Is it possible to expand $\arctan x \ln(1-x^2)$ in series?

Question

I am not sure if its possible to find the Taylor series of $\arctan x \ln(1-x^2)$ so I am giving it a try:

By using the Cauchy product

$$\left(\sum_{n=1}^\infty a_n x^n\right)\left(\sum_{n=1}^\infty b_n x^n\right)=\sum_{n=1}^\infty \left(\sum_{k=1}^n a_k b_{n-k+1}\right)x^{n+1}$$

we have

$$\arctan x \ln(1-x^2)=\left(\sum_{n=1}^\infty \frac{(-1)^{n-1} x^{2n-1}}{2n-1}\right)\left(-\sum_{n=1}^\infty\frac{x^{2n}}{n}\right)$$

$$=\frac1x\left(\sum_{n=1}^\infty \underbrace{\frac{(-1)^{n}}{2n-1}}_{a_n}(x^2)^n\right)\left(\sum_{n=1}^\infty\underbrace{\frac{1}{n}}_{b_n}(x^2)^n\right)$$

$$=\frac1x\sum_{n=1}^\infty\left(\sum_{k=1}^n \frac{(-1)^k}{2k-1}\cdot\frac{1}{n-k+1}\right)(x^2)^{n+1}$$

$$\left\{\text{write $\frac{1}{(2k-1)(n-k+1)}=\frac{1}{2n+1}\left(\frac{2}{2k-1}+\frac{1}{n-k+1}\right)$}\right\}$$

$$=\sum_{n=1}^\infty \frac{2}{2n+1}\left(\sum_{k=1}^n \frac{(-1)^k}{2k-1}\right)x^{2n+1}+\sum_{n=1}^\infty \frac{1}{2n+1}\left(\sum_{k=1}^n \frac{(-1)^k}{n-k+1}\right)x^{2n+1}$$

$$\left\{\text{reverse the order of the terms in the second sum}\right\}$$

$$=\sum_{n=1}^\infty \frac{2}{2n+1}\left(\sum_{k=1}^n \frac{(-1)^k}{2k-1}\right)x^{2n+1}+\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n+1}\left(\sum_{k=1}^n \frac{(-1)^k}{k}\right)x^{2n+1}$$

$$=\sum_{n=1}^\infty\frac{2}{2n+1}\left(\sum_{k=1}^n\frac{(-1)^k}{2k-1}\right)x^{2n+1}+\sum_{n=1}^\infty \frac{(-1)^{n}\overline{H}_n}{2n+1}x^{2n+1}.$$

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Mathematica gives

$$\sum_{k=1}^n\frac{(-1)^k}{2k-1}=\int_0^1\frac{(-x^2)^n-1}{1+x^2}dx=-\frac{\pi}{4}+\frac{(-1)^n}{4}\left(\psi\left(\frac{2n+3}{4}\right)-\psi\left(\frac{2n+1}{4}\right)\right)$$

but this form makes the problem harder I guess. Any idea how to find a better form for $\sum_{k=1}^n\frac{(-1)^n}{2k-1}$ or how to find the whole $\sum_{n=1}^\infty\frac{2}{2n+1}\left(\sum_{k=1}^n\frac{(-1)^k}{2k-1}\right)x^{2n+1}$ or maybe a different way to find the Taylor series of $\arctan x \ln(1-x^2)$?

Answer 1

Here are my results for the coefficients of the Taylor expansion. A possible derivation is shown in §5.

$1 solution in terms of harmonic numbers

Let

$$f(x):=\text{arctan(x)}\log(1-x^2) = \sum_{k=0}^{\infty} c_n x^{2n+3}\tag{1}$$

then the coefficients, expressend completely through harmonic numbers, are

$$\begin{align}c_H(n) = \frac{1}{2 (2 n+3)} \left(-\pi + H_{\frac{n}{2}}-H_{\frac{n+1}{2}}\\ +(-1)^{n+1} \left(H_{\frac{2n+1}{4}}-H_{\frac{2n-1}{4}}+\log (4)\right)\right)\end{align}\tag{2} $$

It is useful to write down the expressions for even and odd $n$ separately

$$\begin{align} c_{H}(n_{even})=&\frac{1}{2 (2 n+3)}\left( H_{\frac{n}{2}}+\left(H_{\frac{1}{4} (2 n-1)}-H_{\frac{1}{4} (2 n+1)}-\pi \right)\\-\left(H_{\frac{n+1}{2}}+\log (4)\right)\right) \end{align}\tag{2a}$$

$$\begin{align}c_{H}(n_{odd})=\frac{1}{2 (2 n+3)}\left(-H_{\frac{n+1}{2}}+\left(-H_{\frac{1}{4} (2 n-1)}+H_{\frac{1}{4} (2 n+1)}-\pi \right)\\+\left(H_{\frac{n}{2}}+\log (4)\right)\right)\end{align}\tag{2b}$$

Notice that the non-rational numbers $\log(4)$ and $\pi$ combine with harmonic numbers with half-integer and quarter-integer indices, resp. leaving rational expressions for $c_{H}(n)$. Specifically, in $(2b)$ and $(2b)$ the brackets in the numerator are rational numbers.

