In a (great) paper "A theorem for the closed-form evaluation of the first generalized Stieltjes constant at rational arguments and some related summations" by Iaroslav V. Blagouchine, the following integral representation of the first Stieltjes constant $\gamma_1$ is given (on page 539): $$\gamma_1=-\left[\gamma-\frac{\ln2}2\right]\ln2+i\int_0^\infty\frac{dx}{e^{\pi x}+1}\left\{\frac{\ln(1-ix)}{1-ix}-\frac{\ln(1+ix)}{1+ix}\right\}.\tag1$$ It's possible to get rid of imaginary numbers in this formula, and rewrite it in terms of only real-valued functions: $$2\int_0^\infty\frac{\arctan x}{1+x^2}\frac{dx}{e^{\pi x}+1}-\int_0^\infty\frac{x\ln(1+x^2)}{1+x^2}\frac{dx}{e^{\pi x}+1}=\gamma_1+\left[\gamma-\frac{\ln2}2\right]\ln2.\tag2$$

Is it possible to find closed forms separately for each integral on the left-hand side of $(2)$?

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    $\begingroup$ I am able to prove the two integrals at once using real method, but not separately, is it okay for you if I post it? $\endgroup$ – Anastasiya-Romanova 秀 Dec 8 '14 at 12:46
  • $\begingroup$ @Anastasiya-Romanova Because we know a value of a linear combination of two integrals, it's enough if we find a value of any of them (we would get the other one for free). Do you know a value of any of them? $\endgroup$ – Vladimir Reshetnikov Dec 8 '14 at 19:28
  • $\begingroup$ @VladimirReshetnikov As I said, I don't know. But I'am able to prove it using a real method as a combination of them, not the separated one. $\endgroup$ – Anastasiya-Romanova 秀 Dec 8 '14 at 19:39
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    $\begingroup$ The first integral is very similar to the remainder of the Stirling formula, written through the second Binet's log-gamma theorem (T). Maybe we can tackle this problem by following the same approach of the proof of (T): an integral representation for $\psi'(z)$ or some related function, followed by the Abel-Plana's formula. $\endgroup$ – Jack D'Aurizio Dec 11 '14 at 18:11

Using contour integration, I am able to evaluate the first integral in terms of a slightly different infinite series which may or may not be able to be evaluated in closed form.

First notice that

$$ \begin{align} \int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \frac{dx}{e^{\pi x}+1} &= \frac{1}{2} \int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \left(1- \tanh \left(\frac{\pi x}{2} \right) \right) \, dx \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \, dx- \frac{1}{4} \int_{-\infty}^{\infty} \frac{\arctan x}{1+x^{2}} \tanh \left(\frac{\pi x}{2} \right) \, dx \\ &= \frac{\pi^{2}}{16} + \frac{1}{4} \, \text{Im} \int_{-\infty}^{\infty} \frac{\log(1-ix)}{1+x^{2}} \tanh \left(\frac{\pi x}{2} \right) \, dx . \end{align} $$

Now let $ \displaystyle f(z) = \frac{\log(1-iz)}{1+z^{2}} \tanh \left(\frac{\pi z}{2} \right).$

Then since $ \displaystyle \int f(z) \ dz$ vanishes along the upper half of the circle $|z|=2N$ as $N \to \infty$ through the positive integers, we get

$$ \begin{align} &\int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \frac{dx}{e^{\pi x}+1} \\ &= \frac{\pi^{2}}{16} + \frac{1}{4} \, \text{Im} \ 2 \pi i \left(\text{Res}[f(z),i] + \sum_{n=1}^{\infty} \text{Res} [f(z), i(2n+1)] \right) \\ &= \frac{\pi^{2}}{16} + \frac{1}{4} \, \text{Im} \ 2 \pi i \left( \frac{\log(2)-1}{2 \pi} - \sum_{n=1}^{\infty}\frac{\log(2n+2)}{2 \pi(n^{2}+n)}\right) \\ &= \frac{\pi^{2}}{16} + \frac{\log (2)}{4} - \frac{1}{4} - \frac{\log (2)}{4} \sum_{n=1}^{\infty} \frac{1}{n(n+1)} - \frac{1}{4} \sum_{n=1}^{\infty} \frac{\log(n+1)}{n(n+1)} \\ &= \frac{\pi^{2}}{16} - \frac{1}{4} - \frac{1}{4} \sum_{n=1}^{\infty} \frac{\log(n+1)}{n(n+1)} . \end{align}$$


In this answer, Olivier Oloa talks about the coefficients of the regular part of the Laurent series expansion at the origin of the poly-Hurwitz zeta function (which he refers to as the poly-Stieltjes constants).

It turns out that we can evaluate the above infinite series in terms of the poly-Stieltjes constants.

