\nIn this post, we will compute the Fourier series expansion of the Log-Gamma function and use it to prove the beautiful Vardi’s integral:\n$$\\int_0^1 \\frac{\\log\\log \\left(\\frac{1}{x}\\right)}{1+x^2}dx = \\frac{\\pi}{2}\\log\\left(\\sqrt{2\\pi}\\frac{\\Gamma\\left(\\frac{3}{4}\\right)}{\\Gamma\\left(\\frac{1}{4}\\right)} \\right)$$\n<\/p>\n\n
Fourier Series of the Log-Gamma Function:<\/b><\/p>\n
For $s\\in (0,1)$, we have\n$$\n\\log \\Gamma(s) = \\left(\\frac{1}{2}-s \\right)(\\gamma + \\log 2)-\\frac{1}{2}\\log(\\sin (\\pi s)) + (1-s) \\log(\\pi) + \\frac{1}{\\pi}\\sum_{n=1}^\\infty \\frac{\\sin(2\\pi n s)\\log n}{n} \\quad \\color{blue}{(1)}\n$$\n<\/p>\n\n\n\n
\nProof: <\/b> It suffices to do the following integrals:\n$$\n\\begin{aligned}\n\\int_0^1 \\log \\Gamma (s) \\; ds &= \\frac{1}{2}\\log(2\\pi) \\\\\n\\int_0^1 \\log \\Gamma(s) \\cos(2\\pi n s)\\; ds &= \\frac{1}{4n} \\quad \\forall n\\in \\mathbb{Z}^+ \\\\\n\\int_0^1 \\log \\Gamma(s) \\sin(2\\pi n s)\\; ds &= \\frac{\\gamma + \\log(2n\\pi )}{2n\\pi}\\quad \\forall n\\in \\mathbb{Z}^+\n\\end{aligned}\n$$\n<\/p>\n\n\n\n
\nThe first integral can be evaluated with the help of Euler’s reflection formula:\n$$\n\\begin{aligned}\n\\int_0^1 \\log \\Gamma(s) \\; ds &= \\frac{1}{2} \\int_0^1 \\log \\Gamma(s) \\; ds + \\frac{1}{2}\\int_0^1 \\log \\Gamma(1-s) \\; ds \\\\\n&= \\frac{1}{2}\\int_0^1 \\log\\left(\\frac{\\pi}{\\sin(\\pi s)} \\right)\\; ds \\\\\n&= \\frac{\\log (\\pi)}{2} – \\frac{1}{2}\\int_0^1 \\log(\\sin(\\pi s))\\; ds \\\\\n&= \\frac{\\log(2\\pi)}{2}\n\\end{aligned}\n$$\n<\/p>\n
\nThe second integral can also be dealt in a similar way. We have:\n$$\n\\begin{aligned}\n\\int_0^1 \\log \\Gamma(s) \\cos(2\\pi n s)\\; ds &= \\frac{1}{2}\\int_0^1 \\log \\Gamma(s) \\cos(2\\pi n s)\\; ds + \\frac{1}{2}\\int_0^1 \\log \\Gamma(1-s) \\cos(2\\pi n (1-s))\\; ds \\\\\n&= \\frac{1}{2}\\int_0^1 \\log\\left(\\frac{\\pi}{\\sin(\\pi s)} \\right)\\cos(2\\pi n s)\\; ds \\\\\n&= -\\frac{1}{2}\\int_0^1 \\log(2\\sin(\\pi s))\\cos(2\\pi n s)\\; ds \\\\\n&= \\frac{1}{2}\\int_0^1 \\left(\\sum_{k=1}^\\infty \\frac{\\cos(2\\pi k s)}{k} \\right)\\cos(2\\pi n s)\\; ds \\\\\n&= \\frac{1}{2}\\sum_{k=1}^\\infty \\frac{1}{k}\\int_0^1 \\cos(2\\pi k s)\\cos(2\\pi n s)\\; ds \\\\\n&= \\frac{1}{4n}\n\\end{aligned}\n$$\n\nIn the above calculation, we used the well known Fourier series:\n$$\\log(2\\sin (\\pi s)) = -\\sum_{k=1}^\\infty \\frac{\\cos(2\\pi k s)}{k} \\quad \\forall s\\in (0,1) \\quad\\quad \\color{blue}{(2)}$$\n<\/p>\n\n\n\n