You can see this from the following relations (here $n$ is a non negative integer)

$$H_x = H_ {x - 1} + \frac {1} {x}$$ $$H_{n-\frac{1}{2}}=\sum _{k=1}^n \frac{1}{k-\frac{1}{2}}+H_{-\frac{1}{2}}\text{ and } H_{-\frac{1}{2}}=-\log (4)$$ $$H_{n\pm \frac{1}{4}}=\sum _{k=1}^n \frac{1}{k\pm \frac{1}{4}}+H_{\pm \frac{1}{4}}$$ $$H_{-\frac{1}{4}}=\frac{\pi }{2}-\log (8)\text{ and }H_{\frac{1}{4}}=-\frac{\pi }{2}+4-\log (8)$$

The results in $(2)$ were obtained from §2 in two steps

(1) write digamma functions in terms of harmonic numbers using

$$\psi(x) \to H_{x-1} - \gamma$$

(2) transform arguments of $H_x$ in the negative range to positive values using the reflexion formula

$$H_{1-\alpha }=\frac{1}{1-\alpha }-\frac{1}{\alpha }+\pi \cot (\pi \alpha )+H_{\alpha }$$

My reviously obtained equivalent results are:

§2. solution containing limits of digamma functions

Playing around with double integral representations I came up with the following form of the coefficients

$$\begin{align}c_n=\lim_{k\to n} \,\left( \frac{1}{2 (2 k+3)}\left((-1)^{k+1} \left(\pi (-1)^k+2 \pi (-1)^k \csc (\pi k)-2 \pi \sec (\pi k)\\ -(-1)^k \psi ^{(0)}\left(-\frac{k}{2}\right)+(-1)^k \psi ^{(0)}\left(-\frac{k}{2}-\frac{1}{2}\right)-\psi ^{(0)}\left(\frac{1}{4}-\frac{k}{2}\right)\\ +\psi ^{(0)}\left(-\frac{k}{2}-\frac{1}{4}\right)+\log (4)\right)\right)\right)\end{align}\tag{3}$$

Despite the horrible expression the values of $c(n)$ are rational for $n=0, 1, 2...$.

The simplicity of the formula lies in the fact that it needs only digamma functions.

§3. solution with the limits in §2 carried out

Here is the result with the limit carried out in two equivalent forms:

$$\begin{align} c(n_\text{odd}) = \frac{1}{2(2n+3)}\left(\pi + (-1)^n \left(H_{\left\lfloor \frac{n-1}{2}\right\rfloor +1}-\gamma -\log (4)\right)+\left(-\psi ^{(0)}\left(-\frac{1}{4} (2 n-1)\right)+\psi ^{(0)}\left(-\frac{1}{4} (2 n+1)\right)+\psi ^{(0)}\left(-\frac{1}{4} (2 n)\right)\right)\right)\end{align} \tag{4a}$$

$$\begin{align}c(n_\text{odd}) = \frac{1}{2(2n+3)}\left(\pi + (-1)^n \left(H_{\frac{n+1}{2}}-\gamma -\log (4)\right)+\left(-\psi ^{(0)}\left(-\frac{1}{4} (2 n-1)\right)+\psi ^{(0)}\left(-\frac{1}{4} (2 n+1)\right)+\psi ^{(0)}\left(-\frac{1}{4} (2 n)\right)\right)\right)\end{align} \tag{4b}$$

$$\begin{align}c(n_\text{even}) =\frac{1}{2(2n+3)}\left(\pi+(-1)^n \left(H_{\left\lfloor \frac{n-1}{2}\right\rfloor +1}-\gamma -\log (4)\right)+\left(\psi ^{(0)}\left(-\frac{1}{4} (2 n-1)\right)-\psi ^{(0)}\left(-\frac{1}{4} (2 n+1)\right)-\psi ^{(0)}\left(-\frac{1}{4} (2 n+2)\right)\right) \right)\end{align}\tag{5a}$$

$$\begin{align}c(n_\text{even}) =\frac{1}{2(2n+3)}\left(\pi+(-1)^n \left(H_ {\frac{n}{2}}-\gamma -\log (4)\right)+\left(\psi ^{(0)}\left(-\frac{1}{4} (2 n-1)\right)-\psi ^{(0)}\left(-\frac{1}{4} (2 n+1)\right)-\psi ^{(0)}\left(-\frac{1}{4} (2 n+2)\right)\right) \right)\end{align}\tag{5b}$$

