Semicircle contour for integrating $t^2/(t^2+a^2)^3$ > Let $a\in\mathbb{R}$. Evaluate $$\int_0^{\infty}\dfrac{t^2}{(t^2+a^2)^3}dt$$ The function is even, so the value of the integral is half of $\int_{-\infty}^{\infty}\dfrac{t^2}{(t^2+a^2)^3}dt$ I'm going to use countour integration along the semicircle in upper-half plane with radius $R$. The integral along the real line is $\int_{-R}^{R}\dfrac{t^2}{(t^2+a^2)^3}dt$. Now for the curved part of the semicircle, I parametrize $z=Re^{it}$ for $0\leq t\leq 2\pi$. Then the integral becomes $$\int \frac{z^2}{(z^2+a^2)^3}dz=\int_0^{2\pi}\frac{R^2e^{2it}}{(R^2e^{2it}+a^2)^3}\cdot Rie^{it}dt$$ And the absolute value of this integral is bounded from above by $\left|\dfrac{2\pi R^3}{(R^2e^{2it}+a^2)^3}\right|$. Why does this go to zero as $R\rightarrow \infty$? It seems like it's because the denominator has larger power of $R$, but how to make that rigorous?

Standard calculus result. If $p(x),q(x)$ are two polynomials with $\deg(q(x))>\deg(p(x))$, then $$\lim_{x\rightarrow\pm \infty} \frac{p(x)}{q(x)} = 0.$$ You can prove it by dividing top and bottom by the highest power of $x$ appearing in $p$.

In your particular case, the numerator is degree 3 in $R$, and the denominator is degree 6 in $R$.

Also note that $|R^2 e^{2it}+a^2| \geq R^2-a^2$.