Talk:Primitive of Reciprocal of x squared plus a squared/Arctangent Form

Okay, so I figured out Function is Odd Iff Inverse is Odd, which means that:


 * $ \dfrac 1 {-a} \arctan \dfrac x {-a} = \dfrac 1 a \arctan \dfrac x a$

so presumably it shouldn't matter if $a > 0$ or $a < 0$?? And either way $\frac a {a^2} = \frac 1 a$, right? So it should seem to hold as long as $a \ne 0$. Does it? --GFauxPas 08:13, 19 December 2011 (CST)

If it doesn't matter, perhaps the reason that I've seen $a > 0$ in literature is that since $a^2$ has two square roots, $a$ is ambiguous. Perhaps it would be better to say the theorem this way:


 * $\displaystyle \int \frac 1 {x^2 + a} \ \mathrm dx = \frac 1 {\sqrt a} \arctan{\frac x {\sqrt a}} + C$

--GFauxPas 08:26, 19 December 2011 (CST)
 * I agree on this presentation. It is perhaps a little less clear, but at least incorporates all desired cases. I will think of generalising this theorem to all degree 2 polynomials in the denominator (by completing the square and a simple substitution). --Lord_Farin 09:18, 19 December 2011 (CST)


 * Sorry but really really really don't like this presentation. It doesn't add anything extra: $\sqrt a$ is just as two-cased as $\sqrt {a^2}$ and it makes it look really ugly. The whole point about $x^2 + a^2$ is that $a^2 > 0$ always, and you don't need to make that messy specification that $a > 0$. --prime mover 16:21, 19 December 2011 (CST)


 * PM you have my permission to alter the proof as you please, as I said before I don't feel a particular attachment to the layout I write proofs in. I just aim to find some compromise between people. --GFauxPas 16:24, 19 December 2011 (CST)

Adapted my opinion on this one. With a reference to the equality at the top of this page, the original presentation is the most clear. Sorry to turn like a leaf. --Lord_Farin 16:38, 19 December 2011 (CST)


 * I may well get the lead out and write a neat exposition of this theorem in due course, which ought to cover all our various bases. In the meantime don't feel you have to get too hung up on detail of existing pages if you want to crack on with other results. --prime mover 17:06, 19 December 2011 (CST)

Explain
So, regarding what's in the explain template. Where should I put that? In the bottom as a comment? --GFauxPas (talk) 10:42, 13 May 2014 (UTC)


 * Do you know, I'm not even sure what I meant now -- it doesn't make big sense. I probably meant "for $a = 0$" (in which case it would be an "also see" for the integral of $1 / x^2$).


 * The investigation as to what happens to the usual integral when $a \to 0$ is to see whether $\dfrac 1 a \arctan (x/a)$ does actually tend to $-1/x + C$ (I'd expect it would) and if so, to see what the constant $C$ would be in this circumstance. I think I may just have thrown this out as "something interesting to investigate", or some such. --prime mover (talk) 11:12, 13 May 2014 (UTC)


 * How about this? --GFauxPas (talk) 11:36, 13 May 2014 (UTC)


 * ... except not in a "Comment" section -- we're trying to remove such sections -- but as part of the main proof. Then the statement of the proof does not need to make a special case for "$a \ne 0$". If you're not sure what I mean, I'll have a go at doing this later when I'm back at home base. --prime mover (talk) 12:40, 13 May 2014 (UTC)