What is wrong with this Taylor series $f(x)=\int_0^x\frac{t \arctan^2 t}{1+t^2}\ dt $

Question

We have $$f(x)=\int_0^x\frac{t \arctan^2 t}{1+t^2}\ dt $$ and I'm trying to evaluate the first terms of its Taylor series at zero. Wolfram output shows: $$ f(x)= \frac{3\zeta(3)}{8}+\frac{x^4}{4}+o(x^4)$$ I can't see how to get that $\zeta(3)$ factor, since the first derivative is $$f'(x)=\frac{x \arctan^2 x}{1+x^2}$$ which is $0$ when $x=0$. And so is the second derivative $$f''(x)=\arctan x\frac{2x+\arctan x-x^2\arctan x}{(1+x^2)^2} $$ The term of $\frac{3\zeta(3)}{8}$ should be the linear term of the expansion right? But it is $0$ so how can this be?

Answer 1

I'm trying to evaluate the first terms of its Taylor series at zero, which should be $$ f(x)= \frac{3\zeta(3)}{8}+\frac{x^4}{4}+o(x^4)$$

Why do you think this is the right form? If you plug in $x=0$, you get

$$ f(0)= \frac{3\zeta(3)}{8}$$

But contradicts with the fact

$$f(0)=\int_0^0\frac{t \arctan^2 t}{1+t^2}\ dt=0 $$

So your Taylor series is wrong.

Regarding the Wolfram output, note at the bottom it has the polylog term,

$$-\frac{1}2\text{Li}_3\left(\frac{i+x}{-i+x}\right)$$

If you set $x=0$, it becomes

$$-\frac{1}2\text{Li}_3(-1)=\frac{3}8\zeta(3)$$

which cancels your constant term.

Answer 2

If you use the complex representation of the arctangent function $$\tan^{-1}(t)=\frac{i}{2} \log{\left ( \frac{i+t}{i-t}\right )}$$ the antiderivative is not too difficult.

Expanding the result around $x=0$

$$f(x)=\frac{x^4}{4}-\frac{5 x^6}{18}+\frac{49 x^8}{180}-\frac{409 x^{10}}{1575}+O\left(x^{12}\right)$$ which you could have obtained directly by the Taylor expansion of the integrand $$\frac{t}{1+t^2} \big[\tan ^{-1}(t)\big]^2=\sum_{n=1}^\infty a_n\, t^{2n+1}$$ the first coefficients being $$\left\{1,-\frac{5}{3},\frac{98}{45},-\frac{818}{315},\frac{517}{1 75},-\frac{169819}{51975}\right\}$$