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Introduction:
With the long introduction I want to provide the context, include the source and motivation for my question to the community.

Interest in this question began with curiosity about a different form of a new identity stocha

\begin{align} \mathcal{I}\left( \beta \right) = \int_{0}^{\infty }\frac{\beta}{1+\beta^2\,x^{2}}~\theta _{4}^{2}\left( \exp \left( -\pi \,x\right)\right) \,dx\text{,} \qquad \mathcal{I}\left(1\right) = 1\text{,} \label{eq:rzwei} \end{align}

using the "evaluating integrals over the positive real axis" property of the Laplace transform WIKIPEDIA and e.g. Glasser. This property, as will be seen, can be used to derive the following different form of the same identity:

\begin{align} \mathcal{I}\left( \beta \right) = \frac{\pi }{2}+\int_0^{\infty } (\psi (1+i\, x)+\psi (1-i\,x))\, \left(\text{sech}\left(\frac{\pi\, x}{\beta }\right)-2\,\text{sech}\left(\frac{\pi\, x}{\beta }\right)\right) \, dx-\beta\log(2)\text{,} \label{eq:rdrei} \end{align}

which is surprisingly related to the sum over the Laplace transform of the complex valued digamma function:

\begin{align} P_{C} = \int_0^{\infty }sech(\lambda \,x) \, \psi (1\pm i \,x) \, dx = 2 \sum_{k=0}^{\infty}(-1)^k \times \label{eq:rvier} \end{align}

\begin{equation*} \times\int_0^{\infty }e^{-\lambda\,(2\,k+1)\, x} \, \psi (1\pm i \,x) \, dx \quad\text{for}\quad \lambda=\frac{2 \,\pi \, x}{\beta }\, \text{,} \qquad \lambda=\frac{\pi \, x}{\beta }\text{,} \end{equation*} where we used the known series expansion of $sech(x)$ and two different $\lambda$.

An explicit formula for the Laplace transform of the real valued digamma function is discussed in Dixit in the context of the famous integral OLIVIER OLOA

\begin{equation} \int_0^{\frac{\pi }{2}} \frac{\theta^2}{\theta^2+\log (2\, \cos (\theta ))} \, d\theta = \frac{1}{8} \pi \, (1+\log (2 \,\pi )-\gamma ) \end{equation}

Laplace transform of the complex valued digamma function: In the present post we give an explicit expression for the Laplace transform of a complex valued digamma function (the long prove is skipped here):

Propositon: Let $a > 0$, $\gamma$ the Euler-Mascheroni constant and $\psi\left(i\,x\right)$ the complex valued digamma function, then \begin{align} L_{C}(a)= \int_0^{\infty } e^{-a z} \psi (1+i\,x) \, dx = 2\,a \sum _{n=1}^{\infty } \frac{\log (n) }{a^2-4 n^2 \pi ^2} -\frac{\gamma +\log\left(a\right) }{a} + \label{eq:rfuenfzehn} \end{align} \begin{equation*} \left(\frac{1}{2} \cot \left(\frac{a}{2}\right)-\frac{1}{a}\right) \log \left(\frac{2 \pi }{a}\right)+\frac{\pi }{4}+ \frac{i}{2}\left(\log \left(\frac{a}{2 \pi }\right)-\psi \left(\frac{a}{2 \pi }\right)-\frac{\pi }{a}\right) \end{equation*}

The expression for $L_{C}(a)$ is explicitly given in the form of a real - and imaginary part: \begin{align} L_{C}(a)= \Re(L_{C}(a))+i\,\Im(L_{C}(a)) \label{eq:rsechzehn}\text{.} \end{align}

$\Im(L_{C}(a))$ - the last term describes the imaginary part of and is explicitly given in a closed form: \begin{align} \Im(L_{C}(a))= \frac{1}{2}\left(\log \left(\frac{a}{2 \pi }\right)-\psi \left(\frac{a}{2 \pi }\right)-\frac{\pi }{a}\right) \label{eq:rsiebzehn} \end{align}

A further important property is given:

Conjecture: For a complex number $a=i\,y$ we have: \begin{align} \Re(L_{C}(i\,y)) = 0 \text{,}\qquad \underset{a\to \,i\,y}{\Re\left(\Re(L_{C}(a))\right)}=\Re\left(\Im(L_{C}(i\,y))\right) \label{eq:rvierundfuenfzig} \end{align}

This property ends up in the next Propositon:

Propositon: If $z > 0$, then \begin{align} \Im(\psi(i\, z))=\frac{\pi}{2}\left(\coth (\pi \, z)+\frac{1}{\pi \, z}\right)\text{,}\qquad \Im(\psi(1+i\, z))=\frac{\pi}{2}\left(\coth (\pi \, z)-\frac{1}{\pi \, z}\right) \label{eq:reinundzwanzig} \end{align}