§4 Recursion

The difficulties with the recursion of the coefficients encountered in https://math.stackexchange.com/a/4474186/198592 can easily be overcome when we consider the coefficients multiplied by $2(3+2n)$

$$d_e(n) = 2(3+2n) c(n_\text{even})\tag{6a}$$ $$d_o(n) = 2(3+2n) c(n_\text{odd})\tag{6b}$$

From $(2a,2 b)$ we find easily ($n$ is even here)

$$d_e(n) - d_e(n-2) =\left( -\frac{2}{n+1}+\frac{4}{2 n-1}-\frac{4}{2 n+1}+\frac{2}{n}\right)\tag{7a}$$ $$d_o(n+1)-d_0(n-1) = \left(-\frac{2}{n+2}-\frac{4}{2 n+1}+\frac{4}{2 n+3}+\frac{2}{n+1}\right)\tag{7b}$$

The initial conditions are

$$d_e(0) = -6, d_o(1) = -\frac{5}{3}$$

These linear recursion relations are easily solved by Mathematica. The resulting expressions come in terms of the Lerch transcendent. They are, of course, equvalent to our solutions (because we have derived the recursion relations from them).

Having calculated $d(n)$ we have simply to invert $(6)$ to find the coefficients themselves.

§5 Derivation of coefficients using integral representations

We write the two factors of $f(x)$ as follows

$$\tan ^{-1}(x)=\int_0^1 \frac{x}{s^2 x^2+1} \, ds\tag{5.1}$$ $$-\log \left(1-x^2\right)=\int_0^1 \frac{x^2}{1-t x^2} \, dt\tag{5.2}$$

$f(x)$ is then given by the double integral

$$f(x) = -x^3 \int _0^1\int _0^1 \frac{1}{\left(s^2 x^2+1\right) \left(1-t x^2\right)}dtds\tag{5.3}$$

Decomposing the fraction

$$\frac{\frac{s^2}{s^2 x^2+1}-\frac{t}{t x^2-1}}{s^2+t}$$

and expanding into geometric series gives the coefficient $a(k)$ of $x^{3+2k}$ as the double integral

$$\int _0^1\int _0^1\frac{(-1)^k s^{2 k+2}+ t^{k+1}}{s^2+t}dsdt\tag{5.4}$$

To do the integrals we can successfully employ Mathematica.

The t-integral gives

$$i_t =\int_0^1 \frac{(-1)^k s^{2 k+2}+t^{k+1}}{s^2+t} \, dt=\frac{\, _2F_1\left(1,k+2;k+3;-\frac{1}{s^2}\right)+(-1)^k (k+2) s^{2 k+4} \log \left(\frac{1}{s^2}+1\right)}{(k+2) s^2}\tag{5.5}$$

here $_2F_1\left(a,b;c;z\right)$ is the hypergeometric function.

The subsequent s-integral gives

$$\begin{align}a(k) = & \int_0^1 i_t \, ds\\ & =\frac{(-1)^{k+1} \Phi \left(-1,1,k+\frac{5}{2}\right)}{2 k+3}+\frac{2 (-1)^k}{(2 k+3)^2}\\ & +\frac{\pi }{4 k+6}+\frac{(-1)^k \log (2)}{2 k+3}\\ &+\frac{\psi\left(\frac{k+3}{2}\right)}{4 k+6}-\frac{\psi\left(\frac{k}{2}+1\right)}{4 k+6}\end{align}\tag{5.6}$$

Here $\Phi \left(z,s,a\right)$ is the Lerch transcendent and $\psi$ is the digamma function.

The first few values are

$$a(k)_{k=1}^{k=6} = \left\{1,\frac{1}{6},\frac{11}{30},\frac{121}{1260},\frac{281}{1260},\frac{929}{13860}\right\}$$

We see that $a(k) = c(k)$.

Answer 2

A bit tricky but it makes things simpler.

