Prove $\int_{0}^{\infty} \psi^{(2)} (1+x) \ln (x) \, dx = \zeta'(2) + \zeta(2)$

Question

I'm looking for alternate methods of proving the following result:

$$\int_{0}^{\infty} \psi^{(2)} (1+x) \ln (x) \, dx = \frac{\pi^2}{6} \left( \gamma + \ln (2\pi)-12 \ln A +1\right) = \zeta'(2) + \zeta(2)$$ Where $\psi$ is the polygamma function, $A$ is the Glaisher-Kinkelin constant and $\gamma$ is the Euler-Mascheroni constant.

Here is how I originally proved it:

Begin with the known result $$\int_{0}^{\infty} \frac{x \ln x}{e^{2 \pi x}-1} \, dx = \frac{1}{2} \zeta'(-1)=\frac{1}{24}-\frac{1}{2} \ln A$$

then use the following well-known property of the Laplace transform $$\int_{0}^{\infty} f(x) g(x) \, dx = \int_{0}^{\infty} \left( \mathcal{L} f\right) (y) \left( \mathcal{L}^{-1} g \right) (y) \, dy$$

Let $f(x) = \frac{x^2}{e^{2 \pi x} - 1}$ and $g(x) = \frac{\ln x}{x}$ then: $$\left(\mathcal{L} f\right) (y) = -\frac{1}{8 \pi^3} \psi^{(2)} \left(1+\frac{y}{2\pi}\right)$$ $$\left(\mathcal{L}^{-1} g\right)(y) = -\gamma - \ln (y)$$

$$\implies \frac{1}{24} - \frac{1}{2} \ln A = \frac{1}{8\pi^3}\int_{0}^{\infty} \gamma \, \psi^{(2)} \left(1+\frac{y}{2\pi}\right) \, dy + \frac{1}{8\pi^3} \int_{0}^{\infty} \psi^{(2)} \left(1+\frac{y}{2\pi}\right) \ln (y) \, dy$$

Proceed with the substitution $y \mapsto 2\pi x \implies dy = 2 \pi\,dx$ $$\implies \frac{1+\gamma}{24} - \frac{1}{2} \ln A = \frac{1}{4\pi^2} \int_{0}^{\infty} \psi^{(2)} \left(1+x\right) \ln (2 \pi x) \, dx$$ Now separate $\ln (2\pi x) = \ln (2 \pi) + \ln (x)$ and the result quickly follows.

I was wondering if there are any alternative methods that don't rely on the initial $\frac{1}{2} \zeta'(-1)$ result.

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Motivation:

I've been experimenting with the following identity: $$\int_{0}^{\infty} \frac{x^{2n-1}}{e^{2\pi x}-1}\, dx = \frac{(-1)^{n-1} B_{2n}}{4n}\,\text{ for } n \in \mathbb{N}$$

I noticed that if one takes the partial derivative with respect to $n$ and then proceeds to take the limit as $n \to 1$, then we arrive at our $-\frac{1}{2} \zeta'(-1)$ result. This means it is possible for one to define the 'derivative' of the Bernoulli numbers at $n=1$.

More precisely: $$B_{2}^{\prime} = \frac{1}{4} - 2 \ln A$$ is this result already known, and if so, are any other Bernoulli derivatives definable in a similar way using the Glaisher-Kinkelin constant? (I know it is possible to do it with Glaisher-Kinkelin-like constants such as the Bendersky-Adamchik constants due to their similar connection to hyperfactorials, $\zeta'(-2)$ and $\zeta(3)$ for example). I'm not sure what meaning one can give to the 'derivative' of the Bernoulli numbers, but figure 2 of this paper seems to roughly agree with my result.

