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I have been recently looking at integrals of the following form: $$\int_0^{1} \frac{\arctan^n{ \left( x \right)}}{1+x}dx, n \in \mathbb{N}$$ Some interesting patterns seem to be emerging for the cases I have calculated: $$\int_0^{1} \frac{\arctan{ \left( x \right)}}{1+x}dx = \frac{\pi}{8} \log \left( 2 \right)$$ $$\int_0^{1} \frac{\arctan^2{ \left( x \right)}}{1+x}dx = \frac{\pi}{4} G + \frac{\pi^2}{32} \log \left( 2 \right) - \frac{21}{32} \zeta \left( 3 \right)$$ $$\int_0^{1} \frac{\arctan^3{ \left( x \right)}}{1+x}dx = \frac{3 \pi^2}{38} G + \frac{\pi^3}{128} \log \left( 2 \right) - \frac{63 \pi}{256} \zeta \left( 3 \right)$$ $$\int_0^{1} \frac{\arctan^4{ \left( x \right)}}{1+x}dx = \frac{\pi^3}{32} G + \frac{\pi^4}{512} \log \left( 2 \right) - \frac{63 \pi^2}{512} \zeta \left( 3 \right) + \frac{1395}{512} \zeta \left( 5 \right) + \frac{ \pi}{2048} \left( \psi^{(3)} \left( \frac{3}{4} \right) - \psi^{(3)} \left( \frac{1}{4} \right) \right)$$ The case for $n = 4$ is a bit more complex than the previous cases, but I still think there could be a general formula. Note that when calculating specific cases Wolfram Alpha refuses to give closed-form answers to these integrals directly. Also, for odd $n > 1$ one can perform the variable change $$x = \frac{1-u}{1+u}$$ to rewrite the integral as $$\int_0^{1} \frac{ \left( \frac{\pi}{4} - \arctan{ \left( u \right)}\right)^n}{1+u}du$$ Expanding out the numerator gives some variation of each previous integral down to $n=0$ which can be immediately substituted with those values to find the current case. For both even cases above I integrated by parts once and was left with a number minus an integral of this form $$\int_0^{1} \frac{\arctan^{n-1} \left( x \right) \log \left( 1+x \right)}{1+x^2}dx$$ which can be evaluated with the substitution $x= \tan \left( \theta \right)$ along with the identity $1 + \tan \left( \theta \right) = \sqrt{2} \sec \left( \theta \right) \sin \left( \theta + \frac{\pi}{4} \right)$ and Fourier series.

So my question is mainly whether there is a closed form of each integral for all $n \in \mathbb{N}$, and whether it can be generalized to a single formula.

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  • $\begingroup$ Does an $n$th derivative expression satisfy the question? $\endgroup$ Oct 19, 2023 at 15:50

1 Answer 1

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The 7th power yields

 Subtract @@ 
  Integrate[ArcTan[x]^7/(1 + x), x] /. {{x -> 1}, {x -> 0}} // 
   FullSimplify // TraditionalForm

$$\begin{align}&\frac{4 \pi \left(-40325040 \zeta (7)+\pi \left(21 \left(\psi ^{(5)}\left(\frac{1}{4}\right)-\psi ^{(5)}\left(\frac{3}{4}\right)\right)+4 \pi \left(448 \pi ^3 C-3528 \pi ^2 \zeta (3)+390600 \zeta (5)+2159 i \pi ^5+16 \pi ^4 \log (2)+35 \pi \left(\psi ^{(3)}\left(\frac{3}{4}\right)-\psi ^{(3)}\left(\frac{1}{4}\right)\right)\right)\right)\right)-\psi ^{(7)}\left(\frac{1}{4}\right)+\psi ^{(7)}\left(\frac{3}{4}\right)}{8388608}-\\& \frac{315 i \text{Li}_8(i)}{8}-\frac{127 i \pi ^8}{30720}\end{align}$$

$$\int_0^1 \frac{{\tan^{(-1)}}^n(x)}{1+x^2} dx$$ is much more simple.

Perhaps on can steer the integration by parts though a manageable tree

 Integrate[ArcTan[x]^7/(1 + x), x] == 
    Log[1 + x] ArcTan[x]^6 - Integrate[ 
        Log[1+x] 6 ArcTan[x]^5*1/(1 + x^ 2),x]

since then

  Integrate[Log[x]/(1 + x^2), x]//Simplify//ExpToTrig

yields

$$-\frac{1}{2} i \text{Li}_2(-i x)+\frac{1}{2} i \text{Li}_2(i x)+\log (x) \tan ^{-1}(x)$$

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