Parametric equation of $x^y=y^x$ curve What is the easiest/most natural way to parametrize the following curve?
$$\{(x,y)\in\Bbb R^{+2}\mid x^y=y^x, x\neq y\}\cup\{(e,e)\}$$
The best I could do was taking it apart, and for $x>y$ use $y(x)=\sqrt[x]{x}^{\sqrt[x]{x}^{\sqrt[x]{x}^{.....}}}$, but I see no theoretical reason why I should break it into parts, since the curve looks so nice.
Maybe the niceness of the curve is just an illusion, and in fact the $x<y$ and $x>y$ cases must be considered separately?
EDIT:$$(x,y)=(e^{t/(e^t-1)},e^{t/(1-e^{-t})})\quad t \in\Bbb R$$ This is a very nice parametrization, but still doesn't include $(e,e)$.
 A: Suppose you take $x=e^{r\cos\theta}$ and $y=e^{r\sin\theta}$ for $\theta\in(0,\pi/2)$.  Then you want to solve
$$
e^{r\cos\theta e^{r\sin\theta}}=e^{r\sin\theta e^{r\cos\theta}}
$$
for $r$.  Taking the logarithm of both sides and dividing through by $r$ (losing the $r=0$ solution in the process) gives
$$
\cos\theta e^{r\sin\theta} = \sin\theta e^{r\cos\theta},
$$
hence
$$
\tan\theta = e^{r(\sin\theta - \cos\theta)},
$$
or
$$
r(\sin\theta - \cos\theta) = \ln\tan\theta.
$$
This is satisfied for $\theta=\pi/4$, for any $r$, which gives the line $y=x$.  The piece you're interested in is given by
$$
r=\frac{\ln\tan\theta}{\sin\theta - \cos\theta}.
$$
Plugging this back in, you have
$$
(x,y)=\left(\exp\left(\frac{\ln\tan\theta}{\tan\theta - 1}\right), \exp\left(\frac{\ln\tan\theta}{1-\cot\theta}\right) \right) \\
=\left(\left(\tan\theta\right)^{\frac{1}{\tan\theta-1}}, \left(\tan\theta\right)^{\frac{1}{1-\cot\theta}}\right),
$$
or simply
$$
(x,y)=\left(t^{1/(t-1)}, t^{t/(t-1)}\right)
$$
for $t \in (0,\infty)$.  (Yes, there is a hole at $t=1$.)
A: I've done this here before, but it's quicker 
to write it again
than to try to find it.
Let $y = rx$.
$x^y = y^x$
becomes
$x^{rx} = (rx)^x$
or
$x^r = rx$
or
$x^{r-1} = r$
or
$x = r^{1/(r-1)}$
and
$y = rx = r^{1+1/(r-1)}
=r^{r/(r-1)}
$.
If $r=1$,
all values of $x$ work
since the condition is
$x^{r-1} = r$.
A: Edit: I just see that you already mention this one in your question...  But unlike you state there, it does include $(e,e)$ namely for $t=0$.
Basically a reparameterisation of other answers but $$ x(t) = \exp\left(\dfrac{t}{e^t-1}\right) \quad y(t)=\exp\left(\dfrac{t}{1-e^{-t}}\right) $$ for $t\in \mathbb{R}$ seems to be a nice and symmetric one.  Note that these expressions extend nicely over $t=0$ and $$\left(x(0), y(0)\right) = (e,e)$$ and $$ \left(x(-t), y(-t)\right) = \left(y(t), x(t)\right).$$
A: I can't really answer your question, but I think I can see why it might split like that:
$x^y=y^x$ generally means that
$e^{y\log x}=e^{x\log y}$, which occurs when $y\log x = x\log y$, so 
$\frac {\log y}y=\frac{\log x}x$.
If you graph $f(x)=\frac{\log x}x$, you will see that the only region where $f$ is not injective is $(1,\infty)$, in which $f$ first rises rapidly from $0$ to a maximum of $1/e$ at $x=e$ and then declines gradually forever, approaching $0$.
So giving $y$ in terms of $x$ or $x$ in terms of $y$ means giving one point in terms of the other, and it's not surprising that odd things would happen when they cross.
A: As stated in the edit of my question, and then poined out by WimC, the best result is
$$x(t) = \exp\left(\dfrac{t}{e^t-1}\right) \quad y(t)=\exp\left(\dfrac{t}{1-e^{-t}}\right)$$For the simplest way to obtain this, take marty cohen's answer and substitute $r=e^t$. Or, alternatively, see mjqxxxx's post.
For $t=0$ we can holomorphically extend this function by defining $x(0)=y(0)=e$.
