Maclaurin series for $\arctan^{2}(x)$

Question

I have a question here that requires me to find the Maclaurin series expansion of $\arctan^{2}(x)$. Now I know how to find it for $\arctan(x)$, by taking the derivative, expanding it into a series, and integrating back (given x is in the interval of uniform convergence), But applying that here leaves me with $$\frac{df}{dx}=2\arctan(x)\frac{1}{(x^2+1)}$$ I am not sure If I can pick out parts of the product (like $\arctan(x)$) and differentiate, expand , then integrate them. Even if I did, how would I go about multiplying two series? Is there an easier way to do this? Thanks.

Answer 1

Using the Cauchy Product Formula, $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\arctan^2(x) &=2\frac{\arctan(x)}{1+x^2}\\ &=2\left(x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+\cdots\right)\left(1-x^2+x^4-x^6+\cdots\right)\\ &=2\left((1)x-\left(1+\tfrac13\right)x^3+\left(1+\tfrac13+\tfrac15\right)x^5-\left(1+\tfrac13+\tfrac15+\tfrac17\right)x^7+\cdots\right)\\ &=2\sum_{k=1}^\infty(-1)^{k-1}\left(H_{2k}-\tfrac12H_k\right)x^{2k-1} \end{align} $$ Therefore, $$ \arctan^2(x)=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\left(H_{2k}-\tfrac12H_k\right)x^{2k} $$

Answer 2

We can try to obtain the series in the following way:

$$f(x)=\arctan^2 x=x^2 \int_0^1 \int_0^1 \frac{du~dv}{(1+x^2u^2)(1+x^2v^2)}$$

It's easier to consider:

$$g(x)=\int_0^1 \int_0^1 \frac{du~dv}{(1+x^2u^2)(1+x^2v^2)}$$

Let's use partial fractions:

$$\frac{1}{(1+x^2u^2)(1+x^2v^2)}=\frac{u^2}{(u^2-v^2)(1+x^2u^2)}-\frac{v^2}{(u^2-v^2)(1+x^2v^2)}$$

We obtain a sum of two singular integrals, which however, can both be formally expanded into a series:

$$g(x)=\int_0^1 \int_0^1 \left(\frac{u^2}{(u^2-v^2)(1+x^2u^2)}-\frac{v^2}{(u^2-v^2)(1+x^2v^2)} \right) du ~dv=$$

$$g(x)=\sum_{n=0}^\infty (-1)^n x^{2n} \int_0^1 \int_0^1 \left(\frac{u^{2n+2}}{u^2-v^2}-\frac{v^{2n+2}}{u^2-v^2} \right) du ~dv$$

Now:

$$g(x)=\sum_{n=0}^\infty (-1)^n x^{2n} \int_0^1 \int_0^1 \frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} du ~dv$$

Obviously, every integral is finite now, and we can write:

$$\int_0^1 \int_0^1 \frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} dv ~du=2 \int_0^1 \int_0^u \frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} dv ~du= \\ = 2 \sum_{k=0}^\infty \int_0^1 \int_0^u u^{2n} \left(1-\frac{v^{2n+2}}{u^{2n+2}} \right) \frac{v^{2k}}{u^{2k}} dv ~du =2 \sum_{k=0}^\infty \int_0^1 \int_0^1 u^{2n+1} \left(1-t^{2n+2} \right) t^{2k} dt ~du= \\ = 2 \sum_{k=0}^\infty \int_0^1 \left(\frac{1}{2k+1}-\frac{1}{2k+2n+3} \right) u^{2n+1}~du= 2\sum_{k=0}^\infty \frac{1}{(2k+1)(2k+2n+3)}$$

So we get:

$$g(x)=\frac{1}{2} \sum_{n=0}^\infty (-1)^n x^{2n} \sum_{k=0}^\infty \frac{1}{(k+\frac{1}{2})(k+n+\frac{3}{2})}$$

The inner series converges for all $n$, and we can formally represent it as a difference of two divergent harmonic series, which, after some manipulations, should give us the same result as robjohn obtained.

I suppose, a kind of closed form for the general term can also be given in terms of digamma function:

$$g(x)=\frac{1}{2} \sum_{n=0}^\infty (-1)^n \frac{\psi \left(n+\frac32 \right)-\psi \left(\frac12 \right)}{n+1} x^{2n}$$

Which makes:

$$\arctan^2 x=\frac{x^2}{2} \sum_{n=0}^\infty (-1)^n \frac{\psi \left(n+\frac32 \right)-\psi \left(\frac12 \right)}{n+1} x^{2n}$$

Which is essentially the same as robjohn's answer.

Answer 3

For the sake of an alternative method, here's a double series.

