Is the curve determined by the function $\displaystyle{ f(x) = \frac{x}{2} \cdot (\sin(\ln{x}) - \cos(\ln{x}))}$ a fractal?
This function is the result of the integral $\displaystyle{\int{\sin{(\ln{x})}dx} }.$
It seems to repeat itself from $0$ to $+\infty.$
The difference between the magnitude of the repetitions is very big between each other.
The function zeroes are on the form: $\displaystyle{ e^{\pi n - 3\pi/4} }.$
On
We have $\sin y - \cos y=\sqrt 2(\sin (y-\frac \pi 4)$ so your function is $\frac x{\sqrt 2}(\sin (\ln x-\frac \pi 4)).$ As $x$ approaches $\frac \pi 4$ the term $\frac x{\sqrt 2}$ goes to a constant, so we can define $y=x-\frac \pi 4$are considering $g(y)= \sin (\ln(y))$ as $y \to 0$ The $\sin $ term will go through zero every time the argument is $k\pi$ for $k$ an integer. These are the points $y=e^{k\pi}$, which get closer together by a factor $e^{-\pi}$ each step. I couldn't get Alpha to give a nice plot. It is similar to the topologist's sine curve. It is a like a fractal in that the wavelength gets smaller and smaller. I think the Hausdorff dimension will get larger than $1$, which is one definition of a fractal.
In fact, the behavior you're seeing will happen for curves of the form $y=f(x) = x{\mathcal P}(\ln x)$ for any periodic function $\mathcal{P}()$; assuming that the period of ${\mathcal P}$ is $C$, then $f(e^{C}x) = e^Cx{\mathcal P}(\ln(e^Cx))$ $=e^Cx{\mathcal P}(\ln(e^C)+\ln x))$ $= e^Cx{\mathcal P}(C+\ln x)$ $=e^Cx{\mathcal P}(\ln x)$ (by the periodicity of ${\mathcal P}$) $=e^Cf(x)$. In other words, scaling $x$ and $y$ both by $e^C$ leaves the curve invariant: it's self-similar.
We can show the converse, too: any function with a self-similar graph — that is, any function that satisfies a non-trivial functional equation $f(Cx)=Cf(x)$ — is of the form $f(x)=x{\mathcal P}(\ln x)$ for some periodic function ${\mathcal P}$. Note first that without loss of generality we can take $C\gt 1$: if $f(Cx)=Cf(x)$ with $C\lt 0$ then $f(C^2x) = Cf(Cx)=C^2f(x)$, so $f()$ satisfies $f(Dx)=Df(x)$ with $D=C^2\gt 0$. Likewise, if $f(Cx)=Cf(x)$ with $0\lt C\lt 1$, then $f(x)=f(C\cdot C^{-1}x)=Cf(C^{-1}x)$ or $f(C^{-1}x)=C^{-1}f(x)$, so $f()$ satisfies $f(Ex)=Ef(x)$ with $E=C^{-1}\gt 1$.
Now, consider $g(x)=f(x)/x$ (so $f(x)=xg(x)$): then $g(Cx)=f(Cx)/(Cx)$ $=(Cf(x))/(Cx)$ $=f(x)/x$ $=g(x)$; in other words, $g()$ is 'multiplicatively periodic'. Take ${\mathcal P}()$ such that ${\mathcal P}(t)=g(e^t)$ (in other words, $g(x)={\mathcal P}(\ln x)$), and let $\omega=\ln C$. Then ${\mathcal P}(\omega+t)=g(e^{\omega+t})$ $=g(e^\omega e^t)$ $=g(Ce^t)$ $=g(e^t)={\mathcal P}(t)$; that is, ${\mathcal P}()$ is a periodic function of period $\omega$, and we have $f(x)=xg(x)=x{\mathcal P}(\ln x)$.
However, this self-similarity doesn't make the curve a 'fractal' in the more usual sense of having a fractional dimension; the short version of the explanation is that the curve 'stays smooth' as $x\to 0$ and this keeps it of dimension $1$. In slightly more detail, the total arc length (from $x=1$ to $x=0$, say) is bounded: let $a_0$ be the arc length of the curve from $x=e^{-C}$ to $x=1$, $a_1$ be the arc length from $x=e^{-2C}$ to $x=e^{-C}$, and generally $a_n$ the arc length from $x=e^{-(n+1)C}$ to $x=e^{-nC}$. Then the self-similarity of the curve means that these lengths satisfy $a_{n+1}=e^{-C}a_n$, since all lengths scale linearly under uniform scale of $x$ and $y$. This implies the total arc length $A$ from $x=1$ to $x=0$ as given by $A=\sum_{n=0}^\infty a_n$ is finite, because it's the sum of a geometric series: $A=\frac1{1-e^{-C}}a_0$. The boundedness of the total length of the curve towards zero then means that it has dimension $1$ by most usual measures (e.g., it can be covered by $O(1/\epsilon)$ balls of size $\epsilon$, etc).