A question related with propagators in position space

[sufive]

Dear Colleagues,

How to prove Fourier transformation rules of the following function

\int k^a * exp[i*k*x] * dk = x^{-1-a} * Gamma[1+a] * sin[a*pi/2] ------------(1)

I need this equation to prove some conclusions in translating propagators
in momentum space to position-space. I wish those know solutions to this
question give me hints or references urgently.

The following is the information I know, but I cannot get to the final goal.
By variable changes, k*x -> z, it's easy to get the factor k^{-1-a}, i.e.

l.h.s -> (\int z^a * exp[i*z] * dz) / x^{1+a} -------------------------------------(2)

but the remaining integration seems very difficult.
We know,

\int z^a * exp[-z] * dz \propto Gamma[1+a] --------------------------------------(3)

So, my key question is, how to do integrations in eq(2) whose exponential
argument is imaginary instead negative?

Thanks to everyone!

 

[eljose79]

Dear colleagues,

Thank you for your question. The Fourier transformation rules for a function can be proven using integration techniques and properties of the Fourier transform. In this case, we can use the following steps to prove the equation given in (1):

1. Start by rewriting the original equation as:

\int k^a * exp[i*k*x] * dk = x^{-1-a} * Gamma[1+a] * sin[a*pi/2] ------------(4)

2. Use the substitution k*x = z, which gives us dk = dz/x and transforms the integral into:

\int z^a * exp[i*z] * dz / x^{1+a} --------------------------------------------(5)

3. To simplify the integral, we can use the property of the Fourier transform that states that:

\int f(x) * exp[i*k*x] * dx = F(k) ---------------------------------------------(6)

where F(k) is the Fourier transform of the function f(x). In this case, we can apply this property to the integral in (5), which gives us:

F(k) = \int z^a * exp[i*z] * dz, where F(k) is the Fourier transform of x^{1+a} ----(7)

4. Now, we can use the inverse Fourier transform to express F(k) in terms of x^{1+a}. The inverse Fourier transform states that:

f(x) = (1/2pi) * \int F(k) * exp[-i*k*x] * dk -------------------------------------(8)

Applying this to (7), we get:

x^{1+a} = (1/2pi) * \int z^a * exp[i*z] * dz * exp[-i*k*x] * dk -------------------(9)

5. Rearranging (9), we get:

\int z^a * exp[i*z] * dz = (2pi) * x^{1+a} * exp[i*k*x] * dk ----------------------(10)

6. Substituting (10) into (5), we get:

\int z^a * exp[i*z] * dz / x^{1+a} = (2pi) * exp[i*k*x] * dk ----------------------(11)

7. Finally, we can use the property of the Fourier transform that states that:

F(k) = \int f(x) * exp[-i*k*x] * dx ------------------------------------------------(12)