Binet's Log Gamma Formulas

Today we will prove the famous Binet´s formulas for Log Gamma function, namely:


\begin{aligned} &\ln \Gamma(z)=\frac{1}{2} \ln 2 \pi+\left(z-\frac{1}{2}\right) \ln z-z+\int_{0}^{\infty}\left(\frac{1}{2}-\frac{1}{x}+\frac{1}{e^{x}-1}\right) \frac{e^{-z x}}{x} d x \\ &\\ &\ln \Gamma(z)=\frac{1}{2} \ln 2 \pi+\left(z-\frac{1}{2}\right) \ln z-z+2 \int_{0}^{\infty} \frac{\arctan \left(\frac{x}{z}\right)}{e^{2 \pi x}-1} d x \end{aligned}


Let´s start by computing the integral


\begin{aligned} I(z)&=\int_{0}^{\infty}\left(\frac{1}{2}-\frac{1}{x}+\frac{1}{e^{x}-1}\right) \frac{e^{-z x}}{x} d x \qquad \text{differentiating w.r. to z}\\ & \\ I^{\prime}(z)&=-\int_{0}^{\infty}\left(\frac{1}{2}-\frac{1}{x}+\frac{1}{e^{x}-1}\right) e^{-z x} d x \\ &=-\frac{1}{2}\int_0^\infty e^{-zx}dx-\int_0^\infty \left(\frac{1}{e^x-1}-\frac{1}{x} \right) e^{-zx}dx \qquad e^{-x} \mapsto w \,\, \text{in the second integral}\\ &=-\frac{1}{2}\left( \frac{e^{-zx}}{-z}\Big|_0^\infty\right)-\int_1^0 \left(\frac{1}{\frac{1}{w}-1}-\frac{1}{- \ln w} \right) w^{z}\frac{(-dw)}{w}\\ &=-\frac{1}{2}\left( \frac{1}{z} \right)-\int_0^1 \left(\frac{w^z}{1-w}+\frac{w^{z-1}}{ \ln w} \right) dw\\ &=-\frac{1}{2z}-\left(\ln z -\psi(z+1)\right)\\ &=-\frac{1}{2z}-\left(\ln z -\frac{1}{z}-\psi(z)\right)\\ &=\frac{1}{2z}+\psi(z)-\ln z \\ \end{aligned}


Where we used the result proved here


\begin{aligned} \int_0^1 \left(\frac{x^{z-1}}{ \ln x}+\frac{x^{w-1}}{1-x} \right) dx=\ln z - \psi(w) \end{aligned}

Then,


\begin{aligned} I(z)&=\int \frac{dz}{2z} + \int\psi(z) \, dz-\int\ln z \,dz + C\\ \end{aligned}


                  \begin{aligned} \int_{0}^{\infty}\left(\frac{1}{2}-\frac{1}{x}+\frac{1}{e^{x}-1}\right) \frac{e^{-z x}}{x} d x&=\frac{1}{2} \ln z + \ln \Gamma(z) -z\ln z+z + C\\ &=-\left(z-\frac12 \right)\ln z+z+\ln \Gamma(z)+C \end{aligend}        (1)                              

Now recall Stirling’s approximation for the Gamma function


\Gamma (z) = z^z\sqrt{\frac{2 \pi}{z}} e^{-z} \qquad \text{as} \,\, z \rightarrow \infty(2)


Taking logarithms in both sides of (2)


\ln \Gamma (z) = \frac{1}{2}\ln 2 \pi+ \left(z-\frac{1}{2}\right)\ln z-z \qquad \text{as} \,\, z \rightarrow \infty(3)


Plugging (3) in (1) and taking the limit z \longrightarrow \infty


\int_{0}^{\infty}\left(\frac{1}{2}-\frac{1}{x}+\frac{1}{e^{x}-1}\right) \frac{e^{-z x}}{x} d x &=-\left(z-\frac12 \right)\ln z+z+C+\frac{1}{2}\ln 2 \pi+ \left(z-\frac{1}{2}\right)\ln z-z


