Show that $\int_0^\infty \frac{x\log(1+x^2)}{e^{2\pi x}+1}dx=\frac{19}{24} - \frac{23}{24}\log 2 - \frac12\log A$

Contour Integration

Integrate $\frac{z\log{z}}{e^{2\pi z}+1}$ along a rectangular contour with a quarter-circle indent around $0$ and with vertices $0$, $R$, $R+i$ and $i$. Taking limits, one can check that the contributions from the line segment $[R,R+i]$ and the indent tend to zero. By the Residue Theorem, $$\operatorname{Re}\pi i \operatorname*{Res}_{z=\frac{i}{2}}\frac{z\log{z}}{e^{2\pi i z}+1}=\operatorname{Re}\left\{\int^\infty_0\frac{x\log{x}-(x+i)\log(x+i)}{e^{2\pi x}+1}\ dx+PV\int^1_0\frac{y\log(iy)e^{-\pi iy}}{2\cos(\pi y)}dy\right\}$$ Simplifying this, \begin{align} -\frac{\log{2}}{4} &=\color{red}{\int^\infty_0\frac{x\log{x}}{e^{2\pi x}+1}\ dx}-\frac{1}{2}\int^\infty_0\frac{x\log(1+x^2)}{e^{2\pi x}+1}\ dx+\color{blue}{\int^\infty_0\frac{\arctan\left(\frac{1}{x}\right)}{e^{2\pi x}+1}\ dx}\\ &\ \ \ \ \ +\color{green}{\frac{1}{2}\int^1_0x\log{x}\ dx}+\color{purple}{\frac{\pi}{4}PV\int^1_0x\tan(\pi x)\ dx} \end{align} Let us evaluate these integrals individually.

The Red Integral

We have \begin{align} \int^\infty_0\frac{x\log{x}}{e^{2\pi x}+1}\ dx &=\int^\infty_0\frac{x\log{x}}{e^{2\pi x}-1}\ dx-2\int^\infty_0\frac{x\log{x}}{e^{4\pi x}-1}\ dx\\ &=\frac{1}{4}\int^\infty_0\frac{x\log\left(\frac{x}{2}\right)}{e^{\pi x}-1}\ dx-\frac{1}{8}\int^\infty_0\frac{x\log\left(\frac{x}{4}\right)}{e^{\pi x}-1}\ dx\\ &=\frac{1}{8}\int^\infty_0\frac{x\log{x}}{e^{\pi x}-1}\ dx\\ &=\frac{1}{8}\left.\frac{d}{ds}\int^\infty_0\frac{x^{s-1}}{e^{\pi x}-1}\ dx\right|_{s=2}\\ &=\frac{1}{8}\left.\frac{d}{ds}\sum^\infty_{n=0}\int^\infty_0 x^{s-1}e^{-\pi(n+1)x}\ dx\right|_{s=2}\\ &=\frac{1}{8}\left.\frac{d}{ds}\sum^\infty_{n=0}\frac{\Gamma(s)}{\pi^{s}(n+1)^{s}}\right|_{s=2}\\ &=\frac{1}{8}\left.\frac{d}{ds}\pi^{-s}\Gamma(s)\zeta(s)\right|_{s=2}\\ &=\left.\frac{d}{ds}\frac{2^{s-4}\zeta(1-s)}{\cos\left(\frac{\pi s}{2}\right)}\right|_{s=2}\\ &=\left.\frac{2^{s-4}\left[-\zeta'(1-s)+\zeta(1-s)\left(\log{2}+\frac{\pi}{2}\tan\left(\frac{\pi s}{2}\right)\right)\right]}{\cos\left(\frac{\pi s}{2}\right)}\right|_{s=2}\\ &=\color{red}{\frac{\zeta'(-1)}{4}+\frac{\log{2}}{48}} \end{align}

The Blue Integral

Integrate $\frac{\log{z}}{e^{2\pi z}+1}$ along the same contour as we did for the integral in question. This yields $$\operatorname{Im}\pi i \operatorname*{Res}_{z=\frac{i}{2}}\frac{\log{z}}{e^{2\pi z}+1}=\operatorname{Im}\left\{\int^\infty_0\frac{\log{x}-\log(x+i)}{e^{2\pi x}+1}\ dx-i\ PV\int^1_0\frac{\log(iy)e^{-\pi iy}}{2\cos(\pi y)}dy\right\}$$ Simplifying, \begin{align} \int^\infty_0\frac{\arctan\left(\frac1x\right)}{e^{2\pi x}+1}\ dx &=-\frac{\log 2}{2}-\frac{1}{2}\int^1_0\log x\ dx-\frac{\pi}{4}PV\int^1_0\tan(\pi x)\ dx\\ &=-\frac{\log 2}{2}-\frac{1}{2}\left[x\log{x}-x\right]^1_0-\frac{\pi}{4}(0)\\ &=\color{blue}{-\frac{\log 2}{2}+\frac{1}{2}} \end{align}

