Etydhv

I have a simple Fourier transform problem, originating from mathematical physics (system of linear PDEs), which reduces to taking the integral
$$
I(alpha)equivint_-infty^infty dk e^ikr cfracalpha^2 + beta k^2k(k^2+alpha^2)
$$
where $alpha$ and $beta$ are some free real-valued parameters. I need to compute $I(0)$. It turns out that if I simply set $alpha=0$ in the integral above, I get an absolutely different answer than if I first compute the integral and set $alphato 0$ in the final expression.

My question is as follows: why do these two procedures lead to different answers? From the physical viewpoint this means that a massless field behaves in a totally different way than a massive with infinitesimal mass, which seems unreasonable.

My attempt is as follows.

I lift the pole at $k=0$ to the upper half-plane:
$$
I_varepsilon(alpha) equiv int_-infty^infty dk e^ikr cfracalpha^2 + beta k^2(k-ivarepsilon)(k+ialpha)(k-ialpha)equiv int_-infty^infty dk cfracg(k)h(k)
$$
where
$$
h(k) = (k-ivarepsilon)(k+ialpha)(k-ialpha)=k^3-ivarepsilon k^2+a^2k +i varepsilon a^2,
$$
$$
h'(k)=3k^2-2ikvarepsilon+alpha^2
$$
I take the integral making use of the Jordan's lemma and Cauchy theorem: I choose a contour in the upper half-plane $mathbb H$, so that the integral reduces to the sum of residues at $k=ivarepsilon$ and $k=ialpha$:
$$
I_1(alpha)=2pi i lim_varepsilon to 0left[cfracg(ivarepsilon)h'(ivarepsilon)+cfracg(ialpha)h'(ialpha)right]
$$
$$
=2pi ilim_varepsilonto 0left[ cfracalpha^2 + beta (ivarepsilon )^23(ivarepsilon )^2-2i(ivarepsilon )varepsilon +alpha^2,e^-varepsilon r+
cfracalpha^2 + beta (ialpha)^23(ialpha)^2-2i(ialpha)varepsilon +alpha^2,e^-alpha r right]
$$
$$
=2pi i left[ 1+
cfrac1 - beta-3+1 right]=2pi i cfrac1+beta2=pi i(1+beta).
$$
Thus, $I_1(alpha) = pi i(1+beta)$. Clearly then, $lim_alphato0I_1(alpha) = pi i(1+beta)$.

However, if I consider
$$
I_2equiv I(alpha=0)=
int_-infty^infty dk e^ikr cfracbetak = lim_varepsilon rightarrow 0int_-infty^infty dk e^ikr cfracbetak-ivarepsilon = lim_varepsilon rightarrow 02pi i beta e^-varepsilon r = 2pi i beta.
$$
Hence, $I_2neq lim_alphato0I_1$!!! Please, give a hint why this sort of thing happens. I clearly understand that in my reasoning there is a flaw -- but it escapes me.

Thank you for any help!

Care should be taken because of the pole at $k=0$, let me first take the principal value of the integral. I note that $I(alpha,-r)=barI(alpha,r)$ (complex conjugate), for convenience I will restrict myself to $r>0$.

The principal value integral evaluates to
$$I(alpha,r)=int_-infty^infty dk, e^ikr cfracalpha^2 + beta k^2k(k^2+alpha^2)=ipi+ipi(beta-1)e^alpha.$$
So for $alpha=0$ the result is $I(0,r)=ipibeta$. There is no discontinuity at $alpha=0$, but there is a discontinuous derivative. The same result would have been obtained if we would have set $alpha=0$ before carrying out the integral, because the principal value integral $int dk e^ikrk^-1=ipi$ for $r>0$.

Alternatively, you could shift the pole off the real axis, still taking $r>0$ the answer then becomes
$$I(alpha,r)=lim_epsilondownarrow 0int_-infty^infty dk, e^ikr cfracalpha^2 + beta k^2(k-iepsilon)(k^2+alpha^2)=2ipi+ipi(beta-1)e^alpha.$$
So now $I(0,r)=ipi(beta+1)$, still continuous and with a discontinuous derivative.

We have recoved the result $I_1$, where the limit $epsilondownarrow 0$ is taken before the limit $alpharightarrow 0$. These two limits do not commute, which is why the result $I_2$ in the OP differs from $I_1$.