2

For real numbers $0 < a < b$ we let \begin{align*} F(a,b) = 2\int_a^b \frac{dt}{\sqrt{(t^2-a^2)(b^2-t^2)}} \end{align*}

Using some substitutions, we can show that \begin{align*} F(a,b) = \int_{a^2}^{b^2} \frac{dt}{\sqrt{t(t-a^2)(b^2-t)}} = \int_{-\infty}^0 \frac{dt}{\sqrt{t(t-a^2)(b^2-t)}}. \end{align*} (For the first one take $u = t^2$, we get the second from the first by substituting $u = \frac{t-b^2}{t-a^2}$)

I would like to show that \begin{align*} F(a,b) = \int_0^{\pi/2} \frac{d\theta}{\sqrt{a^2\cos^2(\theta) + b^2\sin^2(\theta)}}, \end{align*} but was unable to do so. From \begin{align*} F(a,b) = \int_{-\infty}^0 \frac{dt}{\sqrt{t(t-a^2)(b^2-t)}}, \end{align*} I substituted $\theta = -\arctan(t)$. Then $d\theta = -(t^2 + 1)^{-1}dt$ and the bounds become $\pi/2$ and $0$. \begin{align*} F(a,b) &= \int_{\pi/2}^0 \frac{-(t^2 + 1)d\theta}{\sqrt{\tan(-\theta)(\tan(-\theta)-a^2)(b^2-\tan(-\theta))}}\\ &= \int_0^{\pi/2} \frac{\sec^2(-\theta)d\theta}{\sqrt{\tan(-\theta)(\tan(-\theta)-a^2)(b^2-\tan(-\theta))}}\\ &= \int_0^{\pi/2} \frac{d\theta}{\sqrt{\cos^4(-\theta)\tan(-\theta)(\tan(-\theta)-a^2)(b^2-\tan(-\theta))}} \end{align*} I am stuck here as I could not figure out how to show that this becomes \begin{align*} \int_0^{\pi/2} \frac{d\theta}{\sqrt{a^2\cos^2(\theta) + b^2\sin^2(\theta)}}. \end{align*}

I am fairly certain it should be possible given this and this.

SescoMath
  • 1,909
  • 1
    Have a look at Jack d'Aurizio book https://cel.archives-ouvertes.fr/cel-01813978 chapter 11 – Claude Leibovici May 06 '20 at 04:23
  • 3
    Gauss proved this using the transformation $$ \sin\theta = \frac{2a \sin\phi}{a + b + (a - b)\sin^{2}\phi}$$ in the integral $$F(a, b) =\int_{0}^{\pi/2}\frac{d\theta} {\sqrt{a^2\cos^2\theta+b^2\sin^2\theta} } $$ More details are available in this post. – Paramanand Singh May 09 '20 at 10:44

0 Answers0