In order to proove Gauss theorem on elliptic integrals it's written :
For any couples $(a, b$), let us defines : $$\displaystyle \mathcal{T}(a, b) = \int_{0}^{+\infty} \frac{1}{\sqrt{(a^2+t^2)\cdot(b^2+t^2)}} \cdot \mathrm{d}t$$
we may notice that Lagrange's identity implies : $$ (a^2+t^2)(b^2+t^2) = (at+bt)^2+(t^2-ab)^2$$ hence by setting $u = \frac{1}{2} (t-ab/t)$ we have :
$$ \displaystyle \mathcal{T}(a, b) = \int_{0}^{+\infty} \frac{1}{\sqrt{(a+b)^2 + (t-\frac{ab}{t})^2}} \cdot \frac{\mathrm{d}t}{t} = \int_{-\infty}^{+\infty} \frac{\mathrm{d}u}{\sqrt{((a+b)^2+4u^2)(ab+u^2)}} = \mathcal{T}(\frac{a+b}{2}, \sqrt{ab})$$
I must say that I don't understand at all the last part, I mean : how do you get from : $$\displaystyle \mathcal{T}(a, b) = \int_{0}^{+\infty} \frac{1}{\sqrt{(a+b)^2 + (t-\frac{ab}{t})^2}} \cdot \frac{\mathrm{d}t}{t}$$ to $$\int_{-\infty}^{+\infty} \frac{\mathrm{d}u}{\sqrt{((a+b)^2+4u^2)(ab+u^2)}}$$ ? And how do you get that is is also equals to : $\mathcal{T}(\frac{a+b}{2}, \sqrt{ab})$.
This demonstrations can be found here : https://drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view page $142$