I came across the following integral:
$$\int_{- \infty}^{\infty} \frac{x^2}{((x-t)^2 + \delta^2 )^2((x + t)^2 + \delta^2)^2} \textrm{d}x.$$
I understand that if we turn this into a complex integral along a semi-circular contour in the upper half of the complex plane, denoting the function in the integrand as $f(x)$, we get
$$\oint f(z) \textrm{d}z = 2 \pi i \sum \text{Res} (f(z)) = \int_{\text{arc}} f(z) \textrm{d}z + \int_{- \infty}^{\infty} f(x) \textrm{d}x $$
where $\int_{- \infty}^{\infty} f(x) \textrm{d}x$ is what I'm trying to calculate. The residues can be easily calculated, however, I'm struggling to show that the integral over the arc of the semicircle vanishes as the radius of the semicircle tends to infinity. I understand that you can use the estimation lemma and show that $$ |f(z)| < M $$ for some non-negative constant $M$ for all of the points on the arc of the semicircle. The absolute value of the numerator will be $z = R^2$, then I was hoping to show that the denominator is bounded from below using the triangle inequality $|a - b | \geq ||a|-|b||$.
$$| z^2 - 2tz + t^2 + \delta^2 |^2|z^2 + 2tz + t^2 + \delta^2|^2 = |z^2 - (2tz - t^2 - \delta^2)|^2 |z^2 - (- 2tz - t^2 - \delta^2)|^2$$
$$ \geq ||z|^2 - |2tz - t^2 - \delta^2||^2 ||z|^2 - |- 2tz - t^2 - \delta^2||^2$$
can I split the modulus signs in the denominator up any further so as to obtain an expression in terms of $R$, without any $z$ terms?
It's much easier to observe that the integrand is asymptotically equivalent to $z^{-6}$ as $z\to\infty$; thus $z\cdot f(z)$ is asymptotically equivalent to $z^{-5}$ and the arc integral vanishes.
We make the last step rigorous as follows; there is $R>0$ so large that for $|z|\ge R$ we have $|z^6\cdot f(z)|<2$. That is, $R\cdot |f(z)|<\frac{2}{|z|^5}<\frac{2}{R^5}$ on the region $|z|\ge R$. By the ML lemma the arc integral is bounded by $\pi R\cdot\sup_{|z|=R}|f(z)|<\frac{2\pi}{R^5}$ which vanishes as $R\to\infty$.
(asymptotic equivalence is not necessary, we only need a Big-$O$ bound)
The trick here is that known properties of limits make evaluating the behaviour of $f$ easy, and then this translates back into concrete bounds. Manually trying to bound the integrand is harder as you have to do all the basic grunt work again (show a product of continuous functions is continuous, etc.) with the absolute values.