# Calculate and provide some justification to get an approximation of $\Re(\int_0^1(\log x)^{-\operatorname{li}(x)}dx)$

I was playing with Wolfram Alpha online calculator about possible variations of an integral, the so-called Sophomore's dream. I'm agree that my example is artificious and isn't related with previous nice integral, but I'm curious to know how we can obtain and justify a good approximation of next integral.

Question. Can you provide me a justification for a good approximation (I say a few right digits, four or six) of $$\Re\left(\int_0^1\frac{1}{(\log x)^{\operatorname{li}(x)}}dx\right),$$ where $\operatorname{li}(x)$ is the logarithmic integral (this MathWorld). Thanks in advance.

I prefer some strategy using analysis, but if you want explain me how get it using a good strategy from the numerical analysis (or combining a numerical method with some facts deduced from the analysis of our integral) it also is welcome.

I add also the plot of our function that you can see using this Wolfram Alpha online calculator

plot Re ((log(x))^(-li(x))), from x=0 to 1

• About the approach using numerical analysis, I am asking about an appropriate method for our problem (if this is your choice, please explain why such method provide us good results for this problem, if it is possible). Many thanks.
– user243301
Mar 6 '18 at 19:21
• Many thanks @VonNeumann
– user243301
Mar 6 '18 at 19:24
• For the records, the Real Part of the integral results to be $$0.384494$$. Now let's focus on the approximation. Mar 6 '18 at 19:28

Solution:

This combination of integrals $$\boxed{A=\int_0^{0.6266}\left(2-e^x+\frac29\sin(6x)\right)\,dx\\B=-\frac9{25}\int_{0.5829}^{0.8685}\cos\left(11x-1.7\right)\,dx\\C=-\frac1{40}\int_{0.8736}^{0.9297}\sin(56x-1.8)\,dx}$$ gives the answer of $$A+B+C=0.3846\cdots$$ which does the job with an error of around $1.3\times10^{-4}$.

For a visualisation see here.

Strategy:

We split the function into three parts, one between $0$ and around $0.58$ (call this $A$), one between $0.58$ to around $0.87$ (call this $B$) and finally one between $0.87$ to around $0.93$ (call this $C$).

We see that for $A$, the function starts at $1$, rises a little and drops to $0$ at $x=0.58$.

For $B$, the function starts at $0$ at $x=0.58$, reaches a minimum of about $0.36$ and rises to $0$ at $0.87$.

For $C$, the function starts at $0$ at $x=0.87$, rises to around $0.02$ and falls back down to $0$ at $0.93$.

These little rises and falls can be modelled simply by either quadratic or trigonometric functions.

Experimenting with functions in Desmos show that the trig ones are better, and further trial and error gives the coefficients shown in the Solution.

• Many thanks for your strategy, now and tomorrow in the morning I am going to study your answer.
– user243301
Mar 6 '18 at 20:42
• Ahhh awesome! They are indeed simple functions! Maths is beautiful. Mar 6 '18 at 21:29

I am going to give you a pretty eccentric answer, which may be the springboard for other future answers, or modification (improvement) of this one.

Every consideration of mine arose from a simple look of the plot above, which we can split into three parts:

• The first one is the curve that runs from $x = 0$ to $x = 0.5$.

Remark

I know it's not really $x = 0.5$, but since this has been made by a trivial and fast look, please grant me this, for the spirit of approximation.

• The second part runs from $x = 0.5$ to $x = 0.9$.

• The third part is the final negligible streak, which indeed I'll neglect.

Observations

Let's start from the second part: that curve resemble a parabola. Since the two points $x = 0.5$ and $x = 0.9$ are the zeroes, we can easily build the function that describes that trait:

$$f_1(x) = (x - 0.5)(x - 0.9) = x^2 - 1.4x + 0.45$$

The idea of my analysis

It's to build a completely observative curve, continuous but piecewise.

The first piece is then:

$$f_1(x) = x^2 - 1.4x + 0.54 ~~~~~~~ x\in[0.5, 0.9]$$

Which plotted gives us, roughly, that curve.

Now let's proceed with the first one, which is a bit more challenging (not much, actually). In observing it, I noticed the trend (very very approximative) of an ellipse. A quarter of an ellipse, nay.

In particular, one can describe that piece through a square root like

$$f_2(x) = \sqrt{1 - x^2} ~~~~~~~ x\in [0, 0.5]$$

Our sperimental function is then created:

$$F(x) = \begin{cases} \sqrt{1 - x^2} & x\in [0, 0.5]\\ x^2 - 1.4x + 0.45 & x\in [0.5, 0.9] \end{cases}$$

The integration of this function is very trivial, yet it will give you a pretty stoned result:

$$\int F(x)\ dx \approx 0.467639$$

With an error of $21.62$ % from the true result which is $0.384494$.

Corollary

• What I did was purely observative;

• One may improve my answer in a very big fat way, especially for what concerns the ellipse part;

• Probably the neglected part has a weight, for ow little it might be, so one can also look at it in details.

That being said, and now I leave this to posterity.

Notice that I tried to analyze this in the simplest way possible. Probably someone will come with big monster function or unspeakable methods far better than mine, but still.

I had fun with this, so thank you so much!

• Many thanks to you, now and tomorrow in the morning I am going to study your answer.
– user243301
Mar 6 '18 at 20:42