Consider the following integral $$ f = \int_0^1 \frac{1}{\sqrt{-\frac{1}{2} \, t^{2} + 1} \sqrt{-t^{2} + 1}} \mathrm \,dt. $$ If we change variable by letting $x^2=t^2/(2-t^2)$, then we have $$ f = \int_0^1 \sqrt{2} \cdot\sqrt{\frac{1}{1-x^{4}}} \mathrm \,dx, $$ which is a simpler form. I read this in a book and wonder how can we come up with this sort of simplification? Is it just experience or is there systematic way to do it?

Note: The integral is the complete elliptic integral of the first kind $K(1/\sqrt{2})$.

  • $\begingroup$ this integral leads to an elliptic one $\endgroup$ – Dr. Sonnhard Graubner Jun 30 '17 at 12:02

$\int\frac{dt}{\sqrt{1-t^2}}=\arcsin(t)$, $\int\frac{dt}{\sqrt{1-\frac{t^2}{2}}}=\sqrt{2}\arcsin\left(\frac{t}{\sqrt{2}}\right)$, hence two candidate substitutions for simplifying things are $t=\sin\theta$ and $t=\sqrt{2}\sin\frac{\theta}{\sqrt{2}}$. Let us try the first one:

$$ I = \int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-\frac{1}{2}\sin^2\theta}} $$ followed by the substitution $\theta=\arctan u$: $$ I = \int_{0}^{+\infty}\frac{du}{(1+u^2)\sqrt{1-\frac{1}{2}\cdot\frac{u^2}{1+u^2}}}=\sqrt{2}\int_{0}^{+\infty}\frac{du}{\sqrt{(u^2+1)(u^2+2)}}=\frac{\pi}{\text{AGM}(2,\sqrt{2})}$$ The AGM allows an efficient numerical evaluation (it immediately tells us that $I\geq\frac{2\pi}{2+\sqrt{2}}$, for instance) and the identity $\text{AGM}(a,b)=\text{AGM}\left(\frac{a+b}{2},\sqrt{ab}\right)$ gives many equivalent integrals.
We may notice that

$$ \int_{0}^{1}\frac{dx}{\sqrt{1-x^4}}\stackrel{x\mapsto\sqrt{z}}{=}\frac{1}{2}\int_{0}^{1}\frac{dz}{\sqrt{z(1-z^2)}}\stackrel{z\mapsto t^{-1}}{=}\frac{1}{2}\int_{1}^{+\infty}\frac{dt}{\sqrt{t(t-1)(t+1)}}$$ leads to $$ \int_{0}^{1}\frac{dx}{\sqrt{1-x^4}}=\frac{1}{2}\int_{0}^{+\infty}\frac{dt}{\sqrt{t(t+1)(t+2)}}=\int_{0}^{+\infty}\frac{du}{\sqrt{(u^2+1)(u^2+2)}} $$

This completely explains how to find the useful substitution by underlying the relation between the initial elliptic integral, the lemniscate constant and the AGM. We may also add a fourth actor on the scene, since by the substitution $x=w^{1/4}$ the integral $\int_{0}^{1}\frac{dx}{\sqrt{1-x^4}}$ is related with the Beta function, hence with the $\Gamma$ function. Here it is a complete summary:

$$\boxed{ \int_{0}^{1}\frac{dx}{\sqrt{1-x^4}} = \frac{\pi}{\text{AGM}(1,\sqrt{2})}=\frac{1}{\sqrt{2}}\,K\left(\frac{1}{\sqrt 2}\right)=\frac{1}{4}B\left(\frac{1}{4},\frac{1}{2}\right)=\frac{\Gamma\left(\frac{1}{4}\right)^2}{4\sqrt{2\pi}}. }$$

This excursion gives as a by-product an efficient numerical approach for computing $\Gamma\left(\frac{1}{4}\right)$, proving $\Gamma\left(\frac{1}{4}\right)=\frac{(2\pi)^{3/4}}{\sqrt{\text{AGM}(1,\sqrt{2})}}$.



$$t=x \sqrt{\frac{2}{x^2+1}}$$

$$dt = \sqrt{2} \sqrt{\frac{1}{x^2+1}} \left(1-\frac{x^2}{x^2+1}\right) dx$$ So the integrand becomes is $$\frac{\sqrt{2} \left(\sqrt{\frac{1}{x^2+1}} \left(1-\frac{x^2}{x^2+1}\right)\right)}{\sqrt{1-\frac{2 x^2}{x^2+1}} \sqrt{1-\frac{x^2}{x^2+1}}}$$ and finally $$\frac{\sqrt{2}}{\sqrt{1-x^4}}$$

  • 1
    $\begingroup$ This is the correct changing variable. But I am actually asking how can we come up with what to change in the first place. $\endgroup$ – ablmf Jun 30 '17 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.