The L.H.S. goes to zero, and we conclude that 


                                                          C=-\frac{1}{2}\ln 2 \pi

Therefore we get


\boxed{\ln \Gamma(z)=\frac{1}{2} \ln 2 \pi+\left(z-\frac{1}{2}\right) \ln z-z+\int_{0}^{\infty}\left(\frac{1}{2}-\frac{1}{x}+\frac{1}{e^{x}-1}\right) \frac{e^{-z x}}{x} d x}


Now for the second Binet´s relation, consider the Integral


\begin{aligned} I(z)&=2 \int_{0}^{\infty} \frac{\arctan \left(\frac{x}{z}\right)}{e^{2 \pi x}-1} d x \\ &=2 \int_{0}^{\infty} \frac{1}{e^{2 \pi x}-1}\left\{\int_{0}^{\infty} \frac{\sin (x t) e^{-z t}}{t} d t\right\} d x \\ &=2 \int_{0}^{\infty} \frac{e^{-z t}}{t}\left\{\int_{0}^{\infty} \frac{\sin (x t)}{e^{2 \pi x}-1} d x\right\} d t \\ &=2 \int_{0}^{\infty} \frac{e^{-z t}}{t}\left\{\int_{0}^{\infty} \frac{e^{-2 \pi x} \sin (x t)}{1-e^{-2 \pi x}} d x\right\} d t \\ &=2 \int_{0}^{\infty} \frac{e^{-z t}}{t}\left\{\int_{0}^{\infty} \sin (x t) \sum_{k=1}^{\infty} e^{-2 \pi x k} d x\right\} d t \\ &=2 \int_{0}^{\infty} \frac{e^{-z t}}{t}\left\{\sum_{k=1}^{\infty} \int_{0}^{\infty} \sin (x t) e^{-2 \pi x k} d x\right\} d t \\ &=2 \int_{0}^{\infty} \frac{e^{-z t}}{t}\left\{\frac{\operatorname{coth}\left(\frac{t}{2}\right)}{4}-\frac{1}{2 t}\right\} d t \end{aligned}

Where we have used

\arctan \left(\frac{x}{z}\right)=\int_{0}^{\infty} \frac{\sin (x t) e^{-z t}}{t} d t

in the second line. Now make the following substitution, \frac{t}{2}=y  to get:


I(z)=\frac{1}{2} \int_{0}^{\infty} \frac{e^{-2 z y}}{y}\left\{\operatorname{coth}(y)-\frac{1}{y}\right\} d y(4)

Differentiating (4) w.r. to z


\begin{aligned} I'(z)&=-\int_{0}^{\infty}{e^{-2zy}}\left\{\coth\left({y}\right)-\frac{1}{y}\right\} dy\\$ &=\int_{0}^{\infty}{e^{-2zy}}\left\{\frac{1}{y}-\coth\left({y}\right)\right\} dy \end{aligned}


This last integral we have already computed here, it´s value is


I'(z)=\frac{1}{2z}+\psi(z)-\ln z

Then

\begin{aligned} I(z)&=\int \frac{dz}{2z} + \int\psi(z) \, dz-\int\ln z \,dz + C\\ \end{aligned}

\ln \Gamma(z)=\left(z-\frac{1}{2} \right)\ln z+z \ln z -z + C^{\prime}+2 \int_{0}^{\infty} \frac{\arctan \left(\frac{x}{z}\right)}{e^{2 \pi x}-1} d x(5)


Following the same procedure as before we obtain


\boxed{\ln \Gamma(z)=\frac{1}{2} \ln 2 \pi+\left(z-\frac{1}{2}\right) \ln z-z+2 \int_{0}^{\infty} \frac{\arctan \left(\frac{x}{z}\right)}{e^{2 \pi x}-1} d x}


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