The Green Integral

This is elementary. Integrate by parts once to get $$\frac{1}{2}\int^1_0x\log{x}\ dx=\frac{1}{2}\left[\frac{x^2}{2}\log{x}-\frac{x^2}{4}\right]^1_0=\color{green}{-\frac{1}{8}}$$

The Purple Integral

Splitting the limits of integration, \begin{align} \frac{\pi}{4}PV\int^1_0x\tan(\pi x)\ dx &=\frac{\pi}{4}\lim_{\epsilon\to0}\left[\left(\int^{\frac{1}{2}-\epsilon}_0+\int^1_{\frac{1}{2}+\epsilon}\right)x\tan(\pi x)\ dx\right]\\ &=\frac{\pi}{4}\lim_{\epsilon\to0}\left[\int^{\frac{1}{2}-\epsilon}_0x\tan(\pi x)\ dx-\int^{\frac{1}{2}-\epsilon}_0(1-x)\tan(\pi x)\ dx\right]\\ &=\frac{\pi}{4}\int^\frac{1}{2}_0(2x-1)\tan(\pi x)\ dx\\ &=\frac{1}{2}\int^\frac{1}{2}_0\ln(\cos(\pi x))\ dx\\ &=\color{purple}{-\frac{\log{2}}{4}} \end{align} where the last step follows from a classic result.

The Final Result

Therefore, we conclude that \begin{align} \int^\infty_0\frac{x\log(1+x^2)}{e^{2\pi x}+1}\ dx &=2\left(\frac{\log{2}}{4}+\color{red}{\frac{\zeta'(-1)}{4}+\frac{\log 2}{48}}\color{blue}{-\frac{\log{2}}{2}+\frac12}\color{green}{-\frac{1}{8}}\color{purple}{-\frac{\log{2}}{4}}\right)\\ &=\frac{\zeta'(-1)}{2}-\frac{23}{24}\log{2}+\frac{3}{4}\approx0.00302338011316028053\cdots \end{align}

$\textbf{Note:}$ As pointed out by @Sangchul Lee, there was originally a discrepancy between the value in my answer and that provided in the question due to a computational fault in WolframAlpha / Mathematica. The value given in the question has now been corrected.

The following approach uses the Abel-Plana formula, which was mentioned in a comment by the user tired.

By applying the Abel-Plana formula to $f(\frac{x}{2})$ and then subtracting the result from the Abel-Plana formula applied to $f(x)$, we get $$\begin{align} &\sum_{n=0}^{\infty} f(n) - \sum_{n=0}^{\infty} f \left(\frac{n}{2} \right) \\ &= \int_{0}^{\infty} f(x) \, dx - \int_{0}^{\infty} f \left(\frac{x}{2} \right) \, dx + i \int_{0}^{\infty} \frac{f(ix) - f(-ix)}{e^{2 \pi x}-1} \, dx - i \int_{0}^{\infty} \frac{f\left(i \frac{x}{2}\right) - f\left(-i\frac{x}{2}\right)}{e^{2 \pi x}-1} \, dx \\ &= \int_{0}^{\infty} f(x) \, dx - 2 \int_{0}^{\infty} f(u) \, du + i \int_{0}^{\infty} \frac{f(ix) - f(-ix)}{e^{2 \pi x}-1} \, dx - 2 i \int_{0}^{\infty} \frac{f(iu) - f(-iu)}{e^{4 \pi u}-1} \, du \\ &= -\int_{0}^{\infty} f(x) \, dx + i \int_{0}^{\infty} \frac{f(ix) - f(-ix)}{e^{2 \pi x}+1} \, dx. \end{align} $$

Let's apply the above formula to the function $$ f(x)= \frac{1}{(1+x)^{s}} , \quad \text{Re}(s) >1.$$

Doing so, we get $$\sum_{n=0}^{\infty} \frac{1}{(1+n)^{s}} - 2^{s} \sum_{n=0}^{\infty}\frac{1}{(2+n)^{s}} = \frac{1}{1-s} + 2 \int_{0}^{\infty} \frac{\sin \left(s \arctan x \right)}{(1+x^{2})^{s/2}(e^{2 \pi x}+1)} \, dx,$$ which implies that

$$2 \int_{0}^{\infty} \frac{\sin \left(s \arctan x \right)}{(1+x^{2})^{s/2}(e^{2 \pi x}+1)} \, dx = (1-2^{s}) \zeta(s) + 2^{s} + \frac{1}{s-1}. \tag{1}$$

By analytic continuation, $(1)$ should hold for all complex values of $s$. (For $s=1$, the right side of the equation should be treated as a limit.)

