MAMLSTEN INTEGRALS - PART II

In continuation of our previous post about Malmsten integrals, we will today prove the following four integrals:


\begin{aligned} \int_{0}^{1} \frac{\ln \ln \frac{1}{x}}{1+x^{2}} d x &=\frac{\pi}{2} \ln\left( \frac{\Gamma\left(\frac{3}{4}\right)^{2}}{\sqrt{\pi}}\right) \\ \\ \int_{0}^{1} \frac{\ln \ln \frac{1}{x}}{(1+x)^{2}} d x &=\frac{1}{2}\left(\ln \frac{\pi}{2}-\gamma\right) \\ \\ \int_{0}^{1} \frac{\ln \ln \frac{1}{x}}{1-x+x^{2}} d x &=\frac{2 \pi}{\sqrt{3}} \ln \left(\frac{\sqrt[6]{32 \pi^{5}}}{\Gamma\left(\frac{1}{6}\right)}\right) \\ \\ \int_{0}^{1} \frac{\ln \ln \frac{1}{x}}{1+x+x^{2}} d x &=\frac{\pi}{3} \ln \left(\frac{\Gamma\left(\frac{2}{3}\right)\sqrt[3]{2 \pi}}{\Gamma\left(\frac{1}{3}\right)} \right) \end{aligned}


For the First Integral

\begin{aligned} I&=\int_{0}^{1} \frac{\ln \ln \left(\frac{1}{x}\right)}{1+x^{2}} d x\\ &=\int_0^\infty \frac{\ln\left(x\right)e^{-x}}{1+e^{-2x}}\,dx \qquad (x \to e^{-x})\\ &=\int_0^\infty \frac{e^x}{e^x}\cdot\frac{\ln\left(x\right)e^{-x}}{1+e^{-2x}}\,dx\\ &=\frac22\int_0^\infty \frac{\ln\left(x\right)}{e^x+e^{-x}}\,dx\\ &=\frac12\int_0^\infty \frac{\ln\left(x\right)}{\cosh(x)}\,dx\\ &=\frac{\pi}{4} \ln \left( \frac{4 \pi^3}{\Gamma^4 \left(\frac{1}{4}\right) }\right) \\ &=\frac{\pi}{2} \ln \left( \frac{2 \pi^{3/2}}{\Gamma^2 \left(\frac{1}{4}\right) }\right) \\ &=\frac{\pi}{2} \ln \left( \frac{2 \pi^{3/2}\Gamma^2 \left(\frac{3}{4}\right)}{2 \pi^2 }\right) \\ &=\frac{\pi}{2} \ln \left(\frac{\Gamma\left(\frac{3}{4}\right)^{2}}{\sqrt{\pi}}\right) \qquad \blacksquare \\ \end{aligned}


Where We used Vardi´s integral:


\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)


and that

\Gamma\left( \frac34\right)=\frac{\sqrt{2\pi^2}}{\Gamma\left( \frac14\right)}

To evaluate the second integral, we first consider the following integral


\begin{aligned} I&=\int_0^\infty \, \frac{e^{-x} \,x^s}{(1+e^{-x})^2} \, dx\\ &=\sum_{k=1}^\infty k(-1)^{k-1} \, \int_0^\infty x^s \, e^{-kx}\, dx\\ &=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^s} \, \int_0^\infty x^s \, e^{-x}\, dx\\ &= \eta(s)\, \Gamma(s+1) \end{aligned}


If we differentiate the above result w.r. to s we obtain


\int_0^\infty \, \frac{e^{-x} \,x^s \, \ln x}{(1+e^{-x})^2} \, dx=\eta^{\prime}(s)\, \Gamma(s+1)+\eta(s)\, \Gamma^{\prime}(s+1)


Letting s \to 0


\begin{aligned} \int_0^\infty \, \frac{e^{-x} \, \ln x}{(1+e^{-x})^2} \, dx&=\eta^{\prime}(0)\, \Gamma(1)+\eta(0)\, \Gamma^{\prime}(1)\\ &=\frac{1}{2}\left(\ln \frac {\pi}{2}-\gamma \right) \qquad \blacksquare \end{aligned}


Then to compute our integral


\begin{aligned} I&=\int_0^1 \frac{\ln \, \ln \, \tfrac{1}{x}}{(1+x)^2} \, dx \\ &=\int_0^\infty \, \frac{e^{-x} \,\ln x}{(1+e^{-x})^2} \, dx \qquad ( x \mapsto e^x)\\ &=\frac{1}{2}\left(\ln \frac {\pi}{2}-\gamma \right) \qquad \blacksquare \end{aligned}

Where we used that


\eta(s)=\left(1-2^{1-s}\right) \zeta(s)(1)


Differentiating  (1)  w.r. to s


\eta^{\prime}(s)=2^{1-s} \ln 2 \zeta(s)+\left(1-2^{1-s}\right) \zeta^{\prime}(s)(2)


We have also previously proved here that

\zeta(0)=-\frac{1}{2}(3)

and

\zeta^{\prime}(0)=-\frac{1}{2} \log 2 \pi(4)

