Variation on Binet´s second formula

In this post we will prove the following result


\int_0^{\infty}\frac{x^2\arctan \left(x\right)}{e^{2 \pi x}-1}\,dx=-\frac{11}{36}+\frac{\ln(2 \pi)}{4}+\zeta^\prime(-1)+\frac{\zeta(3)}{8 \pi^{2}}


We start by recalling Hermite´s integral representation of the Hurwitz zeta function (proved here)


\begin{align} \zeta(s,u)=\frac{u^{-s}}{2}+\frac{u^{1-s}}{s-1}+2\int_0^{\infty}\frac{\sin\left( s \arctan \left( \frac{x}{u}\right)\right)}{\left( u^2+x^2\right)^{s/2}\left(e^{2 \pi x}-1\right)}\,dx \notag \end{align}(1)



We may rewrite it in a slightly different way which facilitates some calculations, to this end we start with the following lemma:

Lemma:

\begin{align} \frac{1}{(u+ix)^s}-\frac{1}{(u-ix)^s}=\frac{2 \sin\left( s \arctan \left( \frac{x}{u}\right)\right)}{i\left( u^2+x^2\right)^{s/2}} \notag \end{align}

Proof:

\begin{align} \frac{1}{(u+ix)^s}-\frac{1}{(u-ix)^s}&=\frac{(u-ix)^s-(u+ix)^s}{(u+ix)^s(u-ix)^s} \notag\\ &=\frac{(re^{-i \theta})^s-(re^{i \theta})^s}{(u^2+x^2)^s} \notag\\ &=\frac{-r^s(e^{i s \theta}-e^{-i s \theta})}{(u^2+x^2)^s} \notag\\ &=\frac{-2ir^s(\sin(s \theta))}{(u^2+x^2)^s} \notag\\ &=\frac{2r^s(\sin(s \theta))}{i(u^2+x^2)^s} \notag\\ &=\frac{2(u^2+x^2)^{s/2}(\sin(s \tan^{-1}(u/x)))}{i(u^2+x^2)^s} \notag\\ &=\frac{2 \sin\left( s \arctan \left( \frac{x}{u}\right)\right)}{i\left( u^2+x^2\right)^{s/2}} \notag \end{align}

We can therefore rewrite (1) as


\begin{align} \zeta(s,u)=\frac{u^{-s}}{2}+\frac{u^{1-s}}{s-1}+i\int_0^{\infty}\frac{\left(u+ix \right)^{-s}-\left(u-ix \right)^{-s}}{e^{2 \pi x}-1}\,dx \notag \end{align}(2)


Differentiating (2) w.r. to s we obtain


\begin{align} \zeta^{\prime}(s,u)&=-\frac{u^{-s}\ln(u)}{2}-\frac{u^{1-s}\ln(u)}{s-1}-\frac{u^{1-s}}{(s-1)^2}+ \notag\\& \qquad \qquad \qquad +i\int_0^{\infty}\frac{\left(u-ix \right)^{-s}\ln(u-ix)-\left(u+ix \right)^{-s}\ln(u+ix)}{e^{2 \pi x}-1}\,dx \notag \end{align}(3)

Letting s=-2 and using the logarithmic representation of arctan


\begin{align} \arctan{z} = \frac{i}{2} \log{\left ( \frac{i+z}{i-z}\right )} \notag \end{align}

we obtain


\begin{align*} \zeta^{\prime}(-2,u)&=-\frac{u^{2}\ln(u)}{2}+\frac{u^{3}\ln(u)}{3}-\frac{u^{3}}{9}+ \notag\\& \qquad \qquad \qquad +i\int_0^{\infty}\frac{\left(u-ix \right)^{2}\ln(u-ix)-\left(u+ix \right)^{2}\ln(u+ix)}{e^{2 \pi x}-1}\,dx \\ &=-\frac{u^{2}\ln(u)}{2}+\frac{u^{3}\ln(u)}{3}-\frac{u^{3}}{9}+ \notag\\& \qquad \qquad \qquad +i\int_0^{\infty}\frac{\left(u^2-x^2-2iux \right)\ln(u-ix)-\left(u^2-x^2+2iux \right)\ln(u+ix)}{e^{2 \pi x}-1}\,dx \\ &=-\frac{u^{2}\ln(u)}{2}+\frac{u^{3}\ln(u)}{3}-\frac{u^{3}}{9}+ \notag\\& \qquad \qquad \qquad +i\int_0^{\infty}\frac{\left(u^2-x^2 \right)\left[\ln(u-ix)-\ln(u+ix)\right]-2iux \ln(u+ix)\ln(u-ix)}{e^{2 \pi x}-1}\,dx \\ &=-\frac{u^{2}\ln(u)}{2}+\frac{u^{3}\ln(u)}{3}-\frac{u^{3}}{9}+2u^2\int_0^{\infty}\frac{\arctan \left(\frac{x}{u}\right)}{e^{2 \pi x}-1}\,dx \\& \qquad \qquad \qquad-2\int_0^{\infty}\frac{x^2\arctan \left(\frac{x}{u}\right)}{e^{2 \pi x}-1}\,dx +2u\int_0^{\infty}\frac{x \ln(u^2+x^2)}{e^{2 \pi x}-1}\,dx \\ \end{align*}


rearranging terms we get:


\begin{align*} \int_0^{\infty}\frac{x^2\arctan \left(\frac{x}{u}\right)}{e^{2 \pi x}-1}\,dx&=-\frac{u^{2}\ln(u)}{4}+\frac{u^{3}\ln(u)}{6}-\frac{u^{3}}{18}+u^2\int_0^{\infty}\frac{\arctan \left(\frac{x}{u}\right)}{e^{2 \pi x}-1}\,dx \\& \qquad \qquad \qquad +u\int_0^{\infty}\frac{x \ln(u^2+x^2)}{e^{2 \pi x}-1}\,dx - \frac{\zeta^{\prime}(-2,u)}{2}\\ \end{align*}

Now let u=1 to obtain


\begin{align*} \int_0^{\infty}\frac{x^2\arctan \left(x\right)}{e^{2 \pi x}-1}\,dx&=-\frac{1}{18}+\left(\frac12-\frac{\ln(2 \pi)}{4} \right) +\left(\zeta^\prime(-1)-\frac34+\frac{\ln(2 \pi)}{2} \right) - \frac{\zeta^{\prime}(-2,1)}{2}\\ &=-\frac{11}{36}+\frac{\ln(2 \pi)}{4}+\zeta^\prime(-1)- \frac{\zeta^{\prime}(-2)}{2}\\ &=-\frac{11}{36}+\frac{\ln(2 \pi)}{4}+\zeta^\prime(-1)+\frac{\zeta(3)}{8 \pi^{2}} \quad \blacksquare \end{align*}

Which concludes the proof.


Note: In the proof above we used the following previously established results


\int_0^\infty\frac{x \ln(1+x^2)}{e^{2\pi x}-1}\,dx=\zeta^{\prime}(-1)-\frac{3}{4}+\frac12 \ln 2 \pi

Proved here

\int_{0}^{\infty}\frac{\arctan(x)}{e^{2\pi x}-1}dx=\frac12-\frac{\ln 2 \pi}{4}

Proved here

&\zeta^{\prime}(-2)=-\frac{\zeta(3)}{4 \pi^{2}}

Proved here

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