I have been trying to find the arc-length of $\sin^{-1}(x)$ over $[0,1]$.

Of course, it is given by the integral $$J=\int_0^1\sqrt{1+\frac1{1-x^2}}\ dx=\int_0^1 \sqrt{\frac{2-x^2}{1-x^2}}\, dx$$ To compute this, I used $x\mapsto \cos x$: $$J=\int_0^{\pi/2}\sqrt{\frac{1+1-\cos^2x}{1-\cos^2x}}\, \sin(x)dx=\int_0^{\pi/2}\sqrt{1+\sin^2x}\,dx=\mathrm{E}(-1)$$ where $$\mathrm{E}(k)=\int_0^{\pi/2}\sqrt{1-k\sin^2x}\,dx$$ is the complete elliptic integral of the second kind.

I am usually one to accept values of $\mathrm{E}$ as closed forms by themselves, but by chance, Wolfram kindly provided the explicit evaluation $$\mathrm{E}(-1)=\frac{\pi\sqrt{2\pi}}{\Gamma^2(1/4)}+\frac{\Gamma^2(1/4)}{4\sqrt{2\pi}}\tag{1}$$ Which I would like to know how to prove.

I was able to immediately recognize that $$\frac{\Gamma^2(1/4)}{4\sqrt{2\pi}}=\frac{1}{4\sqrt{2}}\int_0^1\frac{dx}{x^{3/4}(1-x)^{3/4}}$$ But I cannot seem to find an integral representation for the other chunk, namely $\frac{\pi\sqrt{2\pi}}{\Gamma^2(1/4)}$. I suspect that the solution involves the Jacobi elliptic functions and the Jacobi theta functions, but I have no idea how to use them. Could I have some help proving $(1)$? Thanks.

  • $\begingroup$ This should be exactly the same as the length of $\sin$ over $[0,\pi/2]$, that is, $\int_0^{\pi/2}\sqrt{1+\cos^2x}dx$. Just to shorten your first computations. $\endgroup$ – ajotatxe Jun 6 at 16:52

This is conventionally $\mathrm{E}(i)$, not $\mathrm{E}(-1)$. Its evaluation begins with $$\mathrm{E}(i)=\int_0^1\sqrt\frac{1+x^2}{1-x^2}\,dx=\int_0^1\frac{1+x^2}{\sqrt{1-x^4}}\,dx=\frac{1}{4}\mathrm{B}\Big(\frac{1}{4},\frac{1}{2}\Big)+\frac{1}{4}\mathrm{B}\Big(\frac{3}{4},\frac{1}{2}\Big)=\ldots$$

  • $\begingroup$ okay that was way easier than I thought. $\endgroup$ – clathratus Jun 6 at 17:13

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.