It doesn't stand neither when a is negative. Hence, suppose a > 0.

Notice that, if k=0, we are left with ∫ dx / (x²+a²), which is π/a (change of variables, and use of tan⁻¹).

notice also that we are free to change k into -k, without altering the result.

Now k > 0.

Let I = ∫ cos²(kx) dx / (x²+a²) and J = ∫ sin²(kx) dx / (x²+a²)

I + J = π/a, as see before.
I - J = ∫ cos(2kx) dx / (x²+a²)

Let f (z) = e^(2ikz) / (z²+a²) for every complex number z except ia and -ia.

Consider K a semi-circle of center O and radius M, on the upper complex half-plane.
ia is the only pole of f in K :

Res(f ; ia) = lim (z→ia) f(z)*(z-ia) = lim e^(2ikz) / (z+ia) = e^(-2ka) / (2ia)

and ∫ f over ∂K = 2iπ * e^(-2ka) / (2ia) = (π/a) * e^(-2ka).

On the other hand,
∫(∂K) f = ∫[-M;M] f + ∫ f as z = M.e^(iθ), θ∈[0;π].

Prove that |f(z)| < 1 / (M²-a²) when z = M.e^(iθ).
It follows that the second integral tends to 0 as M tends to +∞.

We are left with : ∫ f over R = (π/a) * e^(-2ka).

Since ∫ cos(2kx) dx / (x²+a²) is the real part of it, we can see that : I - J = (π/a) * e^(-2ka).

Now solve for I :

I + J = π/a
I - J = (π/a) * e^(-2ka)

im dead lmfao