VARDI´S INTEGRAL

         Today´s post we will evaluate the following three interrelated integrals


\begin{aligned} &\int_{0}^{\infty} \frac{ \ln x}{\cosh x} \, d x=\frac{\pi}{2} \ln \left( \frac{4 \pi^3}{\Gamma^4 \left(\frac{1}{4}\right) }\right) \\ & \\ &\int_{\pi/4}^{\pi/2}\ln \left(\ln \, \tan x \right)\,dx=\frac{\pi}{4} \ln \left( \frac{4 \pi^3}{\Gamma^4 \left(\frac{1}{4}\right) }\right)\\ & \\ & \int_{0}^{\pi/4}\ln \left(\ln \, \cot x \right)\,dx=\frac{\pi}{4} \ln \left( \frac{4 \pi^3}{\Gamma^4 \left(\frac{1}{4}\right) }\right)\\ \end{aligned}

Dirichlet Beta function

We should first recall Dirichlet´s Beta function

\begin{aligned} \beta(s)&=\sum_{k=0}^\infty\,\frac{(-1)^k}{(2k+1)^s}\\ &=\sum_{k=0}^\infty\,(-1)^k(2k+1)^{-s}\\ &=\sum_{k=0}^\infty\,(-1)^k\,e^{-s\ln(2k+1)}\\ \end{aligned}

Then, if we differentiate it w.r. to s we obtain

\begin{aligned} \beta^{\prime}(s)&=\frac{d}{ds}\left(\sum_{k=0}^\infty\,(-1)^k\,e^{-s\ln(2k+1)} \right)\\ &=-\sum_{k=0}^\infty\,\frac{(-1)^k\,\ln(2k+1)}{(2k+1)^s}\\ &=\sum_{k=1}^\infty\,\frac{(-1)^{k+1}\,\ln(2k+1)}{(2k+1)^s}\\ \end{aligned}

Letting s \to 1 we obtain

\beta^{\prime}(1)=\sum_{k=1}^\infty\,\frac{(-1)^{k+1}\,\ln(2k+1)}{2k+1}(1)

Now, recall Kummer´s expansion for the LogGamma function:

\ln \Gamma(x) = \left(\tfrac{1}{2}-x \right)(\gamma+\ln2)+(1-x)\ln \pi - \frac{1}{2} \ln\sin(\pi x) + \frac{1}{\pi} \sum_{k=1}^{\infty} \frac{\ln k \, \sin(2\pi kx)}{k}(2)

Rearranging terms we get


\frac{1}{\pi} \sum_{k=1}^{\infty} \frac{\ln k \, \sin(2\pi kx)}{k}= \ln \Gamma(x)-\left(\frac{1}{2}-x \right)(\gamma+\ln2)+(x-1)\ln \pi +\frac{1}{2} \ln \sin(\pi x)(3)


If we let   x=\frac{1}{4}    in (3)   we obtain



\begin{aligned} \frac{1}{\pi} \sum_{k=1}^{\infty} \frac{\ln k \, \sin\left(\frac{\pi }{2}k\right)}{k}&= \ln \Gamma\left(\frac{1}{4}\right)-\frac{1}{4} (\gamma+\ln2)-\frac{3}{4}\log \pi +\frac{1}{2} \ln \sin\left(\frac{\pi}{4}\right)\\ & \\ -\sum_{k=1}^\infty \frac{(-1)^{k+1} \ln(2k+1)}{2k+1} &=\pi\ln \Gamma \left(\frac{1}{4} \right) - \frac{\pi}{4} (\gamma+\ln 2) -\frac{3\,\pi}{4}\ln \pi -\frac{\pi}{4} \ln 2 \\ &=-\frac{\pi}{4}\left(\gamma +2\ln 2+3\ln\pi-4\ln \Gamma \left(\frac{1}{4}\right) \right)\\ \end{aligned}

Or


\sum_{k=1}^\infty \frac{(-1)^{k+1} \ln(2k+1)}{2k+1}=\frac{\pi}{4}\left(\gamma +2\ln 2+3\ln\pi-4\ln \Gamma \left(\frac{1}{4}\right) \right)(4)


