Confusion regarding a second order pole on the contour

Question

Let us define a function

$$f(x) = \frac{1}{(x^2 + A^2) (x^2 - B^2)^2}$$ where, $A,B > 0$. We need to calculate

$$G = \int_{-\infty}^{\infty} f(x) dx $$

I am confused regarding whether $G$ diverges or not.

Argument 1: $G$ does not diverge. By complex analysis, we can deform the contour around the pole on the real line via a small semicircle and in the limit that the semi-circle's radius goes to zero, we can get a non-divergent contribution from the 2nd order pole, zero in this case.

Argument 2: $G$ diverges because there is a second order pole on the real line. This divergence is actually there and one cannot complex analyze one's way out of it.

Any insight regarding this confusion will be really helpful.

Answer 1

Argument 2 is correct, the integral diverges because of the second-order poles at $x=B$ and $x=-B$.

Argument 1 is wrong because the integral of $f$ along a small semicircle around this singularities does not converge to zero if the radius goes to zero.

Near $z=B$ is $$ \begin{align} f(z) &= \frac{1}{(z-B)^2} \cdot \frac{1}{(z^2+A^2)(z+B)^2} \\ &= \frac{1}{(z-B)^2} \left( \frac{1}{(B^2+A^2)4B^2} + \cdots\right) \end{align} $$ and integration over the semicircle $\gamma_r(t) = B + re^{it}$, $0 \le t \le \pi$, gives $$ \int_{\gamma_r} f(z) \, dz = \frac{2}{r} \cdot \frac{1}{(B^2+A^2)4B^2} + O(1) $$ and that diverges to $\infty$ for $r \to 0^+$. For the semicircle $\gamma_r'(t) = -B + re^{it}$, $0 \le t \le \pi$ one gets the same asymptotics.

Now let $\Gamma$ be a the contour which goes from $+R$ to $-R$ in a positively oriented “large” semicircle, and then from $-R$ to $+R$ along the real axis, only with $[-B-r, -B+r]$ and $[B-r, B+r]$ replaced by “small” negatively oriented semicircles. Then $$ \int_\Gamma f(z) \, dz = 2 \pi i \operatorname{Res}(f, iA) = \frac{\pi}{A(B^2-A^2)} $$ for sufficiently large $R$ and sufficiently small $r$. The contribution of the “large” semicircle with radius $R$ is $O(1/R^5)$, and one gets $$ \int_{-R}^{-B-r} f(x) \, dx + \int_{-B+r}^{B-r}f(x) \, dx + \int_{B+r}^R f(x) \, dx = \frac{4}{r} \cdot \frac{1}{(B^2+A^2)4B^2} + O(1) + O(1/R^5) $$ for $r \to 0^+$ and $R \to +\infty$, which is consistent with the result from argument 2.

Answer 2

Since $$\frac1{4B^2(A^2+B^2)}\int_0^B\frac{dx}{(x-B)^2}dx\leq\int_0^{B} f(x)dx\leq\int_{-\infty}^\infty f(x)dx,$$ it is enough to show that $\int_0^B\frac{dx}{(x-B)^2}dx$ diverges: $$\int_0^B\frac{dx}{(x-B)^2}dx=\lim_{t\to B^{-}}\int_0^t\frac{dx}{(x-B)^2}dx\\ =\left.\lim_{t\to B^{-}}\frac{1}{B-x}\right\vert_0^t=\lim_{t\to B^{-}}\left(\frac{1}{B-t}-\frac1B\right)=\infty $$