Integrating with volume element (d^3)x

[captain]

i'm at a loss about how to do this type of integration. can some one show me how to evaluate the integral of (d^3)k exp[ik*(x1-x2)]/[(k^2+m^2)(2pi)^3], where "*" is the dot product between the 3 vector k and (x1-x2), which are both 3 vectors. this come from the energy equation used to get the inverse square law (1/r^2).

 

[G01]

The notation d^3x is usually shorthand for a volume integral with differential elements dx, dy, dz (In Cartesian coordinates.) So:

$$\int_V f(x,y,z) d^3x = \int\int\int f(x,y,z) dxdydz$$

So, in your case, $$d^3k$$ probably stands for $$dk_x dk_y dk_z$$, where k_x, etc. are the components of the k vector. (Again, in Cartesian Coordinates)

 

[m2003]

Integrating with a volume element can be a challenging task, but with the right approach, it can be solved. In this case, we have an integral of the form (d^3)x, which represents the volume element in three-dimensional space. This means that we need to integrate over all three dimensions (x, y, z) to get the total volume.

To evaluate the given integral, we can use the method of integration by parts. This involves breaking down the integral into smaller parts and using the product rule to evaluate each part separately. In this case, we can break down the integral into two parts: the exponential term and the denominator term.

Starting with the exponential term, we can use the substitution u = ik*(x1-x2) to simplify the integral. This will result in du = ik*(dx1-dx2). We can then rewrite the integral as ∫exp(u)/(k^2+m^2)(2pi)^3 du. Using the product rule, we can evaluate this integral as [exp(u)/(k^2+m^2)(2pi)^3] - ∫exp(u)/[(k^2+m^2)(2pi)^3] du.

Next, we can focus on the denominator term. We can use the substitution v = (k^2+m^2)(2pi)^3 to simplify the integral. This will result in dv = 2k*(dk)/(k^2+m^2). We can then rewrite the integral as ∫1/v dv. Using the product rule, we can evaluate this integral as ln(v). Substituting back in for v, we get ln((k^2+m^2)(2pi)^3).

Bringing both parts together, we get the final integral as [exp(u)/(k^2+m^2)(2pi)^3] - ln((k^2+m^2)(2pi)^3) + C, where C is the constant of integration. Substituting back in for u and v, we get [exp(ik*(x1-x2))/(k^2+m^2)(2pi)^3] - ln((k^2+m^2)(2pi)^3) + C.

This is the general solution to the given integral. To evaluate it for specific values of x1, x2, k, and m, we can plug in the values and solve for the final result. I hope this