According to Theorem 2,

$$ \sum_{n=1}^{\infty} \frac{\log(n+1)}{n(n+1)} = \gamma_{1}(1,0) - \gamma_{1}(1,1)$$ which, if I understand correctly, reduces to $$\gamma_{1}(1,0) - \gamma_{1}.$$

As a side note, Wolfram Alpha does not return a very good approximation of the value of that infinite series. And if you ask for more digits, it will return a different result.

But it appears that the value of the infinite series is approximately $ 1.2577468869$.

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    $\begingroup$ I have a doubt that these integrals can't be evaluated separately, I mean each integral most likely has no closed-form. I hope I am wrong & there's a user who can provide the closed-form. $\endgroup$ – Anastasiya-Romanova 秀 Dec 12 '14 at 15:52
  • $\begingroup$ I wonder if the identity $\ln{(z)}=-\frac{1}{2z}+\psi{\left(z+1\right)}+\sum_{k=1}^{\infty}\frac{B_{2k}}{2k\,z^{2k}}$ could improve our prospects of finding a closed form for that last series. At the very least, changing the order of summation improves convergence. $\endgroup$ – David H Dec 12 '14 at 17:07
  • $\begingroup$ @DavidH The series which you wrote (the Stirling series) is divergent and holds only asymptotically. The reason is that Bernoulli numbers grow very quickly: ${B}_{2k}\sim(-1)^{k-1}2^{1-2k}\pi^{-2k}(2k)!$ as $k\to\infty$. $\endgroup$ – Iaroslav Blagouchine Dec 12 '14 at 17:16
  • $\begingroup$ @IaroslavBlagouchine I'm pretty sure the series I wrote converges for all $z>1$. $\endgroup$ – David H Dec 12 '14 at 17:27
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    $\begingroup$ @DavidH The Stirling series diverges for all finite $z$ and holds only asymptotically for a fixed number of term in the last sum as $|z|\to\infty$ in the right half-plane. Just write down several terms of the sum with Bernoulli numbers, and you'll see how rapidly they grow. $\endgroup$ – Iaroslav Blagouchine Dec 12 '14 at 17:33

Just a partial answer for now, but a promising one. Let: $$\begin{eqnarray*} I(z)&\triangleq&\int_{0}^{+\infty}\frac{\arctan\frac{t}{z}}{(t^2+1)(e^{\pi t}+1)}\,dt\\ &=&\int_{0}^{+\infty}\frac{\arctan\frac{t}{z}}{(t^2+1)(e^{\pi t}-1)}\,dt+2\int_{0}^{+\infty}\frac{\arctan\frac{t}{z}}{(t^2+1)(e^{2\pi t}-1)}\,dt\\&=&2\int_{0}^{+\infty}\left(\frac{\arctan\frac{2t}{z}}{4t^2+1}+\frac{\arctan\frac{t}{z}}{t^2+1}\right)\frac{dt}{e^{2\pi t}-1}.\end{eqnarray*}$$ We have: $$ I'(z) = -2\int_{0}^{+\infty}\left(\frac{2}{(4t^2+1)(4t^2+z^2)}+\frac{1}{(t^2+1)(t^2+z^2)}\right)\frac{t\,dt}{e^{2\pi t}-1}$$ that is treatable through the Abel-Plana's formula with $$ f(u) = \frac{1}{(u+1)(u+z)}.$$ We have: $$ \sum_{n\geq 0}f(n) = \frac{\psi(z)-\psi(1)}{z-1},\qquad \int_{0}^{+\infty}f(u)\,du = \frac{\log z}{z-1},$$ and since for any $z>0$ we have: $$I(z)\leq \frac{1}{z}\int_{0}^{+\infty}\frac{t\,dt}{e^{\pi t}+1}=\frac{1}{12 z}$$ $I(1)$ just depends on: $$\int_{1}^{+\infty}\frac{\log z}{z^2-1}\,dz = \frac{\pi^2}{8}$$ and: $$ J = \int_{1}^{+\infty}\frac{\psi(z)-\psi(1)}{z^2-1}\,dz = \sum_{n\geq 0}\frac{1}{n+1}\int_{1}^{+\infty}\frac{dz}{(z+1)(z+n)}=\sum_{n\geq 0}\frac{\log\frac{n+1}{2}}{n^2-1}.$$ Since: $$ J = \log 2 +\frac{1}{4}-\sum_{n\geq 2}\frac{\log 2}{n^2-1}+\sum_{n\geq 2}\frac{\log(n+1)}{n^2-1}=\frac{1+\log 2}{4}+\sum_{n\geq 2}\frac{\log(n+1)}{n^2-1}$$ it is sufficient to compute: $$ \sum_{n\geq 2}\frac{\log(n+1)}{n^2-1}=\frac{d}{d\alpha}\left.\sum_{n\geq 2}\frac{(n+1)^{\alpha-1}}{(n-1)}\right|_{\alpha=0}.$$

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