Now if we differentiate under the integral sign and then let $s=-1$, we get

$$\int_{0}^{\infty} \frac{x \log(1+x^{2})}{e^{2 \pi x}+1} \, dx + 2 \int_{0}^{\infty} \frac{\arctan x}{e^{2 \pi x}+1} \, dx = - \frac{\log 2}{2} \underbrace{\zeta(-1)}_{- \frac{1}{12}}+\frac{\zeta'(-1)}{2} + \frac{\log 2}{2}-\frac{1}{4}. $$

But using Binet's second formula for the log gamma function, we have $$ \begin{align} 2 \int_{0}^{\infty} \frac{\arctan x}{e^{2 \pi x}+1} \, dx &= 2 \int_{0}^{\infty} \frac{\arctan x}{e^{2 \pi x}-1} \, dx - 4 \int_{0}^{\infty} \frac{\arctan x}{e^{4 \pi x}-1} \, dx \\ &=2 \int_{0}^{\infty} \frac{\arctan x}{e^{2 \pi x}-1} \, dx -2 \int_{0}^{\infty} \frac{\arctan \left(\frac{u}{2}\right)}{e^{2 \pi u}-1} \, du \\ &= 1- \frac{\log (2 \pi)}{2} + \frac{3}{2} \log 2 -2 + \frac{\log(2 \pi)}{2} \\ &= \frac{3 \log 2}{2} -1 .\end{align}$$

Therefore, $$ \begin{align}\int_{0}^{\infty} \frac{x \log(1+x^{2})}{e^{2 \pi x}+1} \, dx &= \frac{\log 2}{24} + \frac{\zeta'(-1)}{2} + \frac{\log 2}{2} - \frac{1}{4} + 1 - \frac{3 \log 2}{2} \\ &= \frac{\zeta'(-1)}{2} - \frac{23}{24} \log 2 + \frac{3}{4}. \end{align}$$

Here is another method which is a mixture of both real-analysis technique and complex-analysis technique (without contour integration). The main idea is similar to my previous answer.

Let $I$ denote the integral and we write

$$ I = \sum_{n=1}^{\infty} (-1)^{n-1} \int_{0}^{\infty} x \log(1+x^2) e^{-2\pi n x} \, dx. $$

In order to proceed, we claim the following:

$$ \int_{0}^{\infty} x \log(1+x^2)e^{-sx} \, dx = \frac{2}{s^2} - \frac{2}{s} \int_{0}^{\infty} \frac{\sin(st)}{t+1} \, dt + \frac{2}{s^2} \int_{0}^{\infty} \frac{\cos(st)}{t+1} \, dt. \tag{1} $$

We postpone the proof to the end and discuss the consequence of $\text{(1)}$. Substituting $s = 2\pi n$ and plugging back, $I$ can be simplified as

\begin{align*} I &= \sum_{n=1}^{\infty} (-1)^{n-1} \left( \frac{1}{2\pi^2 n^2} - \frac{1}{\pi n} \int_{0}^{\infty} \frac{\sin(2\pi n t)}{t+1} \, dt + \frac{1}{2\pi^2 n^2} \int_{0}^{\infty} \frac{\cos(2\pi n t)}{t+1} \, dt \right) \\ &= \frac{1}{24} - \int_{0}^{\infty} \left( \sum_{n=1}^{\infty} (-1)^{n-1} \frac{\sin(2\pi n t)}{\pi n} \right) \, \frac{dt}{t+1} + \int_{0}^{\infty} \left( \sum_{n=1}^{\infty}(-1)^{n-1} \frac{\cos(2\pi n t)}{2\pi^2 n^2} \right) \, \frac{dt}{t+1} \end{align*}

Then utilizing the Fourier series of the periodic Bernoulli polynomials $\tilde{B}_n$

$$ \sum_{n=1}^{\infty} \frac{\sin(2\pi n x)}{\pi n} = -\tilde{B}_1(x), \qquad \sum_{n=1}^{\infty} \frac{\cos(2\pi n x)}{\pi^2 n^2} = \tilde{B}_2(x), $$

it follows that

$$ I = \frac{1}{24} - \int_{0}^{\infty} \frac{\tilde{B}_1(t + \frac{1}{2})}{t+1} \, dt - \frac{1}{2}\int_{0}^{\infty} \frac{\tilde{B}_2(t + \frac{1}{2})}{t+1} \, dt. $$

