Approximation of $x\log x$ I was wondering if anyone knows a good approximation for the function 
$$f(x)=x\log x $$
when $x$ goes to infinity. In particular, I would like to get rid of the $\log x$ and so I need a polynomial approximation. (I think Stirling would not work very well.)
 A: There is no adequate polynomial or power law approximation.  For large $x$, $f(x)=x\log x$ grows more slowly than $x^{1+\epsilon}$ and more rapidly than $x^{1-\epsilon}$ for every $\epsilon>0$.
A: Let $x=a+t(b-a)$ for some $t$ in the range $0$ to $1$.  You get :
Let $c=b-a$
$$y = (a+ct)log(a+ct)$$
$$y=(a+ct)log\left( a\left(1+\frac c a t\right) \right)$$
$$y = (a+ct)log(a) + (a+ct)log\left(1+\frac c a t\right)$$
You can now replace the log() in the RHS by any suitable polynomial covering a reasonable range of values.
Note that using the arbitrary choice $a=c$ greatly simplifies things.
Good expressions for $log(1+x)$ are possibly better chosen from Páde approximants than from ordinary polynomials.  This page particularly mentions $log(1+x)$ Páde expressions and nearer home this Q&A on Mathematics SE describes exactly this.
A: By the substitution $u=\frac{1}{x}$ you can convert this to approximating $g(u)=-\displaystyle{\frac{\ln(u)}{u}}$ for $u\to0$.  Then you can use the Taylor series for $g(u)$ centered at $u=1$, for example.