\nThe third integral is the most troublesome of all since the trick involving Euler’s reflection formula does not work. We will instead use the following integral representation of the Log-Gamma function:\n$$\n\\log \\Gamma(s) = \\int_0^\\infty \\left(\\frac{s-1}{t e^t} -\\frac{1-e^{t(1-s)}}{t(e^t-1)}\\right) dt \\quad \\forall s>0\n$$\nOne can easily verify the above equation via the differentiation under the integral technique. Therefore, upon changing the order of integration we get:\n$$\n\\begin{aligned}\n\\int_0^1 \\log \\Gamma(s) \\sin(2\\pi n s)\\; ds &= \\int_0^\\infty \\left(\\frac{1}{t e^t}\\int_0^1 s\\sin(2\\pi ns)\\; ds +\\frac{1}{t(e^t-1)}\\int_0^1e^{t(1-s)}\\sin(2\\pi n s)\\; ds \\right)dt \\\\\n&= \\int_0^\\infty \\left[\\frac{1}{t e^t}\\left(-\\frac{1}{2\\pi n}\\right) +\\frac{1}{t(e^t-1)}\\left(\\frac{2\\pi n (e^t-1)}{t^2 + (2\\pi n)^2} \\right) \\right]dt \\\\\n&= \\int_0^\\infty \\left(-\\frac{1}{2\\pi n t e^t} + \\frac{2\\pi n}{t\\left( t^2 + (2\\pi n)^2 \\right)} \\right)dt \\\\\n&= \\frac{1}{2\\pi n}\\int_0^\\infty\\left(-\\frac{1}{te^t} +\\frac{1}{t}-\\frac{t}{t^2 + (2\\pi n)^2} \\right) dt \\\\\n&= \\frac{1}{2\\pi n}\\left(\\gamma + \\left[ \\log t – \\log (t) e^{-t} – \\frac{1}{2}\\log\\left( t^2 + (2\\pi n)^2 \\right)\\right]_{t=0}^\\infty \\right) \\\\\n&= \\frac{\\gamma + \\log(2\\pi n)}{2\\pi n}\n\\end{aligned}\n$$\n<\/p>\n
To get equation (1), one needs to piece together all these calculations. Equation (2) along with the result:\n$$\n\\pi \\left(\\frac{1}{2}- s\\right) = \\sum_{k=1}^\\infty \\frac{\\sin(2\\pi k s)}{k} \\quad \\forall s\\in (0,1)\n$$\nare needed to perform simplifications.\n<\/p>\n\n\n\n
\nNow, we turn our attention to the Vardi’s integral:\n$$\n\\begin{aligned}\nI &= \\int_0^1 \\frac{\\log\\log \\left(\\frac{1}{x}\\right)}{1+x^2}dx \\\\ &= \\int_0^\\infty \\frac{\\log(t) e^{-t}}{1 + e^{-2t}}dt \\\\\n&= \\int_0^\\infty \\log(t)\\sum_{n=0}^\\infty (-1)^n e^{-(2n+1)t} \\; dt\\\\\n&= \\sum_{n=0}^\\infty (-1)^n \\int_0^\\infty \\log (t) e^{-(2n+1)t}\\; dt \\\\\n&= -\\sum_{n=0}^\\infty (-1)^n \\left(\\frac{\\log(2n+1)}{2n+1}+\\frac{\\gamma}{2n+1} \\right) \\\\\n&= -\\frac{\\gamma \\pi}{4}-\\sum_{n=1}^\\infty \\frac{\\sin\\left(\\frac{\\pi n}{2}\\right)\\log n}{n} \\quad \\color{blue}{(3)}\n\\end{aligned}\n$$\nTo get the value of the sum, plug $s=\\frac{1}{4}$ in equation (1). \n$$\n\\sum_{n=1}^\\infty \\frac{\\sin\\left(\\frac{\\pi n}{2}\\right)\\log n}{n} = \\pi \\log \\Gamma\\left(\\frac{1}{4}\\right) – \\frac{\\gamma \\pi}{4} – \\frac{\\pi \\log 2}{2}- \\frac{3\\pi}{4}\\log(\\pi)\n$$ \nSubstituting the above into equation (3) and performing some simplifications gives the desired result.\n<\/p>\n","protected":false},"excerpt":{"rendered":"
In this post, we will compute the Fourier series expansion of the Log-Gamma function and use it to prove the beautiful Vardi’s integral: $$\\int_0^1 \\frac{\\log\\log \\left(\\frac{1}{x}\\right)}{1+x^2}dx = \\frac{\\pi}{2}\\log\\left(\\sqrt{2\\pi}\\frac{\\Gamma\\left(\\frac{3}{4}\\right)}{\\Gamma\\left(\\frac{1}{4}\\right)} \\right)$$ Fourier Series of the Log-Gamma Function: For $s\\in (0,1)$, we have … Continue reading