y = x \ln x

Graph of the Function $y = x \ln x$

$y = x \ln x$ is the product of a linear function and a logarithm. As $x \to 0^{+}$ the factor $\ln x$ diverges to $-\infty$ while $x$ tends to $0$ , so the first question is which of the two effects wins. The answer is $x$ , and the value converges to $0$ .

Domain and range

The logarithm forces the domain $x > 0$ . As shown below the minimum value is $-\dfrac{1}{e}$ , and the function diverges as $x \to \infty$ , so the range is $y \geq -\dfrac{1}{e}$ .

Behavior at the origin

The limit as $x \to 0^{+}$ follows from l'Hôpital's rule.

\begin{align*} \lim_{x \to 0^{+}} x \ln x &= \lim_{x \to 0^{+}} \frac{\ln x}{1/x} \\ &= \lim_{x \to 0^{+}} \frac{1/x}{-1/x^2} \\ &= \lim_{x \to 0^{+}} (-x) = 0 \end{align*}

The indeterminate form $\dfrac{-\infty}{\infty}$ was rewritten to make it tractable. The graph is drawn into the origin as though sucked in, but $x = 0$ is not in the domain, so that end is open. The derivative $y' = \ln x + 1$ diverges to $-\infty$ there, so the tangent becomes vertical.

Monotonicity and extrema

The derivative $y' = \ln x + 1$ vanishes when $\ln x = -1$ , that is at $x = \dfrac{1}{e}$ . It is negative before that point and positive after, so the function has a minimum there, both local and global, of value $-\dfrac{1}{e}$ . The minimum point $\left( \dfrac{1}{e}, -\dfrac{1}{e} \right) \approx (0.368, -0.368)$ is memorable for having coordinates of equal magnitude.

Concavity and intercept

The second derivative is $y'' = \dfrac{1}{x}$ , positive throughout the domain, so the curve is concave up everywhere and has no inflection point. Since $x \ln x = 0$ only when $\ln x = 0$ , the single $x$ -intercept is $(1, 0)$ .

Range	$\ln x$	$y$
$0 < x < 1$	negative	negative
$x = 1$	$0$	$0$
$x > 1$	positive	positive

Integral

Integration by parts gives the antiderivative.

\int x \ln x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C

From this $\int_0^1 x \ln x\,dx = -\dfrac{1}{4}$ . The improper integral takes a finite value precisely because the integrand settles to $0$ as $x \to 0$ .

Relation to other functions

Since $x^x = e^{x \ln x}$ , this function is the exponent of $y = x^x$ . That $x \ln x$ attains its minimum $-\dfrac{1}{e}$ at $x = \dfrac{1}{e}$ is exactly why $x^x$ attains its minimum $e^{-1/e} \approx 0.692$ at the same place.

Applications

In information theory an event of probability $p$ contributes $-p \ln p$ to the entropy. The shape of this graph, reflected in sign, shows that the contribution vanishes both at $p = 0$ and at $p = 1$ and peaks in between, at $p = \dfrac{1}{e}$ . In complexity theory the lower bound for comparison sorting is $n \log n$ , and this function is precisely that rate of growth.

Graph of the Function y=xln⁡xy = x \ln xy=xlnx