Ytukyg

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
$$frac{df}{dx}=2arctan(x)frac{1}{(x^2+1)}$$ I am not sure If I can pick out parts of the product (like $arctan(x)$) and derivate, 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.

Using the Cauchy Product Formula,
$$
begin{align}
frac{mathrm{d}}{mathrm{d}x}arctan^2(x)
&=2frac{arctan(x)}{1+x^2}\
&=2left(x-frac{x^3}3+frac{x^5}5-frac{x^7}7+cdotsright)left(1-x^2+x^4-x^6+cdotsright)\
&=2left((1)x-left(1+tfrac13right)x^3+left(1+tfrac13+tfrac15right)x^5-left(1+tfrac13+tfrac15+tfrac17right)x^7+cdotsright)\
&=2sum_{k=1}^infty(-1)^{k-1}left(H_{2k}-tfrac12H_kright)x^{2k-1}
end{align}
$$
Therefore,
$$
arctan^2(x)=sum_{k=1}^inftyfrac{(-1)^{k-1}}kleft(H_{2k}-tfrac12H_kright)x^{2k}
$$

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:

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

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.

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

For $|x|<1$, we have
$$arctan x=sum_{kgeq0}frac{(-1)^kx^{1+2k}}{1+2k}$$
As you noted,
$$arctan^2x=2intfrac{arctan x}{x^2+1}mathrm{d}x$$
So, assuming $|x|<1$,
$$arctan^2x=2intfrac1{x^2+1}sum_{kgeq0}frac{(-1)^kx^{1+2k}}{1+2k}mathrm{d}x$$
$$arctan^2x=2sum_{kgeq0}frac{(-1)^k}{1+2k}intfrac{x^{2k}}{1+x^2}xmathrm{d}x$$
Now we focus on
$$I=intfrac{x^{2k+1}}{1+x^2}mathrm{d}x$$
We recall that for $|x|<1$,
$$frac1{1+x^2}=frac12sum_{ngeq0}i^nbig(1+(-1)^nbig)x^n$$
Hence
$$I=frac12sum_{ngeq0}i^nbig(1+(-1)^nbig)int x^{2k+n+1}mathrm{d}x$$
$$I=frac12sum_{ngeq0}frac{i^nbig(1+(-1)^nbig)}{2k+n+2}x^{2k+n+2}$$
Then we have our (pretty inefficient) result:
$$arctan^2x=sum_{kgeq0}frac{(-1)^k}{1+2k}sum_{ngeq0}frac{i^nbig(1+(-1)^nbig)}{2k+n+2}x^{2k+n+2}$$
$$arctan^2x=sum_{n,kinBbb N}frac{(-1)^{k+frac{n}2}big(1+(-1)^nbig)}{(1+2k)(2+2k+n)}x^{2k+n+2}$$