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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 $K(1/\sqrt{2})$.

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  • $\begingroup$ this integral leads to an elliptic one $\endgroup$ – Dr. Sonnhard Graubner Jun 30 '17 at 12:02
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$\int\frac{dt}{\sqrt{1-t^2}}=\arcsin(t)$, $\int\frac{dt}{\sqrt{1-\frac{t^2}{2}}}=\sqrt{2}\arcsin\left(\frac{t}{\sqrt{2}}\right)$, 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 $\Gamma\left(\frac{1}{4}\right)=\frac{(2\pi)^{3/4}}{\sqrt{\text{AGM}(1,\sqrt{2})}}$.

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$$x^2=\frac{t^2}{2-t^2}$$

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

$$dt = \sqrt{2} \sqrt{\frac{1}{x^2+1}} \left(1-\frac{x^2}{x^2+1}\right) dx$$ 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 $$\frac{\sqrt{2}}{\sqrt{1-x^4}}$$

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    $\begingroup$ 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

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