Glasser's master theorem and an application

Glasser’s master theorem (special case): For any $a \ge 0$,

\[\boxed{\int_{-\infty}^{\infty} f \left( x - \frac{a}{x} \right) \, dx = \int_{-\infty}^{\infty} f (x) \, dx.}\]

Proof. Do a change of variable $u = x - \frac{a}{x}$. Then,

\[x = \begin{cases} \frac{ u - \sqrt{u^2 + 4a} }{2}, & x < 0, \\ \frac{ u + \sqrt{u^2 + 4a} }{2}, & x > 0, \end{cases} \quad\text{and hence}\quad \frac{dx}{du} = \begin{cases} \frac{1}{2} \left( 1 - \frac{u}{\sqrt{u^2 + 4a}} \right), & x < 0, \\ \frac{1}{2} \left( 1 + \frac{u}{\sqrt{u^2 + 4a}} \right), & x > 0. \end{cases}\]

Meanwhile, the integration limits need to be split into two cases as well:

\[\begin{alignat*}{3} &\text{When } x < 0: &\quad& x = -\infty \,\,\Rightarrow\,\, u = -\infty, &\quad& x = 0 \,\,\Rightarrow\,\, u = \infty \\ &\text{When } x > 0: &\quad& x = 0 \,\,\Rightarrow\,\, u = -\infty, &\quad& x = \infty \,\,\Rightarrow\,\, u = \infty. \end{alignat*}\]

Therefore,

\[\begin{align*} \int_{-\infty}^{\infty} f \left( x - \frac{a}{x} \right) \, dx &= \int_{-\infty}^{\infty} f(u) \frac{1}{2} \left( 1 - \frac{u}{\sqrt{u^2 + 4a}} \right) \, du \\ &+ \int_{-\infty}^{\infty} f(u) \frac{1}{2} \left( 1 + \frac{u}{\sqrt{u^2 + 4a}} \right) \, du = \int_{-\infty}^{\infty} f(u) \, du. \qquad\square \end{align*}\]

---

Application: We can use Glasser’s master theorem to compute integrals of the form

\[\boxed{I := \int_{-\infty}^{\infty} \frac{ x^2 }{ x^4 + b x^2 + c } \, dx \quad\text{where}\quad c \ge 0.}\]

First, write

\[I = \int_{-\infty}^{\infty} \frac{ dx }{ \left( x - \frac{\sqrt{c}}{x} \right)^2 + b + 2\sqrt{c} } = \int_{-\infty}^{\infty} \frac{ dx }{ x^2 + d } \quad\text{where}\quad d = b + 2\sqrt{c}.\]

If $d > 0$, then

\[I = \left. \frac{\arctan x}{\sqrt{d}} \right|_{-\infty}^{\infty} = \frac{ \pi }{ \sqrt{b + 2\sqrt{c}} }.\]

Otherwise, when $d < 0$, note that $\frac{1}{x^2 + d}$ is even. Therefore, the integral is twice that of the right region:

\[I = 2 \int_{0}^{\infty} \frac{ dx }{ x^2 + d } = \frac{1}{g} \Bigg[ \underbrace{ \int_{0}^{\infty} \frac{dx}{x-g} }_{\text{Part (I)}} - \underbrace{ \int_{0}^{\infty} \frac{dx}{x+g} }_{\text{Part (II)}} \Bigg] \qquad \text{where } g = \sqrt{-d}.\]

We split the integral of Part (I) in three regions:

\[\begin{align*} \text{Part (I)} &= \int_{0}^{g} \frac{dx}{x-g} + \int_{g}^{2g} \frac{dx}{x-g} + \int_{2g}^{\infty} \frac{dx}{x-g} \\ &= \int_{-g}^{0} \frac{dt}{t} - \int_{-g}^{0} \frac{dt}{t} + \int_{g}^{\infty} \frac{dt}{t}, \end{align*}\]

which clearly suggests that $\text{Part (I)} - \text{Part (II)} = 0$. In other words, $I = 0$.

For the special case $d = 0$, the integral $I$ clearly has a principal value $0$.

In summary,

\[\boxed{I = \begin{cases} \frac{ \pi }{ \sqrt{b + 2\sqrt{c}} }, & b + 2\sqrt{c} > 0, \\ 0 & b + 2\sqrt{c} \le 0. \end{cases}}\]