User:Michael Hardy/an integral

Proposition: If


 * $$ c = (\text{geometric mean of }a,b) = \sqrt{ab\, {}} $$

and


 * $$ d = (\text{arithmetic mean of }a,b) = \frac{a+b}{2} $$

then


 * $$ \int_{-\infty}^\infty \frac{dx}{\sqrt{(x^2 + c^2)(x^2 + d^2)}} = \int_{-\infty}^\infty \frac{dx}{\sqrt{(x^2 + a^2)(x^2 + b^2)}} $$

Proof: Write the first integral as


 * $$\int_{-\infty}^\infty \frac{du}{\sqrt{(u^2 + c^2)(u^2 + d^2)}} $$

and then use the substitution


 * $$ u = \frac{x - \frac{ab}{x}}{2}, \qquad du = \frac{1 + \frac{ab}{x^2}}{2}\,dx. $$

First observe that as u goes from &minus;&infin; to +&infin;, then x goes from 0 to &infin;, so the integral becomes



\begin{align} & {} \qquad \int_0^\infty \frac{ \left(\dfrac{x^2 + \frac{ab}{x^2}}{2} \right) }{\sqrt{\left(\left(\frac{x - \frac{ab}{x}}{2} \right)^2 + ab \right) \left(\left( \frac{x - \frac{ab}{x}}{2} \right)^2 + \left( \frac{a+b}{2} \right)^2\right) }} \,dx \\[10pt] & = \int_0^\infty \frac{ 2(x^2 + ab) \, dx}{\sqrt{\Big((x^2 - ab)^2 + 4abx^2\Big) \Big((x^2 - ab)^2 + ((a+b)x)^2\Big)}} \\[10pt] & = \int_0^\infty \frac{2 \, dx}{\sqrt{ (x^2 + a^2)(x^2 + b^2) }} = \int_{-\infty}^\infty \frac{dx}{\sqrt{ (x^2 + a^2)(x^2 + b^2) }}. \end{align} $$