Answer 1

Using the representation $$\psi^{(2)}(x) = - 2 \sum_{n=0}^{\infty} \frac{1}{(n+x)^{3}},$$ we have $$ \begin{align} \int_{0}^{\infty} \psi^{(2)}(x+1) \ln (x) \, \mathrm dx &= -2\int_{0}^{\infty}\ln(x) \sum_{n=0}^{\infty} \frac{1}{(n+x+1)^{3}} \, \, \mathrm dx \\ &= - 2 \sum_{n=0}^{\infty} \int_{0}^{\infty} \frac{\ln (x)}{(n+x+1)^{3}} \, \mathrm dx. \end{align}$$

And from integrating the complex function $$f(z) = \frac{\ln^{2}(z)}{(n+z+1)^{3}}$$ around a keyhole contour with the branch cut along the positive real axis, we get $$ \begin{align} -4 \pi \int_{0}^{\infty} \frac{\ln(x)}{(n+x+1)^{3}} \, \mathrm dx &= \Im \left( 2 \pi i \operatorname{Res}[f(z), -n-1]\right) \\ &= \Im \left(2 \pi i \lim_{z \to -n-1}\frac{1}{2} \frac{\mathrm d^{2}}{\mathrm dz^{2}} \ln^{2}(z) \right)\\ &= \pi \, \frac{2- 2\ln(n+1)}{(n+1)^{2}}. \end{align}$$

Therefore, $$ \begin{align} \int_{0}^{\infty} \psi^{(2)}(x+1) \ln (x) \, \mathrm dx &= - 2 \sum_{n-0}^{\infty} \left( -\frac{1}{2(n+1)^{2}} + \frac{\ln(n+1)}{2(n+1)^{2}} \right) \\ &= \sum_{n=0}^{\infty} \frac{1}{(n+1)^{2}} - \sum_{n=0}^{\infty}\frac{\ln(n+1)}{(n+1)^{2}} \\ &= \zeta(2) + \zeta^{'}(2). \end{align}$$

Answer 2

For another solution using the complete beta function:

Using the same expansion

$$\psi^{(2)}(x) = - 2 \sum_{n=0}^{\infty} \frac{1}{(n+x)^{3}},$$

we have

$$ \begin{align} \int_{0}^{\infty} \psi^{(2)}(x+1) \ln (x) \, \mathrm dx &= -2\int_{0}^{\infty}\ln(x) \sum_{n=0}^{\infty} \frac{1}{(n+x+1)^{3}} \, \, dx \\ &= - 2 \sum_{n=0}^{\infty} \int_{0}^{\infty} \frac{\ln (x)}{(n+x+1)^{3}} \, dx.\\ =& - 2 \sum_{n=0}^{\infty} \frac{1}{(n+1)^3} \int_{0}^{\infty} \frac{\ln (x)}{(1+\frac{x}{(n+1)})^{3}} \, dx\\ =& - 2 \sum_{n=0}^{\infty} \frac{1}{(n+1)^3} \int_{0}^{\infty} \frac{\left(\frac{d}{dt}\Big|_{t=0+}x^t\right)}{(1+\frac{x}{(n+1)})^{3}} \, dx\\ =& \frac{d}{dt}\Big|_{t=0+} - 2 \sum_{n=0}^{\infty} \frac{1}{(n+1)^3} \int_{0}^{\infty} \frac{x^t}{(1+\frac{x}{(n+1)})^{3}}dx \\ =& \frac{d}{dt}\Big|_{t=0+} - 2 \sum_{n=0}^{\infty} \frac{(n+1)^t}{(n+1)^2} \int_{0}^{\infty} \frac{w^t}{(1+w)^{3}}dw \quad \\ =& \frac{d}{dt}\Big|_{t=0+} - 2 \sum_{n=0}^{\infty} \frac{(n+1)^t}{(n+1)^2}B(t+1,2-t)\\ =& \frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty} \frac{(n+1)^t}{(n+1)^2}\pi (t-1)t \csc(\pi t)\\ =& \sum_{n=0}^{\infty} \frac{1}{(n+1)^2} - \sum_{n=0}^{\infty} \frac{\ln(n+1)}{(n+1)^2}\\ =& \zeta(2)+\zeta'(2) \end{align}\\$$