Apply complex residue theory to non even function improper integration

Question

I am considering the integral

$$\int_{-\infty}^{\infty}\frac{t-5}{(t - 3+i)(t - 3-i)(t-2i)(t+2i)} dt.$$

The integrand, say $f(t)$, is readily not even, right? How do we use residues to solve this? The common technique uses the interior of large $[-R,R]$ upper-half plane semicircle and line segment combo. Only two singularities of the integrand are inside that interior. The rest is as usual.

Similar example is Complex analysis: Improper integral of a non-even function.

However, in this question and in that link as well, we will only get the value of Cauchy principle value $\lim_{R\rightarrow\infty} \int_{-R}^{R} f(t)\text{ }dt$, right?

If $f(t)$ is even, then we finally can get the value we want. Since here it is not, how do we continue? Similarly, why in that link, the answer can conclude the value too?

Thanks for any suggestion! Much appreciated!

Answer 1

In this case, you have the quotient of a polynomial function with degree $1$ by a polynomial function with degree $4$, without real roots. So, if $R(t)$ is the quotient, it is automatically true the the integrals $\int_0^\infty R(t)\,\mathrm dt$ and $\int_{-\infty}^0 R(t)\,\mathrm dt$ converge (compare them with $\frac1{t^3}$) and therefore$$\int_{-\infty}^\infty R(t)\,\mathrm dt=\lim_{R\to\infty}\int_{-R}^RR(t)\,\mathrm dt.$$So, go ahead and apply the method of residues.

Answer 2

$$\int_{-\infty}^{\infty}\frac{t-5}{(t - 3+i)(t - 3-i)(t-2i)(t+2i)} dt=2 \pi i \left[\text{res}(f(t),(t,2 i))+\text{res}(f(t),(t,3+i))\right]=2\pi i\left(-\frac{1}{15}+\frac{3 i}{40}+\frac{1}{15}+\frac{i}{20}\right)=2\pi i\left(\frac{i}{8}\right)=-\frac{\pi}{4}$$