Analysis of an Interesting Graph

In my last post I evaluated the integral of the function:$$y= \frac{\sin(\ln x)}{\ln x}$$ between 0 and 1 and arrived at a value of \( \frac{\pi}{4} \) using Feyman's technique. In this post, I want to look at the function more closely in terms of its turning points and intercepts on the \(x\) axis. Figure 1 shows the behaviour of the graph in the region between 0 and 1.

Figure 1: created using GeoGebra

Clearly there is a small negative area that will reduce the largely positive area slightly but the figure of \( \frac{\pi}{4} \approx \) 0.785398163397448 looks pretty right. Clearly, the graph cuts the \(x\) axis around 0.4 but what about further out along the \(x\) axis? Figure 2 shows the graph of the function in the range from 0 to 100:

Figure 2: created using Wolfram Alpha

Figure 3 shows a slightly different view. 

Figure 3: created using Wolfram Alpha

Let's look at the graph in the range from 0 to 1000. Figure 4 shows this:

Figure 4: created using Wolfram Alpha

To see what's going on with these intercepts on the \(x\) axis we just need to set the function equal to zero. Let's do this and solve the resulting equation:$$
\begin{align} \frac{\sin(\ln x)}{\ln x}&=0\\
\implies \sin(\ln x)&=0 \text{ provided that } \ln x \neq 0\\
\implies \ln x &=n\pi \text{ where }n \in \mathbb{Z} \text{ and } n \neq 0\\
\therefore x&=e^{\ \pi \ n} \text{ with } n \neq 0
\end{align}$$Thus we begin to see what's going on:

• The intercept in Figure 1 corresponds to \(n=-1\) and \(e^{-\pi} \approx 0.04 \)
• The intercept in Figure 2 corresponds to \(n=1\) and \(e^{\pi} \approx 23.14 \)
• The intercept in Figure 4 corresponds to \(n=2\) and \(e^{2\pi} \approx 535.49 \)

Clearly as \(n\) grows larger, the values of \(e^{\ \pi \ n}\) become very large and the intercepts more and more spread out along the number line. Similarly as -\(n\) grows larger negatively, the values of \(e^{\ -\pi \ n}\) become very small and the intercepts closer and closer together.

What about the turning points of the graph? To explore this we need to find the first derivative of the function. We find:$$
\begin{align}
\frac{d}{dx} \left (\frac{\sin(\ln x)}{\ln x} \right )&=\frac{\ln x \cos(\ln x)-\sin(\ln x)}{x \ln^2 x}\\
\implies \ln x \cos(\ln x)-\sin(\ln x)&=0 \text{ provided } x \ln^2 x \neq 0
\end{align}$$A partial solution to this is offered by Wolfram Alpha but the computation times out so only three solutions are shown. See Figure 5:

Figure 5: from Wolfram Alpha

From these results we can see the local minimum of \(x \approx 0.01\) in Figure 1 and the local maximum of x=1 in Figure 3. Some further results can be obtained from Wolfram Alpha simply by asking for turning points. These are shown in Figure 6.

Figure 6: from Wolfram Alpha

These results are most interesting because we can see a maximum value of around 0.049 when \(x=0\) but a minimum value of around -0.058 when \(x=0\)! Clearly, the function should be undefined at \(x=0\) even though a limit of 0 seems to exist. The issue of the limit is an interesting one which I may take up in a subsequent post but for the time being that's probably enough for the graph.