A improper integral with complex parameter

For a complex number $\displaystyle z$, How to evaluate

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A related problem. Here is the idea, since the integrand is an even function, then we can write the integral as

$$ \int_0^\infty\frac{\text{d}x}{x^2+(1-z^2x^2)^2}=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\text{d}x}{x^2+(1-z^2x^2)^2}. $$

Expanding the denominator and completing the square, we reach at

$$ \frac{1}{2}\int_{-\infty}^{\infty}\frac{\text{d}x}{x^2+(1-z^2x^2)^2}= \frac{1}{2}\int_{-\infty}^{\infty}\frac{\text{d}x}{(x^2+\alpha^2)^2+\beta^2}, $$

where $\alpha$ and $\beta$ are functions in $z$. Now, recalling the Fourier transform, we can consider the more general integral

$$ \frac{1}{2}\int_{-\infty}^{\infty}\frac{e^{-ixw}\text{d}x}{(x^2+\alpha^2)^2+\beta^2}. $$

I leave it here for you to finish the task.

By the quadratic formula, the denominator of the integrand has roots

r_1,r_2,r_3,r_4 &= \pm\sqrt{\frac{2z^2-1\pm\sqrt{1-4z^2}}{2z^4}}\\

and since the integrand is an even function,
\int_0^\infty \frac{dx}{x^2+(1-z^2x^2)^2} &= \frac{1}{2}\int_{-\infty}^\infty \frac{dx}{x^2+(1-z^2x^2)^2} \\
&= \frac{1}{2}\int_{-\infty}^\infty \frac{1/z^4}{x^4+(1-2z^2)x^2/z^4+1/z^4} dx \\
&= \frac{1}{2z^4}\int_{-\infty}^\infty f(x) dx

where $$f(x) = \frac{1}{(x-r_1)(x-r_2)(x-r_3)(x-r_4)}.$$ Since $f(z)$ is analytic in the upper half complex plane (except for a finite number of poles), and since $f(z)$ vanishes faster than $1/z^2$ for $|z|\rightarrow\infty$, the residue theorem gives $$\frac{1}{2z^4}\int_{-\infty}^\infty f(x)dx = \frac{\pi i}{z^4}\sum\mathrm{res\,\,}f$$ where $\sum\mathrm{res\,\,}f$ is the sum of the residues in the upper-half plane. The trick then is to determine which of the four poles are on the upper half plane. We define the roots as

r_1 &= \frac{1}{2z^2}\left(i+\sqrt{4z^2-1}\right) \\
r_2 &= -r_1 \\
r_3 &= \frac{1}{2z^2}\left(i-\sqrt{4z^2-1}\right) \\
r_4 &= -r_3

If $z$ is purely real, then $r_1$ and $r_3$ are in the upper half plane. Thus
\sum\mathrm{res\,\,}f &= \frac{1}{(r_1-r_2)(r_1-r_3)(r_1-r_4)} + \frac{1}{(r_3-r_1)(r_3-r_2)(r_3-r_4)} \\
&= -i\frac{z^4}{2}

after much algebra. Therefore we have

\int_0^\infty \frac{dx}{x^2+(1-z^2x^2)^2} &= \frac{\pi i}{z^4}\frac{-iz^4}{2} \\
&= \frac{\pi}{2}

which is independent of $z$ (if $z$ is purely real)! In the general case, however, compute the residues of the poles that are in the upper half plane, to be determined by the value of $z$.