Integral $\int_{-\infty}^\infty \frac{\mathrm dx}{(a^2 - x^2)^2 + (bx)^2}$

Question

I'm trying to find the following integral: $$\int_{-\infty}^\infty \frac{\mathrm dx}{(a^2 - x^2)^2 + (bx)^2}$$

The result is $\pi/a^2b$ but somehow I can't recover that.

I tried the residue theorem but it gave me an awfully long square root of square root residue. Mathematica is giving me also an exact integral with tan$^{-1}$ that I can't get close to the expected result and that is equally complex. Any help is appreciated.

Answer 1

$$\int_{-\infty}^\infty \dfrac{dx}{(a^2 - x^2)^2 + (bx)^2}$$

= $$\int_{-\infty}^\infty \dfrac{dx}{(a^4+(-2a^2+b^2)x^2+x^4)}$$

Let $(-2a^2+b^2)$ = $\ p$. Now, dividing and multiplying by $\ 2a^2$, we get:

= $$\frac{1}{2a^2}\int_{-\infty}^\infty \dfrac{(x^2+a^2)-(x^2-a^2)dx}{(a^4+px^2+x^4)}$$

Let $\ I_1$ = $\int_{-\infty}^\infty \dfrac{(x^2+a^2)}{a^4+px^2+x^4}$ and $\ I_2$ = $\int_{-\infty}^\infty \dfrac{(x^2-a^2)}{a^4+px^2+x^4}$

Note that

$$\ I_1 = \int_{-\infty}^\infty \dfrac{(x^2+a^2)}{a^4+px^2+x^4}$$

$$ = \int_{-\infty}^\infty \dfrac{1+\frac{a^2}{x^2}dx}{x^2 + \frac{a^4}{x^2} + p}$$

Now, let $$\ x - \frac{a^2}{x} = t$$

Then $\ dt$ = $\ 1 + \frac{a^2}{x^2}$ $\cdot$ $\ dx$

Substituting t for x, we get

$$\ I_1 = \int_{-\infty}^{\infty} \dfrac{dt}{t^2+p+2a^2}$$

Thus,

$$\ I_1 = 2\cdot \dfrac{\pi}{2 \sqrt{p+2a^2}} = \dfrac{\pi}{b}$$

Proceed in a similar manner for $\ I_2$.

Answer 2

$(a^2+x^2)^2 + (bx)^2 = 0\\ a^2+ x^2 = \pm ibx\\ x^2 \pm ibx + a^2 = 0\\ x = \frac {\pm ib \pm i\sqrt {b^2 + 4a^2}}{2}$

We have 2 roots in the upper half-plane.

The residues appear to be

$2\pi i \left(\frac {1}{(ib)(ib + i\sqrt{b^2+4a^2})(i\sqrt{b^2+4a^2})}+ \frac {1}{(-ib)(-ib + i\sqrt{b^2+4a^2})(i\sqrt{b^2+4a^2})}\right)$

We factor out $i^3$ from the denominators. Both fractions have similar terms in the denominator. The common denominator is

$(b)(\sqrt{b^2+4a^2})(b+\sqrt{b^2+4a^2})(-b+\sqrt{b^2+4a^2}) = 4a^2b\sqrt{b^2+4a^2})$

$\frac {2\pi i}{i^3}(\frac {-b +\sqrt{b^2+4a^2}}{4a^2b\sqrt{b^2+4a^2}} - \frac {b+\sqrt{b^2+4a^2}}{4a^2b\sqrt{b^2+4a^2}})$

$\pi(\frac {1}{a^2\sqrt{b^2+4a^2}})$

And the contour along the semi-circle goes to 0 as the radius goes to infinity. Which is obvious do to the degree of the x terms in the denominator.