Talk:Weierstrass Substitution

Even though PlanetMath imposes $-\pi < \theta < \pi$, other websites do not mention such a limitation. This restriction considerably limits the power of this substitution for solving (indefinite) integrals - is it necessary? --GFauxPas (talk) 01:00, 18 July 2014 (UTC)


 * At $\pm \pi$ the value of $\tan \frac \pi 2$ is undefined. As $\tan \frac \pi 2$ is periodic, it's the same as for $-\pi < \theta < \pi$ as it is for $2 n \pi - \pi < \theta < 2 n \pi + \pi$ for any $n$  You could pick any such interval and it would have the same effect.  The point is sufficiently trivial to make it not worth stating on these various pages, but I suppose there's a case for being explicit here. --prime mover (talk) 07:06, 18 July 2014 (UTC)


 * The issue wasn't the discontinuities, it was the lack of injectivity. But this issue doesn't bother me this morning like it did last night, so I'm good. --GFauxPas (talk) 12:01, 18 July 2014 (UTC)


 * The point about the discontinuities is important. Basically, you need to pick a domain for which $\tan \frac \pi 2$ is bijective. Thus, as I say, you can pick any $n$ and have $2 n \pi - \pi < \theta < 2 n \pi + \pi$ as your interval.  But whatever interval you pick must not include either of the discontinuities. --prime mover (talk) 20:52, 18 July 2014 (UTC)