For $x\in(0,1)$, let \begin{equation} f(x):=\frac1{x\ln^2\frac ex}. \tag{1} \end{equation} Then $f\ge0$ and $\int_0^1f(x)\,dx=1$. On the other hand, $\ln f(x)\sim\ln\frac ex$ as $x\downarrow0$. So, $\ln f(x)\ge\frac12\ln\frac ex$ for some $c\in(0,1)$ and all $x\in(0,c)$. So, $$\int_0^1 f(x)\ln f(x)\,dx\ge \int_0^c \frac1{x\ln^2\frac ex}\frac12\ln\frac ex\,dx=\infty,$$ as desired.

A few words on how this example was found. First, by Jensen's inequality and the condition $\int f=1$, we have $\int f\ln f\ge\int f\,\ln\int f=0$, so that the minimum of $\int f\ln f$ given $\int f=1$ is attained when $f=1$. So, to get a large value of $\int f\ln f$, it makes sense to try letting $f$ be as non-uniform (non-constant) as possible. Taking then $f=f_a:=a\,1_{[0,a]}$ with $a\to\infty$ (so that $f$ "explodes" at $0$ to $\infty$), we have $\int f=1$ and $\int f\ln f=\ln a\to\infty$. Now, instead of a family $(f_a)$ of functions exploding to $\infty$ at $0$, we can try to find a single function exploding to $\infty$ at $0$ as fast as possible, but so that the condition $\int f=1$ holds. This results in the function $f$ in (1), which happens to be the same as the one suggested by Deane Yang.

For $x\in(0,1)$, let \begin{equation} f(x):=\frac1{x\ln^2\frac ex}. \tag{1} \end{equation} Then $f\ge0$ and $\int_0^1f(x)\,dx=1$. On the other hand, $\ln f(x)\sim\ln\frac ex$ as $x\downarrow0$. So, $\ln f(x)\ge\frac12\ln\frac ex$ for some $c\in(0,1)$ and all $x\in(0,c)$. So, $$\int_0^1 f(x)\ln f(x)\,dx\ge \int_0^c \frac1{x\ln^2\frac ex}\frac12\ln\frac ex\,dx=\infty,$$ as desired.

A few words on how this example was found. First, by Jensen's inequality and the condition $\int f=1$, we have $\int f\ln f\ge\int f\,\ln\int f=0$, so that the minimum of $\int f\ln f$ given $\int f=1$ is attained when $f=1$. So, to get a large value of $\int f\ln f$, it makes sense to try letting $f$ be as non-uniform (non-constant) as possible. Taking then $f=f_a:=a\,1_{[0,1/a]}$ with $a\to\infty$ (so that $f$ "explodes" at $0$ to $\infty$), we have $\int f=1$ and $\int f\ln f=\ln a\to\infty$. Now, instead of a family $(f_a)$ of functions exploding to $\infty$ at $0$, we can try to find a single function exploding to $\infty$ at $0$ as fast as possible, but so that the condition $\int f=1$ holds. This results in the function $f$ in (1), which happens to be the same as the one suggested by Deane Yang.

For $x\in(0,1)$, let $$f(x):=\frac1{x\ln^2\frac ex};$$ this is the same function as then one suggested by Deane Yang. Then $f\ge0$ and $\int_0^1f(x)\,dx=1$. On the other hand, $\ln f(x)\sim\ln\frac ex$ as $x\downarrow0$. So, $\ln f(x)\ge\frac12\ln\frac ex$ for some $c\in(0,1)$ and all $x\in(0,c)$. So, $$\int_0^1 f(x)\ln f(x)\,dx\ge \int_0^c \frac1{x\ln^2\frac ex}\frac12\ln\frac ex\,dx=\infty,$$ as desired.

For $x\in(0,1)$, let \begin{equation} f(x):=\frac1{x\ln^2\frac ex}. \tag{1} \end{equation} Then $f\ge0$ and $\int_0^1f(x)\,dx=1$. On the other hand, $\ln f(x)\sim\ln\frac ex$ as $x\downarrow0$. So, $\ln f(x)\ge\frac12\ln\frac ex$ for some $c\in(0,1)$ and all $x\in(0,c)$. So, $$\int_0^1 f(x)\ln f(x)\,dx\ge \int_0^c \frac1{x\ln^2\frac ex}\frac12\ln\frac ex\,dx=\infty,$$ as desired.

A few words on how this example was found. First, by Jensen's inequality and the condition $\int f=1$, we have $\int f\ln f\ge\int f\,\ln\int f=0$, so that the minimum of $\int f\ln f$ given $\int f=1$ is attained when $f=1$. So, to get a large value of $\int f\ln f$, it makes sense to try letting $f$ be as non-uniform (non-constant) as possible. Taking then $f=f_a:=a\,1_{[0,a]}$ with $a\to\infty$ (so that $f$ "explodes" at $0$ to $\infty$), we have $\int f=1$ and $\int f\ln f=\ln a\to\infty$. Now, instead of a family $(f_a)$ of functions exploding to $\infty$ at $0$, we can try to find a single function exploding to $\infty$ at $0$ as fast as possible, but so that the condition $\int f=1$ holds. This results in the function $f$ in (1), which happens to be the same as the one suggested by Deane Yang.