How can I solve this integral with complex number?

Question

$n$ here is a complex number such that $n=n_r+in_i$

How can I solve this integral?

$$\int_{0}^{\infty}\frac{x^4}{|x^2-n^2|^2} d x=? $$

Answer 1

Let's change the notation to a more friendly one: $\;n:= a+bi\;$ , so:

$$x^2-n^2=x^2-(a^2-b^2+(2abi))=(x^2-a^2+b^2)-2abi\implies$$

$$|x^2-n^2|^2=(x^2-a^2+b^2)^2+4a^2b^2$$

Thus your integral is

$$\int_0^\infty\frac{x^4}{(x^2-a^2+b^2)^2+4a^2b^2}dx\stackrel{M=\max\{a,b\}}\ge\frac14\int_M^\infty\frac{x^4}{x^4+M^4}dx=$$

$$=\frac14\int_M^\infty\left(1-\frac{x^4}{x^4+M^4}\right)dx$$

and this last integral diverges.