Rewrite $$\tan ^{-1}(x)\log \left(1-x^2\right) = x \frac{\tan ^{-1}(x)}{x}\log \left(1-x^2\right)$$ and let $x=\sqrt t$ $$\frac{\tan ^{-1}(\sqrt t)}{\sqrt t}=\sum_{n=0}^\infty \frac{(-1)^n }{2 n+1}t^n$$ $$\log \left(1-t\right)=-\sum_{n=1}^\infty \frac {t^n} n$$

Where is see a problem is that the coefficients of the product $$\frac{\tan ^{-1}(\sqrt t)}{\sqrt t}\,\log \left(1-t\right)=-\sum_{n=1}^\infty a_n\,t^n$$ being $$\left\{1,\frac{1}{6},\frac{11}{30},\frac{121}{1260},\frac{281}{1260},\frac{929}{1 3860},\frac{432683}{2702700},\frac{18527}{360360},\frac{84971}{680680},\frac{203 75981}{488864376},\cdots\right\}$$ nothing has been identified by $OEIS$.

Answer 3

Let $f(x)=\sum\limits_{n=0}^\infty a_nx^n=\arctan(x)\ln(1-x^2)$

Multiplying the series of $\arctan$ and $\ln$ with the Cauchy product reveals a bit difficult, but we can separate the two series by derivation. This works because the derivatives of these two functions involve polynomials.

$f'(x)=\dfrac{\ln(1-x^2)}{1+x^2}-\dfrac{2x\arctan(x)}{1-x^2}$

After multiplication by the same common denominator we get the formula:

$$(1-x^4)f'(x)=(1-x^2)\ln(1-x^2)-2x(1+x^2)\arctan(x)$$

Expanding the series gives:

$\begin{align}LHS &=(1-x^4)\sum\limits_{n=1}^\infty na_nx^{n-1}\\ &=\sum\limits_{n=1}^\infty na_nx^{n-1}-\sum\limits_{n=1}^\infty na_nx^{n+3}\\ &=\sum\limits_{n=0}^\infty (n+1)a_{n+1}x^n-\sum\limits_{n=4}^\infty (n-3)a_{n-3}x^n\\ &=a_1+(2a_2)x+(3a_3)x^2+(4a_4)x^3+\sum\limits_{n=4}^\infty \Big((n+1)a_{n+1}-(n-3)a_{n-3}\Big)x^n \end{align}$

$\begin{align}RHS &=(1-x^2)\sum\limits_{n=1}^\infty \dfrac{(-1)}{n}x^{2n}-(2x+2x^3)\sum\limits_{n=0}^\infty \dfrac{(-1)^n}{2n+1}x^{2n+1}\\ &=-\sum\limits_{n=1}^\infty \dfrac{1}{n}x^{2n}+\sum\limits_{n=1}^\infty \dfrac{1}{n}x^{2n+2}-2\sum\limits_{n=0}^\infty \dfrac{(-1)^n}{2n+1}x^{2n+2}-2\sum\limits_{n=0}^\infty \dfrac{(-1)^n}{2n+1}x^{2n+4}\\ &=-\sum\limits_{n=1}^\infty \dfrac{1}{n}x^{2n}+\sum\limits_{n=2}^\infty \dfrac{1}{n-1}x^{2n}+2\sum\limits_{n=1}^\infty \dfrac{(-1)^n}{2n-1}x^{2n}-2\sum\limits_{n=2}^\infty \dfrac{(-1)^n}{2n-3}x^{2n}\\ &=(-1-2)x^2+\sum\limits_{n=2}^\infty \Big(-\dfrac{1}{n}+\dfrac{1}{n-1}+2\dfrac{(-1)^n}{2n-1}-2\dfrac{(-1)^n}{2n-3}\Big)x^{2n}\\ &=-3x^2+\sum\limits_{n=2}^\infty \Big(\dfrac{1}{n(n-1)}+\dfrac{(-4)(-1)^n}{(2n-1)(2n-3)}\Big)x^{2n} \end{align}$

We can now proceed to identify the coefficients of the two sides of the equation to find an induction relation between the $a_n$ terms.

$\begin{cases} a_1=0\\ a_2=0\\ a_3=-1\\ a_4=0\\ \forall n\ge 2 & a_{2n}=0\\ \forall n\ge 2 & (2n+1)a_{2n+1} = (2n-3)a_{2n-3} + \dfrac{1}{n(n-1)}+\dfrac{(-4)(-1)^n}{(2n-1)(2n-3)}\\ \end{cases}$

Also from $f(0)=0$ we have $a_0=0$

Calculated iteratively we can verify that we obtain the same coefficients than Claude:

$$-x^3-\frac{1}{6}x^5-\frac{11}{30}x^7-\frac{121}{1260}x^9-\frac{281}{1260}x^{11}-\frac{929}{13860}x^{13}-\frac{432683}{2702700}x^{15}+O(x^{17})$$

Finding a closed formula seems rather intractable as it is an induction relation with non-constant coefficients, but for practical purposes, you can easily calculate all the $a_n$ up to a given order using this step by step relation.