Proving a closed form of an integral

Question

Is there any proof for this integral?$$\int \limits _0^1\frac{1}{a^2x^2+1}\left [\left (1-\frac{x}{2}\ln \frac{1+x}{1-x}\right )^2+\frac{\pi^2x^2}{4}\right ]^{-1}\,dx=\frac{\arctan a}{a-\arctan a}-\frac{3}{a^2},\quad \operatorname{Re}(a)>0.$$I tried substituting $x=\frac{1-x}{1+x}$ but the integral seems to be harder.

Answer 1

This is not an answer.

This interesting integral did resist to all CAS I have been able to use.

However, there are interesting features which appear if we expand as a series built around $x=0$

$$\left [\left (1-\frac{x}{2}\log \left(\frac{1+x}{1-x}\right)\right )^2+\frac{\pi^2x^2}{4}\right ]^{-1}=\sum_{n=0}^\infty b_n\,x^{2n}$$

Assuming $a>0$ and $n\geq 0$ $$J_n=\int_0^1 \frac {x^{2n}}{a^2x^2+1}\,dx=\frac{\, _2F_1\left(1,n+\frac{1}{2};n+\frac{3}{2};-a^2\right)}{2 n+1}$$ where appears the Gaussian hypergeometric function. It could look complicated but, in fact, $$J_n=(-1)^n\,\, a^{-(2 n+1)} \left(\tan ^{-1}(a)- K_n\right)$$ the first $K_n$ being $$\left( \begin{array}{cc} n & K_n \\ 0 & 0 \\ 1 & a\\ 2 & a-\frac{a^3}{3} \\ 3 & a-\frac{a^3}{3}+\frac{a^5}{5} \\ 4 & a-\frac{a^3}{3}+\frac{a^5}{5}-\frac{a^7}{7} \\ 5 & a-\frac{a^3}{3}+\frac{a^5}{5}-\frac{a^7}{7}+\frac{a^9}{9} \\ 6 & a-\frac{a^3}{3}+\frac{a^5}{5}-\frac{a^7}{7}+\frac{a^9}{9}-\frac{a^{11}}{11} \end{array} \right) $$ which are the expansions of $\tan ^{-1}(a)$ to $O\left(a^{2n+1}\right)$.

The problem is the $b_n$ for which I did not find any formula. Beside $b_0=1$, they are all negative an smaller and smaller $$\left( \begin{array}{cc} n & b_n \\ 1 & 2-\frac{\pi ^2}{4} \\ 2 & \frac{11}{3}-\pi ^2+\frac{\pi ^4}{16} \\ 3 & \frac{32}{5}-\frac{17 \pi ^2}{6}+\frac{3 \pi ^4}{8}-\frac{\pi ^6}{64} \\ 4 & \frac{1136}{105}-\frac{103 \pi ^2}{15}+\frac{23 \pi ^4}{16}-\frac{\pi ^6}{8}+\frac{\pi ^8}{256} \\ 5 & \frac{5632}{315}-\frac{3823 \pi ^2}{252}+\frac{89 \pi ^4}{20}-\frac{29 \pi ^6}{48}+\frac{5 \pi ^8}{128}-\frac{\pi ^{10}}{1024} \\ 6 & \frac{215696}{7425}-\frac{1984 \pi ^2}{63}+\frac{20327 \pi ^4}{1680}-\frac{91 \pi ^6}{40}+\frac{175 \pi ^8}{768}-\frac{3 \pi ^{10}}{256}+\frac{\pi ^{12}}{4096} \end{array} \right)$$

Using $25$ terms with $a=\pi$, this gives a value of $0.368089$ while the formula gives $0.368016$.

Answer 2

Let $f$ be an analytic function defined on $|z|>1$ via $$f(z) = 1-\frac{z}{2}\log \frac{1+z^{-1}}{1-z^{-1}}$$ then it's easy to show:

• $f$ can be analytic continued to $\mathbb{C}-[-1,1]$.
• $f \sim -1/(3z^2)$ for $|z|$ large.
• For $-1<x<1$, $$f_\pm (x) := \lim_{y\to 0^\pm} f(x+yi) = 1-\frac{x}{2}\left(\log\frac{1-x}{1+x} \mp \pi i\right) $$
• $f$ has no zero on $\mathbb{C}-[-1,1]$.

The fourth one is actually an important point: it makes the following discussion invalid if we replace $2$ by $3$ in definition of $f$, for example.

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For $\varepsilon > 0$ small, integrate $I = \int_\gamma \frac{1}{(z+\varepsilon i)(1+a^2 z^2)f(z)} dz$ with $\gamma$ dog-bone contour. Integral along big circle is $-3a^{-2}(2\pi i)$, so $$\begin{aligned} I &= 2\pi i (-\frac{1}{\left(a^2 \epsilon ^2-1\right) f(-i \epsilon )}+\frac{1}{2 f\left(-\frac{i}{a}\right) (a \epsilon -1)}-\frac{1}{2 f\left(\frac{i}{a}\right) (a \epsilon +1)}) \\ &= -\frac{3}{a^2}(2\pi i) + \int_{-1}^1 \frac{1}{(x+\varepsilon i)(1+a^2 x^2)} (\frac{1}{f_+(x)} - \frac{1}{f_-(x)}) dx \\ &= -\frac{3}{a^2}(2\pi i) + \int_{-1}^1 \frac{-\pi i x}{(x+\varepsilon i)(1+a^2 x^2)} \left [\left (1-\frac{x}{2}\ln \frac{1+x}{1-x}\right )^2+\frac{\pi^2x^2}{4}\right ]^{-1} dx\end{aligned}$$

Let $\varepsilon \to 0$ gives $$\int_{-1}^1 \frac{1}{1+a^2 x^2} \left [\left (1-\frac{x}{2}\ln \frac{1+x}{1-x}\right )^2+\frac{\pi^2x^2}{4}\right ]^{-1} dx = \frac{1}{f\left(\frac{i}{a}\right)}+\frac{1}{f\left(-\frac{i}{a}\right)}-2 - \frac{6}{a^2}$$ which you can verify equals $\frac{2\arctan a}{a-\arctan a}-\frac{6}{a^2}$ for $a>0$.