I observed that: $$\int_{0}^{\infty}\dfrac{x^n}{1+x^{2n+2}}\arctan(x) dx=\dfrac{\pi}{4}\int_{0}^{\infty}{\dfrac{x^n}{1+x^{2n+2}} }dx=\dfrac{\pi^2}{4\cdot\left(2n+2\right)}$$
The result is a generalization of my previous post:Proving $\int_{0}^{\infty}{\frac{x}{1+x^4}\arctan(x)}\,dx=\frac{\pi^2}{16}$, in which one of the integrals belonging to this family was evaluated.
Let, $n=0.5$:
This gives us, according to the generalization:$$\int_{0}^{\infty}{\dfrac{\sqrt{x}}{1+x^3}\arctan(x)}dx=\dfrac{\pi}{4}\int_{0}^{\infty}\dfrac{\sqrt{x}}{1+x^3} dx=\dfrac{\pi^2}{12}$$
Therefore:$$\dfrac{\pi^2}{12}=\int_{0}^{\infty}{\dfrac{\sqrt{x}}{1+x^3}\arctan(x)}dx$$
Similarly, $$\dfrac{\pi^2}{20}=\int_{0}^{\infty}{\dfrac{x\sqrt{x}}{1+x^5}\arctan(x)}dx$$
$$\dfrac{\pi^2}{28}=\int_{0}^{\infty}{\dfrac{x^2\sqrt{x}}{1+x^7}\arctan(x)}dx$$And so on…
The values of the RHS evaluated numerically seem to equal the LHS values.
My question is if this identity has been observed before and if it could also be extended to complex values of $n$.
Best Answer
I prove the identity for all $n \in \mathbb{C}$, so instead of $n$ I use $z = \sigma + it$ I divide the proof in three steps :
Step 1
I show that the integral is well defined for $\sigma > 0$ (it should work also for $\sigma > -1$ but for simplicity I assume $\sigma > 0$)
For (say) $x > 2$ I have
$|\frac{x^z}{1+x^{2z+2}}{\arctan(x)}| \leq \frac{\pi}{2}\frac{|x|^\sigma}{|x|^{2\sigma+2}-1} \leq \pi\frac{1}{|x|^{2+\sigma}}$ which is integrable on $(2,\infty)$
while the function is continuos in $[0,2]$ So it's absolutely integrable in $(0,\infty)$
observe that in order for $x^{2+2z} = -1$ for some $x > 0$ one would need $\Re(2+2z) = 0$ which is $\sigma = -1$
Step 2
I show the first Identity
the change of variable $y = 1/x$ yields
$\int_{0}^{\infty}{\frac{x^z}{1+x^{2z+2}}\arctan(x)dx} = \int_{0}^{\infty}{\frac{y^{-z}}{1+y^{-2z-2}}\frac{1}{y^2}\arctan(1/y)dy} = \int_{0}^{\infty}{\frac{y^{z}}{y^{2z+2} + 1}\arctan(1/y)dy}$
Therefore
$2\int_{0}^{\infty}{\frac{x^z}{1+x^{2z+2}}\arctan(x)dx} =\int_{0}^{\infty}{\frac{x^z}{1+x^{2z+2}}\arctan(x)dx} + \int_{0}^{\infty}{\frac{x^z}{1+x^{2z+2}}\arctan(1/x)dx} = \int_{0}^{\infty}{\frac{x^z}{1+x^{2z+2}}(\arctan(x)+\arctan(1/x))dx} = \frac{\pi}{2}\int_{0}^{\infty}{\frac{x^z}{1+x^{2z+2}}dx}$
divide by $2$ to find
$\int_{0}^{\infty}{\frac{x^z}{1+x^{2z+2}}\arctan(x)dx} = \frac{\pi}{4}{ \int_{0}^{\infty}{\frac{x^z}{1+x^{2z+2}}dx} }$
Step 3
Using once again $y = 1/x$ I find
$\int_{1}^{\infty}{ \frac{x^z}{1+x^{2z+2}} dx} = \int_{0}^{1}{\frac{x^z}{1+x^{2z+2}}dx} $
so that
$\frac{\pi}{4}{\int_{0}^{\infty}{ \frac{x^z}{1+x^{2z+2}} dx }} = \frac{\pi}{2}{\int_{0}^1{ \frac{x^z}{1+x^{2z+2}} dx }}$
and now since the integration interval is compact and for any $0 < x < 1$ the function $z \mapsto \frac{x^z}{1+x^{2z+2}}$ is holomorphic one can easily show that the function
$I(z) := {\int_{0}^{1}{ \frac{x^z}{1+x^{2z+2}} dx }}$ is an holomorphic function
Step 4
now suppose $z = t > 0$, then I use $y = x^{t+1}$ to find
$I(t) = \frac{1}{t+1}{\int_{0}^{1}{ \frac{1}{1+y^2} dy }} = \frac{1}{t+1}\frac{\pi}{4}$
Final step
Let now $h(z) := \frac{\pi}{4}\frac{1}{z+1}$, it is an holomorphic function and I have showed that $I(z) = h(z)$ forall $z > 0$, so by the Identity theorem $I(z) = h(z)$
therefore
$\int_{0}^{\infty}{\frac{x^z}{1+x^{2z}} \arctan(x)dx} = \frac{\pi}{2}I(z) = \frac{\pi}{2}\frac{\pi}{4}\frac{1}{z+1} = \frac{\pi^2}{4\cdot(2z+2)}$
this holds for all $\Re(z) > -1$