Next, we introduce the following function

$$f(s) = 2^s - (2^s - 1)\zeta(s) + \frac{1}{s-1}.$$

This is an entire function on $\Bbb{C}$ since all the poles are cancelled out. Then we claim that for $\Re(s) > 0$,

\begin{align*} \int_{0}^{\infty} \frac{\tilde{B}_1(t + \frac{1}{2})}{(t+1)^s} \, dt &= \frac{f(s-1)}{s-1}, \tag{2} \\ \int_{0}^{\infty} \frac{\tilde{B}_2(t + \frac{1}{2})}{(t+1)^s} \, dt &= \frac{2f(s-2)}{(s-1)(s-2)} - \frac{1}{12}\frac{1}{s-1}. \tag{3} \end{align*}

We also postpone the proof to the end. Assuming this, we have

\begin{align*} \int_{0}^{\infty} \frac{\tilde{B}_1(t + \frac{1}{2})}{t+1} \, dt &= \lim_{s\to 1} \frac{f(s-1)}{s-1} = \frac{3}{2}\log 2 - 1, \\ \int_{0}^{\infty} \frac{\tilde{B}_2(t + \frac{1}{2})}{t+1} \, dt &= \lim_{s\to 1} \left(\frac{2f(s-2)}{(s-1)(s-2)} - \frac{1}{12}\frac{1}{s-1}\right) = \frac{7}{12} - \frac{13}{12}\log 2 - \zeta'(-1). \end{align*}

Plugging this back yields the same answer as M.N.C.E.'s:

$$ I = \frac{3}{4} - \frac{23}{24}\log 2 + \frac{1}{2}\zeta'(-1). $$

• Proof of $\text{(1)}$. By the integration by parts, for $s > 0$ we have

  \begin{align*} \int_{0}^{\infty} x \log(1+x^2)e^{-sx} \, dx &= \int_{0}^{\infty} \frac{2x(sx+1)}{s^2(x^2+1)} e^{-sx} \, dx \\ &= \int_{0}^{\infty} \frac{2u(u+1)}{s^2(u^2 + s^2)}e^{-u} \, du \qquad (u = sx) \\ &= \int_{0}^{\infty} \left( \frac{2}{s^2} - \frac{2}{u^2+s^2} + \frac{2u}{s^2(u^2 + s^2)} \right) e^{-u} \, du. \end{align*}

  In order to compute the last integral, we notice that

  $$ \int_{0}^{\infty} \frac{e^{-u}}{u - is} \, du = \int_{0}^{\infty}\int_{0}^{\infty} e^{ist} e^{-ut} e^{-u} \, dt du = \int_{0}^{\infty} \frac{e^{ist}}{t+1} \, dt. $$

  In fact, this is more of a heuristics than a rigorous computation, since the Fubini's theorem does not apply directly. One can regularize both sides by replacing $s$ by $s - i\epsilon$ for $\epsilon > 0$, applying Fubini's theorem to prove the corresponding equailty, and then letting $\epsilon \to 0^+$ to establish this. (Alternatively, this can be also thought as a $\frac{\pi}{2}$-rotation of the contour.) Then the claim follows by utilizing this equality. ////

• Proof of $\text{(2)}$ and $\text{(3)}$. We first prove the following identity

  $$ \int_{0}^{\infty} \frac{1}{(t+1)^s} \, d\tilde{B}_1(t+\tfrac{1}{2}) = f(s), \qquad \Re(s) > 0. $$

  By integration by parts, it is not hard to check that the left-hand side defines a holomorphic function for $\Re(s) > 0$. In view of the principle of analytic continuation, it suffices to prove the identity for $\Re(s) > 1$. Then

  \begin{align*} \int_{0}^{\infty} \frac{1}{(t+1)^s} \, d\tilde{B}_1(t+\tfrac{1}{2}) &= \int_{0}^{\infty} \frac{1}{(t+1)^s} \, d(t - \lfloor t+\tfrac{1}{2}\rfloor) \\ &= \frac{1}{s-1} - \sum_{k=0}^{\infty} \frac{1}{(k+\frac{3}{2})^s}. \end{align*}

  It is easy to check that this coincides with $f(s)$. Now both $\text{(2)}$ and $\text{(3)}$ follows easily. Again, both define holomorphic functions on $\Re(s) > 0$ and the principle of analytic continuation allows us to prove both identities only when $\Re(s)$ is large. Then the claim follows from the following identity:

  \begin{align*} \int_{0}^{\infty} \frac{\tilde{B}_n(t + \frac{1}{2})}{(t+1)^s} \, dt &= \frac{B_n(\frac{1}{2})}{s-1} + \frac{1}{s-1} \int_{0}^{\infty} \frac{1}{(t+1)^{s-1}} \, d\tilde{B}_n(t + \tfrac{1}{2}) \tag{$n \geq 1$} \\ &= \frac{B_n(\frac{1}{2})}{s-1} + \frac{n}{s-1} \int_{0}^{\infty} \frac{\tilde{B}_{n-1}(t + \frac{1}{2})}{(t+1)^{s-1}} \, dt \tag{$n \geq 2$} \end{align*}