For $|x|<1$, we have $$\arctan x=\sum_{k\geq0}\frac{(-1)^kx^{1+2k}}{1+2k}$$ As you noted, $$\arctan^2x=2\int\frac{\arctan x}{x^2+1}\mathrm{d}x$$ So, assuming $|x|<1$, $$\arctan^2x=2\int\frac1{x^2+1}\sum_{k\geq0}\frac{(-1)^kx^{1+2k}}{1+2k}\mathrm{d}x$$ $$\arctan^2x=2\sum_{k\geq0}\frac{(-1)^k}{1+2k}\int\frac{x^{2k}}{1+x^2}x\mathrm{d}x$$ Now we focus on $$I=\int\frac{x^{2k+1}}{1+x^2}\mathrm{d}x$$ We recall that for $|x|<1$, $$\frac1{1+x^2}=\frac12\sum_{n\geq0}i^n\big(1+(-1)^n\big)x^n$$ Hence $$I=\frac12\sum_{n\geq0}i^n\big(1+(-1)^n\big)\int x^{2k+n+1}\mathrm{d}x$$ $$I=\frac12\sum_{n\geq0}\frac{i^n\big(1+(-1)^n\big)}{2k+n+2}x^{2k+n+2}$$ Then we have our (pretty inefficient) result: $$\arctan^2x=\sum_{k\geq0}\frac{(-1)^k}{1+2k}\sum_{n\geq0}\frac{i^n\big(1+(-1)^n\big)}{2k+n+2}x^{2k+n+2}$$ $$\arctan^2x=\sum_{n,k\in\Bbb N}\frac{(-1)^{k+\frac{n}2}\big(1+(-1)^n\big)}{(1+2k)(2+2k+n)}x^{2k+n+2}$$

Answer 4

For $|t|<1$, we have \begin{align*} \arctan t&=\sum_{k=0}^{\infty}(-1)^{k}\frac{t^{2k+1}}{2k+1},\\ \frac{(\arctan t)^2}{2!}&=\sum_{k=0}^{\infty}(-1)^{k}\Biggl(\sum_{\ell=0}^{k}\frac{1}{2\ell+1}\Biggr)\frac{t^{2k+2}}{2k+2}\\ &=\frac{t^2}{2}-\biggl(1+\frac{1}{3}\biggr)\frac{t^4}{4} +\biggl(1+\frac{1}{3}+\frac{1}{5}\biggr)\frac{t^6}{6}-\dotsm,\\ \frac{(\arctan t)^3}{3!}&=\sum_{k=0}^{\infty}(-1)^{k}\Biggl(\sum_{\ell_2=0}^{k}\frac{1}{2\ell_2+2}\sum_{\ell_1=0}^{\ell_2} \frac{1}{2\ell_1+1}\Biggr) \frac{t^{2k+3}}{2k+3}\\ &=\frac{1}{2}\frac{t^3}{3} -\biggl[\frac{1}{2}+\frac{1}{4}\biggl(1+\frac{1}{3}\biggr)\biggr]\frac{t^5}{5} +\biggl[\frac{1}{2}+\frac{1}{4}\biggl(1+\frac{1}{3}\biggr)+\frac{1}{6}\biggl(1+\frac{1}{3}+\frac{1}{5}\biggr)\biggr]\frac{t^7}{7}-\dotsm,\\ \frac{(\arctan t)^4}{4!}&=\sum_{k=0}^{\infty}(-1)^{k}\Biggl(\sum_{\ell_{3}=0}^{k}\frac{1}{2\ell_{3}+3} \sum_{\ell_2=0}^{\ell_3}\frac{1}{2\ell_2+2}\sum_{\ell_1=0}^{\ell_2} \frac{1}{2\ell_1+1}\Biggr)\frac{t^{2k+4}}{2k+4},\\ \frac{(\arctan t)^5}{5!}&=\sum_{k=0}^{\infty}(-1)^{k}\Biggl[\sum_{\ell_{4}=0}^{k}\frac{1}{2\ell_{4}+4} \sum_{\ell_{3}=0}^{\ell_4}\frac{1}{2\ell_{3}+3} \sum_{\ell_2=0}^{\ell_3}\frac{1}{2\ell_2+2}\sum_{\ell_1=0}^{\ell_2} \frac{1}{2\ell_1+1}\Biggr]\frac{t^{2k+5}}{2k+5}, \end{align*} and, generally, \begin{equation} \begin{aligned}\label{arctan-power-series-expansion-gen} \frac{(\arctan t)^n}{n!}&=\sum_{k=0}^{\infty}(-1)^{k} \Biggl(\sum_{\ell_{n-1}=0}^{k}\frac{1}{2\ell_{n-1}+n-1} \sum_{\ell_{n-2}=0}^{\ell_{n-1}}\frac{1}{2\ell_{n-2}+n-2}\dotsm \sum_{\ell_2=0}^{\ell_3}\frac{1}{2\ell_2+2}\sum_{\ell_1=0}^{\ell_2} \frac{1}{2\ell_1+1}\Biggr)\frac{t^{2k+n}}{2k+n}\\ &=\sum_{k=0}^{\infty}(-1)^{k}\Biggl(\prod_{m=1}^{n-1}\sum_{\ell_m=0}^{\ell_{m+1}} \frac{1}{2\ell_{m}+m}\Biggr)\frac{t^{2k+n}}{2k+n} \end{aligned} \end{equation} for all $n\in\mathbb{N}$ with $\ell_n=k$, where the product is understood to be $1$ if the starting index exceeds the finishing index.

References

1. B.-N. Guo, D. Lim, and F. Qi, Maclaurin's series expansions for positive integer powers of inverse (hyperbolic) sine and tangent functions, closed-form formula of specific partial Bell polynomials, and series representation of generalized logsine function, Appl. Anal. Discrete Math. 16 (2022), in press; available online at https://doi.org/10.2298/AADM210401017G.
2. M. Milgram, A new series expansion for integral powers of arctangent, Integral Transforms Spec. Funct. 17 (2006), no. 7, 531--538; available online at https://doi.org/10.1080/10652460500422486 or https://arxiv.org/abs/math/0406337.