Fourier transform is carried out on both sides of the equation at the same time.

f[∫(-∞,+∞)f(τ)/[(x-τ)^2+a^2]dτ]=f[ 1/(x^2+b^2)]

Through convolution theorem, there are

∫ (-∞,+∞) f (τ)/[(x-τ) 2+a 2] d τ = f (x) ※1/(x 2+a 2) where※ stands for convolution symbol.

Get f [∫ (-∞, +∞) f (τ)/[(x-τ) 2+A2] dτ] = f [f (x) ※1/(x2+A2)] = f [f (x)

The solution is f [f (x)] = f [1/(x2+B2)]/f [1/(x2+a2)].

Let's look at the Fourier transform of 1/(x 2+b 2), as can be seen from the definition.

f[ 1/(x^2+b^2)]=∫(-∞,+∞)e^(iwx)/(x^2+b^2)dx=2∫(0,+∞)cos(wx)/(x^2+b^2)dx

Let f (z) = e (iwz)/(z 2+b 2) have a singularity of IB.

When w>0, f [1/(x 2+b 2)] = 2π ires [f (z), ib] = (π/b) e (-WB).

When w

Therefore, f [1/(x2+B2)] = (π/b) e (-| w | b).

f[ 1/(x2+a2)]=(π/a)e(-| w | a)