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$$f(n) = \int_0^1 \frac{n \tan^{-1}(t)}{n^2 + t^2}\mathrm d t \tag{n > 2}$$

Introduction: This is one of the most beautiful and mysterious integrals I've every encountered. It's very simple, but my conjectured closed form is one of the most bizzare I've ever seen.

What I know: All of the closed forms that Mathematica gives have the following form:

$$\frac18 \big(\tan^{-1}(a_n)^2 - 4i \cot^{-1}(n) \log(b_n) + 2 i \tan^{-1}(a_n) \log(b_n) - 2 (\text{Li}_2(c_n - id_n) + \text{Li}_2(c_n + id_n)- \text{Li}_2(\frac{1}{b_n}) \big{)}$$

Pretty ugly, I know. But the really amazing part is what comes next.

The Conjecture:

$$a_n= \frac{\left. \begin{cases} 2n & \text{n even} \\ n & \text{n odd } \end{cases} \right\}}{\text{lcm}(n+1,n-1)} \\ b_n= \frac{\left. \begin{cases} n+1 & \text{n even} \\ \frac{n+1}{2} & \text{n odd } \end{cases} \right\}}{\left. \begin{cases} \frac{n}{2} & \text{n even} \\ n & \text{n odd } \end{cases} \right\} \\} \\ c_n = \frac{\text{(n-1) * largest prime factor of } n-1}{\text{largest odd divisor of } n^2 + 1} $$ And I have no idea what $d_n$ could be.

The conjecture holds for at least the first 20 values of n, as well as 20 other random higher values of n, and I think it's too simple to just be a coincidence. So, my questions are:

  1. What is $d_n$?
  2. How can I prove or disprove this conjecture?

Any help is appreciated, thanks!

  • 0
    It has occured to me that I may have gotten some of the equations wrong when copying them over from Mathematica. If you find an error, please edit and fix it or comment and I will, thanks :D – 2017-01-08
  • 3
    As soon as you know that $\arctan(x)=\frac{1}{2 i}\left(\log(x-i)-\log(x+i)\right)$ and that $\text{Li}_2(x)=-\int_0^xdt\frac{\log(1-t)}{t}$ you should be able to resolve some of the mysteries you discovered :) – 2017-01-08

1 Answers 1

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I'll give you an insight on how to solve the integral

Start by

$$F(a) = \int_0^1 \frac{\tan^{-1}(at)}{n^2 + t^2}\,dt$$

By differentiation

$$F'(a) = \int_0^1 \frac{t}{(n^2 + t^2)((at)^2+1)}\,dt = \frac{\log(1+a^2)+ \log \left( \frac{n^2}{1+n^2}\right)}{(-2 + 2 a^2 n^2)}$$

Integrate

$$\int_0^1 \frac{\tan^{-1}(t)}{n^2 + t^2}\,dt = \frac{1}{2}\int^1_0 \frac{\log(1+a^2)}{(an)^2-1}\,da+\frac{1}{2}\log \left( \frac{n^2}{1+n^2}\right)\int^1_0\frac{1}{(an)^2-1}\,da$$

Then second integral is easy

$$\int_0^1 \frac{\tan^{-1}(t)}{n^2 + t^2}\,dt = \frac{1}{2}\int^1_0 \frac{\log(1+a^2)}{(an)^2-1}\,da-\frac{1}{2}\log \left( \frac{n^2}{1+n^2}\right)\frac{\tanh^{-1}(n)}{n}$$

Then you can use the following properties for the first integral

$$\int^t_0 \frac{\log(1+ax)}{1-x}\, dx = -\log(1-x)\log(1+a)- \text{Li}_2 \left( \frac{a}{a+1} \right) +\text{Li}_2 \left(\frac{a-ax}{a+1}\right)$$

$$ \int^t_0 \frac{\log(1+ax)}{1+x}\, dx = - \text{Li}_2 \left( \frac{t}{t+1} \right) +\text{Li}_2 \left(\frac{t-ta}{t+1}\right)-\text{Li}_2(-at)$$

Note that

$$\frac{\log(1+a^2)}{(an)^2-1} = \frac{\log(1+ix)+\log(1-ix)}{(an-1)(an+1)}$$