Setting s=0 in (1) and (2) we get

\eta(0) = \frac{1}{2}

\eta^{\prime}(0) =\frac{1}{2}\ln \left( \frac{\pi}{2}\right)

For the third integral, recall the previous established result


{\int_0^1 \frac{\ln\left(\ln\left(\frac{1}{x}\right)\right)}{x^2+2x \cos(\theta)+1}\,dx=\frac{\pi}{\sin\left(\theta \right)}\ln\left(\frac{\sqrt{\pi}}{\Gamma\left(\frac12-\frac{\theta}{2 \pi}\right)\sqrt{\cos\left(\frac{\theta}{2 }\right)}}\right)+\frac{\theta}{\sin\left(\theta \right)}\ln\left(\sqrt{2\pi} \right)


If we let  \theta = \frac{2\pi}{3}, then



\begin{aligned} I&=\int_{0}^{1} \frac{\ln \left( \ln \left(\frac{1}{x}\right)\right)}{1-x+x^{2}} d x \\ &=\frac{\pi}{\sin\left(\frac{2\pi}{3} \right)}\ln\left(\frac{\sqrt{\pi}}{\Gamma\left(\frac12-\frac{1}{3}\right)\sqrt{\cos\left(\frac{\pi}{3 }\right)}}\right)+\frac{\frac{2\pi}{3}}{\sin\left(\frac{2\pi}{3} \right)}\ln\left(\sqrt{2\pi} \right)\\ &=\frac{2\pi}{\sqrt{3} }\ln\left(\frac{\sqrt{\pi}}{\Gamma\left(\frac16\right)\sqrt{\frac12}}\right)+\frac{2 \pi}{\sqrt{3} \right)}\ln\left(\sqrt[3]{2 \pi}\right)\\ &=\frac{2\pi}{\sqrt{3} \right)}\ln\left(\frac{\sqrt{2\pi}\sqrt[3]{2 \pi}}{\Gamma\left(\frac16\right)}\right)\\ &=\frac{2\pi}{\sqrt{3} \right)}\ln\left(\frac{\sqrt{2\pi}\sqrt[3]{2 \pi}}{\Gamma\left(\frac16\right)}\right)\\ &=\frac{2 \pi}{\sqrt{3}} \ln \frac{\sqrt[6]{32 \pi^{5}}}{\Gamma\left(\frac{1}{6}\right)} \qquad \blacksquare \end{aligned}


For the last integral let \theta = \frac{\pi}{3} in


{\int_0^1 \frac{\ln\left(\ln\left(\frac{1}{x}\right)\right)}{x^2+2x \cos(\theta)+1}\,dx=\frac{\pi}{\sin\left(\theta \right)}\ln\left(\frac{\sqrt{\pi}}{\Gamma\left(\frac12-\frac{\theta}{2 \pi}\right)\sqrt{\cos\left(\frac{\theta}{2 }\right)}}\right)+\frac{\theta}{\sin\left(\theta \right)}\ln\left(\sqrt{2\pi} \right)


Then


\begin{aligned} I&=\int_{0}^{1} \frac{\ln \left( \ln \left(\frac{1}{x}\right)\right)}{1+x+x^{2}} d x \\ &=\frac{\pi}{\sin\left(\frac{\pi}{3} \right)}\ln\left(\frac{\sqrt{\pi}}{\Gamma\left(\frac12-\frac{1}{6}\right)\sqrt{\cos\left(\frac{\pi}{6 }\right)}}\right)+\frac{\frac{\pi}{3}}{\sin\left(\frac{\pi}{3} \right)}\ln\left(\sqrt{2\pi} \right)\\ &=\frac{2\pi}{\sqrt{3} }\ln\left(\frac{\sqrt{\pi}}{\Gamma\left(\frac13\right)\sqrt{\frac{\sqrt{3}}{2}}}\right)+\frac{2 \pi}{\sqrt{3} \right)}\ln\left(\sqrt[6]{2 \pi}\right)\\ &=\frac{2\pi}{\sqrt{3} \right)}\ln\left(\frac{\sqrt{2\pi}\sqrt[6]{2 \pi}}{\Gamma\left(\frac13\right) \sqrt[4]{3}}\right)\\ &=\frac{\pi}{\sqrt{3} \right)}\ln\left(\frac{2\pi\sqrt[3]{2 \pi}}{\Gamma^2\left(\frac13\right)\sqrt{3}}\right)\\ &=\frac{\pi}{3} \ln \left(\frac{\Gamma\left(\frac{2}{3}\right)\sqrt[3]{2 \pi}}{\Gamma\left(\frac{1}{3}\right)} \right) \qquad \blacksquare \end{aligned}


Where we used the reflection formula

\Gamma\left(x\right)\Gamma\left(1-x\right)=\frac{\pi}{\sin(\pi x)}

To write

\Gamma\left(\frac13\right)\Gamma\left(\frac23\right)=\frac{2\pi}{\sqrt{3}}

Comments

Post a Comment

Popular posts from this blog