$\int_{-\infty}^\infty |(-ax^2 + \mathrm{i}bx + c)^{-1}|^2 \mathrm{d}x$

Question

In D. E. Newland's book, An Introduction to Random Vibrations, Spectral & Wavelet Analysis, there is an integral of the form $$ \int_{-\infty}^\infty \left|\frac{1}{-ax^2 + \mathrm{i}bx + c}\right|^2 \mathrm{d}x, $$ the solution of which is given as $\pi/bc$. The reader is pointed to a reference where a proof is given using Cauchy's residue theorem. Is this the only way to prove it or is there a simpler way to do it?

Answer 1

$\left|(c-ax^2)+i(bx)\right|^2 = (c-ax^2)^2 + b^2x^2$, hence the given integral equals

$$ I(a,b,c)=\int_{-\infty}^{+\infty}\frac{dx}{a^2 x^4 + (b^2-2ac) x^2 + c^2} $$ and since $$ \frac{\partial}{\partial a}\,I(a,b,c) = \int_{-\infty}^{+\infty}\frac{2 c x^2-2 a x^4}{\left(c^2+(b^2-2 a c) x^2+a^2 x^4\right)^2}\,dx = 0 $$ it follows that $$ I(a,b,c)=\lim_{a\to 0^+} I(a,b,c) \stackrel{\text{DCT}}{=} \int_{-\infty}^{+\infty}\frac{dx}{b^2 x^2+c^2}=\frac{\pi}{bc} $$ as claimed. The only non-trivial part is to notice that the given integral does not really depend on $a$.
I am assuming $a,b,c>0$ to avoid sign issues and $\text{DCT}$ stands for Dominated Convergence Theorem.