\,B(t,\vec{x})\frac{1}{|\vec{q}|^2}. $$ In equation $(1)$, $A$ and $B$ are two electron currents.

Now according to the authors, one can rewrite $(1)$ by making use of the Fourier transform

$$\tag{2} \frac{1}{|q|^2} = \int\! d^3x\, e^{i\vec{q}\cdot\vec{x}}\frac{1}{4\pi|\vec{x}|}, $$ to the following $$ \tag{3} T_{fi} = -i\int \!dt\int d^3x_1\int d^3x_2 \, \frac{A(t,\vec{x_1})\,B(t,\vec{x_2})}{4\pi|\vec{x_2}-\vec{x_1}|}. $$

Plugging $(2)$ into $(1)$ is just not enough, there seems to be a delta function missing somewhere. Maybe it's the notation or I have misunderstood their statements, i.e. perhaps $(1)$ and $(2)$ should not be equal to each other, rather just proportional or something?

Anyone got any idea? This is of course not any homework of any kind, just for fun.

I'm sure I've seen this integral (similar problem) in classical electrodynamics, it looks very familiar.

It seems I am stuck with a (at a first sight) trivial problem.

It's from the "Quarks and Leptons" (Halzen, Martin) book page $141$, where one considers the following integral:

$$\tag{1} T_{fi} = -i\int \!d^4x \, A(t,\vec{x})\,B(t,\vec{x})\frac{1}{|\vec{q}|^2}. $$ In equation $(1)$, $A$ and $B$ are two electron currents.

Now according to the authors, one can rewrite $(1)$ by making use of the Fourier transform

$$\tag{2} \frac{1}{|q|^2} = \int\! d^3x\, e^{i\vec{q}\cdot\vec{x}}\frac{1}{4\pi|\vec{x}|}, $$ to the following $$ \tag{3} T_{fi} = -i\int \!dt\int d^3x_1\int d^3x_2 \, \frac{A(t,\vec{x_1})\,B(t,\vec{x_2})}{4\pi|\vec{x_2}-\vec{x_1}|}. $$

Plugging $(2)$ into $(1)$ is just not enough, there seems to be a delta function missing somewhere. Maybe it's the notation or I have misunderstood their statements, i.e. perhaps $(1)$ and $(2)$ should not be equal to each other, rather just proportional or something?

Anyone got any idea? This is of course not any homework of any kind, just for fun.

I'm sure I've seen this integral (similar problem) in classical electrodynamics, it looks very familiar.

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Physics_maths
Physics_maths
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EDIT: I think I've got it, I will post the answer later.

It seems I am stuck with a (at a first sight) trivial problem.

It's from the "Quarks and Leptons" (Halzen, Martin) book page $141$, where one considers the following integral:

$$\tag{1} T_{fi} = -i\int \!d^4x \, A(t,\vec{x})\,B(t,\vec{x})\frac{1}{|\vec{q}|^2}. $$ In equation $(1)$, $A$ and $B$ are two electron currents.

Now according to the authors, one can rewrite $(1)$ by making use of the Fourier transform

$$\tag{2} \frac{1}{|q|^2} = \int\! d^3x\, e^{i\vec{q}\cdot\vec{x}}\frac{1}{4\pi|\vec{x}|}, $$ to the following $$ \tag{3} T_{fi} = -i\int \!dt\int d^3x_1\int d^3x_2 \, \frac{A(t,\vec{x_1})\,B(t,\vec{x_2})}{4\pi|\vec{x_2}-\vec{x_1}|}. $$

Plugging $(2)$ into $(1)$ is just not enough, there seems to be a delta function missing somewhere. Maybe it's the notation or I have misunderstood their statements, i.e. perhaps $(1)$ and $(2)$ should not be equal to each other, rather just proportional or something?

Anyone got any idea? This is of course not any homework of any kind, just for fun.

I'm sure I've seen this integral (similar problem) in classical electrodynamics, it looks very familiar.

It seems I am stuck with a (at a first sight) trivial problem.

It's from the "Quarks and Leptons" (Halzen, Martin) book page $141$, where one considers the following integral:

$$\tag{1} T_{fi} = -i\int \!d^4x \, A(t,\vec{x})\,B(t,\vec{x})\frac{1}{|\vec{q}|^2}. $$ In equation $(1)$, $A$ and $B$ are two electron currents.

Now according to the authors, one can rewrite $(1)$ by making use of the Fourier transform

$$\tag{2} \frac{1}{|q|^2} = \int\! d^3x\, e^{i\vec{q}\cdot\vec{x}}\frac{1}{4\pi|\vec{x}|}, $$ to the following $$ \tag{3} T_{fi} = -i\int \!dt\int d^3x_1\int d^3x_2 \, \frac{A(t,\vec{x_1})\,B(t,\vec{x_2})}{4\pi|\vec{x_2}-\vec{x_1}|}. $$

Plugging $(2)$ into $(1)$ is just not enough, there seems to be a delta function missing somewhere. Maybe it's the notation or I have misunderstood their statements, i.e. perhaps $(1)$ and $(2)$ should not be equal to each other, rather just proportional or something?

Anyone got any idea? This is of course not any homework of any kind, just for fun.

I'm sure I've seen this integral (similar problem) in classical electrodynamics, it looks very familiar.

EDIT: I think I've got it, I will post the answer later.

It seems I am stuck with a (at a first sight) trivial problem.

It's from the "Quarks and Leptons" (Halzen, Martin) book page $141$, where one considers the following integral:

$$\tag{1} T_{fi} = -i\int \!d^4x \, A(t,\vec{x})\,B(t,\vec{x})\frac{1}{|\vec{q}|^2}. $$ In equation $(1)$, $A$ and $B$ are two electron currents.

Now according to the authors, one can rewrite $(1)$ by making use of the Fourier transform

$$\tag{2} \frac{1}{|q|^2} = \int\! d^3x\, e^{i\vec{q}\cdot\vec{x}}\frac{1}{4\pi|\vec{x}|}, $$ to the following $$ \tag{3} T_{fi} = -i\int \!dt\int d^3x_1\int d^3x_2 \, \frac{A(t,\vec{x_1})\,B(t,\vec{x_2})}{4\pi|\vec{x_2}-\vec{x_1}|}. $$

Plugging $(2)$ into $(1)$ is just not enough, there seems to be a delta function missing somewhere. Maybe it's the notation or I have misunderstood their statements, i.e. perhaps $(1)$ and $(2)$ should not be equal to each other, rather just proportional or something?

Anyone got any idea? This is of course not any homework of any kind, just for fun.

I'm sure I've seen this integral (similar problem) in classical electrodynamics, it looks very familiar.

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