Comparing (4) with (1) we conclude that


\beta^{\prime}(1)=\frac{\pi}{4}\left(\gamma +2\ln 2+3\ln\pi-4\ln \Gamma \left(\frac{1}{4}\right) \right)(5)

Now let´s evaluate the first integral, namely:


\begin{aligned} \int_{0}^{\infty} \frac{x^{s-1}}{\cosh x} d x&=2 \int_{0}^{\infty} \frac{x^{s-1}}{e^{x}+e^{-x}} d x \\ &=2 \int_{0}^{\infty} \frac{x^{s-1} e^{-x}}{1+e^{-2 x}} d x \\ &=2 \int_{0}^{\infty} x^{s-1} e^{-x} \sum_{k=0}^{\infty}\left(-e^{-2 x}\right)^{k} d x \\ &=2 \sum_{k=0}^{\infty}(-1)^{k} \int_{0}^{\infty} x^{s-1} e^{-(2 k+1)x} d x \\ &=2 \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{s}} \int_{0}^{\infty} t^{s-1} e^{-t} d t \qquad ((2 k+1)x \mapsto x) \\ &=2 \Gamma(s) \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{s}}\\ &=2 \Gamma(s) \beta(s) \end{aligned}

\int_{0}^{\infty} \frac{x^{s-1}}{\cosh x} d x=2 \Gamma(s) \beta(s)(6)

Differentiating (6) w.r. to s we get



\int_{0}^{\infty} \frac{x^{s-1} \,\ln x}{\cosh x} d x=2 \left(\Gamma^{\prime}(s) \beta(s)+\Gamma(s) \beta^{\prime}(s) \right)(7)


Letting s \to 1 in (7) we obtain


\begin{aligned} \int_{0}^{\infty} \frac{ \ln x}{\cosh x} \, d x &=2 \left(\Gamma^{\prime}(1) \beta(1)+\Gamma(1) \beta^{\prime}(1) \right)\\ &=2\left(-\gamma \sum_{k=0}^\infty\,\frac{(-1)^k}{2k+1} + \beta^{\prime}(1) \right)\\ &=-\gamma\frac{\pi }{4}+ \gamma \frac{\pi}{4}+\frac{2\pi\ln 2}{4}+\frac{3 \pi\ln\pi}{4}- \frac{\pi}{4}\ln \Gamma^4 \left(\frac{1}{4}\right) \\ &=\frac{\pi}{2} \ln \left( \frac{4 \pi^3}{\Gamma^4 \left(\frac{1}{4}\right) }\right) \qquad \blacksquare \end{aligned}

Two related integrals.

The first one is known as Vardi´s integral

\begin{aligned} I&=\int_{\pi/4}^{\pi/2}\ln \left(\ln \, \tan x \right)\,dx\\ &=\int_1^\infty \frac{\ln \left(\ln \, x \right) }{x^2+1}\,dx \qquad (\tan x \to x)\\ &=\int_0^\infty \frac{e^x \,\ln x}{e^{2x}+1}\,dx\\ &=\int_0^\infty \frac{ \,\ln x}{e^{x}+e^{-x}}\,dx\\ &=\frac{1}{2}\,\int_0^\infty \frac{ \,\ln x}{\cosh x}\,dx\\ &=\frac{\pi}{4} \ln \left( \frac{4 \pi^3}{\Gamma^4 \left(\frac{1}{4}\right) }\right) \qquad \blacksquare \end{aligned}

\begin{aligned} I&=\int_0^{\pi/4}\ln \left(\ln \, \cot x \right)\,dx\\ &=\int_1^\infty \frac{\ln \left(\ln \, x \right) }{x^2+1}\,dx \qquad (\cot x \to x)\\ &=\int_0^\infty \frac{e^x \,\ln x}{e^{2x}+1}\,dx\\ &=\int_0^\infty \frac{ \,\ln x}{e^{x}+e^{-x}}\,dx\\ &=\frac{1}{2}\,\int_0^\infty \frac{ \,\ln x}{\cosh x}\,dx\\ &=\frac{\pi}{4} \ln \left( \frac{4 \pi^3}{\Gamma^4 \left(\frac{1}{4}\right) }\right) \qquad \blacksquare \end{aligned}

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