Prove $ I = \int_{-\infty}^\infty dz \frac{1}{(z^2-a^2)^2} \frac{1}{(z-a)^2+b^2} $ diverges I'm trying to analyze the diverging integral
$$ I = \int_{-\infty}^\infty dz \dfrac{1}{(z^2-a^2)^2} \dfrac{1}{(z-a)^2+b^2} $$
where $a$ and $b$ are real, positive numbers. There are vertical asymptotes at $z=\pm a$, so $I$ has no solution. 
Is there some special theorem that would support the statement that $I$ diverges? Just because an integrand has vertical asymptotes doesn't mean that the integral diverges eg. 
$$\int \dfrac{dx}{x} = \ln |x| + C$$
 A: The improper integral of a real function is usually defined for a function $f$ with singularities by splitting the domain at each of $f$'s singularities and taking an improper integral on each of the resulting sections. To be more precise, if we want to integrate $f$ from $a$ to $b$, and it has singularities at $x_1 < x_2 < \ldots < x_n$ (between $a$ and $b$), then we usually define
$$
\int_a^b f(x) \, dx = \int_a^{x_1} f(x) \, dx + \ldots + \int_{x_n}^b f(x) \, dx.
$$
The integral on the LHS is said to converge if and only if all of the (improper) integrals on the RHS do. This is a bit of an abuse of notation, and it's often not made explicit enough that this is what the convention is, so it can often be a source of confusion.
Your example with the integral of $1/x$ is something that gets taught to high school students everywhere, and, while perhaps useful for its simplicity, it just exacerbates this confusion when more rigour is necessary. In fact, if you analyse it by splitting up the domain (as is typically the convention) then we see that the improper integral is not necessarily defined for any pair of endpoints:
$$
\int_{-1}^{1} \frac 1 x \, dx = \int_{-1}^{0} \frac 1 x \, dx + \int_{0}^{1} \frac 1 x \, dx,
$$
and neither of the integrals on the RHS are convergent. Thus it is inaccurate to say that "the integral of $1/x$ with respect to $x$ is $\ln |x| + C$"; this fails when you try to integrate over the singularity. (There are ways to assign values to the integrals that go over 0, such as the Cauchy principal value, but these are not the "default".)
With regards to your original problem, then, we note that the function
$$
f(z) = \frac{1}{(z^2 - a^2)^2} \frac{1}{(z-a)^2 + b^2}
$$
has singularities at $z = \pm a$ and nowhere else, as you observed, so the definition of your integral is
$$
\int_{-\infty}^{\infty} f(z) \, dz = \int_{-\infty}^{-a} f(z) \, dz + \int_{-a}^{a} f(z) \, dz + \int_{a}^{\infty} f(z) \, dz,
$$
and it is not hard to observe that e.g. the middle integral here diverges, because it has positive integrand and
\begin{align*}
\int_{-a}^{a} f(z) \, dz &\geq \frac{1}{4a^2 + b^2} \int_{-a}^{a} \frac{1}{(x^2 - a^2)^2}
 \, dx \\
&\geq \frac{1}{4a^2 + b^2} \int_{\max(a-1,-a)}^{a} \frac{1}{(x^2 - a^2)^2} \, dx \\
&\geq \frac{1}{4a^2 + b^2} \int_{\max(a-1,-a)}^{a} \frac{1}{x^2 - a^2} \, dx
\end{align*}
which can easily be seen to diverge by (for instance) explicitly integrating and noticing that the limit that defines the improper integral,
$$
\int_{\max(a-1,-a)}^{a} \frac{1}{x^2 - a^2} \, dx = \lim_{c \to a^{-}} \int_{\max(a-1,-c)}^{c} \frac{1}{x^2 - a^2} \, dx,
$$
is divergent. Thus the integral in question is divergent.