This identity was already published and proved Raymond Manzoni, but there in the context of the reflection formula. In the following we consider $a=i\,z$ a complex number. We substitute Euler's formula \begin{equation} e^{-i\, t\, z}=\cos (t\, z)-i\, \sin (t\, z) \end{equation} in $L_{C}(a)$ and split it in two integrals. We assumed a close solution for the first integral:

\begin{equation} \int_0^{\infty } \psi (1+i\, t) \cos (t \,z) \, dt = -\frac{i}{2} \left(\frac{1}{\frac{2\, (-i \,z)}{2\, \pi }}-\log \left(-\frac{i\, z}{2\, \pi }\right)+\psi\left(-\frac{i\, z}{2 \pi }\right)\right) \end{equation} The integral involving a real valued digamma function exists, too. It is given in a closed form, too:

\begin{equation} \frac{1}{\pi }\int_0^{\infty } \psi \left(1+\frac{t}{2 \,\pi }\right) \cos (t \,z) \, dt = \frac{1}{2\, z}-\log(z)+\psi (z) \end{equation}

We Integrate over z and evaluate the constant of integration, without proving it: \begin{equation} c = \frac{1}{2}\,(\gamma +\log (2 \pi )) \end{equation}

Finally we get the new log gamma formula:

Propositon: A formula for $\log (\Gamma (z))$ for $\Re((z)>0$ is \begin{align} \log (\Gamma (z))=\left(z-\frac{1}{2}\right) \log(z)-z+\frac{1}{2}\,(\gamma +\log (2 \pi ))+\frac{1}{\pi}\,\int_0^{\infty } \frac{\sin (t z)}{t}\psi\left(1+\frac{t}{2 \pi }\right)\, dt\text{,} \label{eq:reins} \end{align} where $\gamma$ is the Euler-Mascheroni constant.

In other words, the formula means, that we have a closed form solution of the last integral (Proposition).

We note, that in contrast to Binet's Log Gamma Formulas MathWorld, that the Euler-Mascheroni constant $\gamma$ appear like in the solution of the famous OLIVIER OLOA integral.

My attempt was also to start from the Binet's second formula MathWorld \begin{equation} \psi(z)=log (z)-\frac{1}{2\, z}-\int_0^{\frac{\pi }{2}} \frac{t\, (\coth (\pi\, t)-1)}{t^2+z^2} \, dt \end{equation}

and use the "evaluating integrals over the positive real axis" property of the Laplace transform WIKIPEDIA and well known formula: $\psi (x+1)=\frac{1}{x}+\psi (x)$ to get:

\begin{equation} \frac{1}{\pi }\,\int_0^{\infty } \left(\psi \left(1+\frac{t}{2\, \pi }\right)+\gamma \right) \cos (t\, z) \, dt=\frac{1}{2\, z}-\log z)+\psi(z) \end{equation}

I did not finished this way strictly and found the log gamma formula by try and error.

Question:

  1. How can we proof the new formula for the $\log (\Gamma (z))$ explicitely?

  2. Is there a relation between the new integral and the Glasser-Manna-Oloa integral and what is the relation?

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To evaluate the integral term of the formula \begin{equation} I=\frac{1}{\pi}\int_0^{\infty } \frac{\sin (t z)}{t}\psi\left(1+\frac{t}{2 \pi }\right)\, dt \end{equation} we use the representation DLMF \begin{equation} \psi\left(1+\frac{t}{2\pi}\right)=-\gamma+\frac{1}{2\pi}\sum_{n=1}^{\infty}\frac{t}{n(n+t/2\pi)} \end{equation} Then \begin{equation} I=-\frac{1}{2}\gamma+\frac{1}{\pi}\sum_{n=1}^{\infty}\frac{1}{n}\int_0^\infty \frac{\sin (t z)}{2\pi n+t}\,dt \end{equation} Now, using the Laplace transform method for evaluating integrals wikipedia on the real axis as proposed in the OP , with \begin{equation} \mathcal{L}\left[\sin t z\right]=\frac{z}{u^2+z^2}\quad ;\quad \mathcal{L}^{-1}\left[\frac{1}{2\pi n+t}\right]=e^{-2\pi nu} \end{equation} we obtain \begin{align} I&=-\frac{1}{2}\gamma+\frac{1}{\pi}\sum_{n=1}^{\infty}\frac{z}{n}\int_0^\infty \frac{e^{-2\pi nu}}{u^2+z^2}\,du\\ &=-\frac{1}{2}\gamma-\frac{z}{\pi}\int_0^\infty\frac{\ln\left( 1-e^{-2\pi u} \right)}{u^2+z^2}\,du \end{align} Integrating by parts, \begin{equation} I=-\frac{1}{2}\gamma+2\int_0^\infty \frac{\arctan (u/z)}{e^{2\pi u}-1}\,du \end{equation} From the Binet's second formula \begin{equation} \ln\left( \Gamma(z) \right)=\left( z-\frac{1}{2} \right)\ln z-z+\frac{1}{2}\ln\left( 2\pi \right)+2\int_0^\infty \frac{\arctan \left( t/z \right)}{e^{2\pi t}-1}\,dt \end{equation} we have \begin{equation} I=-\frac{1}{2}\gamma-\left( z-\frac{1}{2} \right)\ln z+z-\frac{1}{2}\ln\left( 2 \pi \right)+\ln\left( \Gamma(z) \right) \end{equation} And thus \begin{equation} \left(z-\frac{1}{2}\right) \log(z)-z+\frac{1}{2}\,(\gamma +\log (2 \pi ))+I=\ln\left( \Gamma(z) \right) \end{equation} as expected.

Paul Enta
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