prove $\ln(1+x^2)\arctan x=-2\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{2n+1}x^{2n+1}$

Question

I was able to prove the above identity using 1) Cauchy Product of Power series and 2) integration but the point of posting it here is to use it as a reference in our solutions.

other approaches would be appreciated.

Answer 1

I think this is a much simpler proof.

\begin{align} \tanh^{-1}x\ln(1-x^2)&=\frac12\{\ln(1+x)-\ln(1-x)\}\{\ln(1+x)+\ln(1-x)\}\tag1\\ &=\frac12\ln^2(1+x)-\frac12\ln^2(1-x)\tag2\\ &=\sum_{n=1}^\infty(-1)^n\frac{H_{n-1}}{n}x^n-\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^n\tag3\\ &=-2\sum_{n=1}^\infty\frac{H_{2n-2}}{2n-1}x^{2n-1}\tag4\\ &=-2\sum_{n=1}^\infty\frac{H_{2n}}{2n+1}x^{2n+1}\tag5 \end{align}

Thus $$\tanh^{-1}x\ln(1-x^2)=-2\sum_{n=1}^\infty\frac{H_{2n}}{2n+1}x^{2n+1}\tag6$$

Replace $x$ with $ix$ we get

$$\tan^{-1}x\ln(1+x^2)=-2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}x^{2n+1}\tag7$$

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Explanation:

$(1)$ $\tanh^{-1}x=\frac12\ln\left(\frac{1+x}{1-x}\right)$.

$(2)$ Difference of two squares.

$(3)$ $\frac12\ln^2(1-x)=\sum_{n=1}^\infty\frac{H_n}{n+1}x^{n+1}=\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^n$

$(4)$ $\sum_{n=1}^\infty ((-1)^n-1)a_{n}=-2\sum_{n=1}^\infty a_{2n-1}$

$(5)$ Reindex.

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Bonus:

If we differentiate both sides of $(7)$ we obtain another useful identity

$$\frac{\arctan x}{1+x^2}=\frac12\sum_{n=1}^\infty(-1)^n\left(H_n-2H_{2n}\right)x^{2n-1}\tag8$$

another identity follows from integrating both sides of $(8)$:

$$\arctan^2x=\frac12\sum_{n=1}^\infty\frac{(-1)^n\left(H_n-2H_{2n}\right)}{n}x^{2n}\tag9$$

replace $x$ with $ix$ in $(9)$

$$\text{arctanh}^2x=-\frac12\sum_{n=1}^\infty\frac{\left(H_n-2H_{2n}\right)}{n}x^{2n}\tag{10}$$

Answer 2

actually, you can do the product directly, given well-known series

$$\begin{aligned} \arctan x & = \sum_{n=0}^{\infty} {\frac{(-1)^n x^{2n+1}}{2n+1}}\\ \ln(1+x^2) & = \sum_{n=1}^{\infty} {\frac{(-1)^{n+1} x^{2n}}{n}} \end{aligned}$$

obviously their product has no even order items, set

$$\arctan x \ln (1+x^2) = \sum_{m=0}^{\infty} {a_{2m+1} x^{2m+1}}$$

for item $x^{2m+1}$, it has pair partitions as $(x,x^{2m}),(x^3,x^{2m-2}),\cdots,(x^{2m-1},x^2)$, thus

$$\begin{aligned} a_{2m+1} & = \sum_{n=0}^{m-1} {\frac{(-1)^n}{2n+1} \cdot \frac{(-1)^{m-n+1}}{m-n}} = \sum_{n=0}^{m-1} {\frac{(-1)^{m+1}}{(2n+1)(m-n)}}\\ & = \frac{(-1)^{m+1}}{2m+1} \sum_{n=0}^{m-1} {\frac{2m+1}{(2n+1)(m-n)}} = \frac{(-1)^{m+1}}{2m+1} \sum_{n=0}^{m-1} {\frac{2n+1+2(m-n)}{(2n+1)(m-n)}}\\ & = \frac{(-1)^{m+1}}{2m+1} \left( \sum_{n=0}^{m-1} {\frac1{m-n}} + \sum_{n=0}^{m-1} {\frac2{2n+1}} \right)\\ & = \frac{(-1)^{m+1}}{2m+1} \left( H_{m} + 2\left( \sum_{n=1}^{2m} {\frac1{n}} - \sum_{n=1}^{m} {\frac1{2n}} \right) \right)\\ & = \frac{(-1)^{m+1} (H_{m} + 2H_{2m} - H_{m})}{2m+1} = \frac{(-1)^{m+1} \cdot 2H_{2m}}{2m+1} \end{aligned}$$

Answer 3

knowing that fact that $$2\sum_{n=1}^\infty f(2n)=\sum_{n=1}^\infty f(n)(1+(-1)^n)$$ then \begin{align} 2\sum_{n=1}^\infty (-1)^nx^{2n}H_{2n}&=2\sum_{n=1}^\infty (i)^{2n}x^{2n}H_{2n}\\ &=\sum_{n=1}^\infty (ix)^nH_{n}+\sum_{n=1}^\infty (-ix)^nH_{n}\\ &=-\frac{\ln(1-ix)}{1-ix}-\frac{\ln(1+ix)}{1+ix}\\ &=-\frac{\ln(1-ix)+\ln(1+ix)+ix(\ln(1-ix)-\ln(1+ix))}{1+x^2}\\ &=-\frac{\ln(1+x^2)+ix(-2i\arctan x)}{1+x^2}\\ &=-\frac{\ln(1+x^2)}{1+x^2}-\frac{2x\arctan x}{1+x^2} \end{align} integrate both sides from $x=0$ to $z$ \begin{align} 2\sum_{n=1}^\infty (-1)^nH_{2n}\int_0^zx^{2n}\ dx&=2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}z^{2n+1}\\ &=-\int_0^z\left(\frac{\ln(1+x^2)}{1+x^2}+\frac{2x\arctan x}{1+x^2}\right)\ dx\\ &=-\int_0^zd(\ln(1+x^2)\arctan x)\\ &=-\ln(1+z^2)\arctan z \end{align}