MAMLSTEN INTEGRALS - PART I

We will today prove the following integral that belongs to a family of log-log integrals known as Malmsten integrals which Vardi´s integral is a particular case:


\int_{0}^{1} \frac{\ln \ln \frac{1}{x}}{1+2 x \cos 2 \pi \alpha+x^{2}} d x=\frac{\pi}{\sin 2 \pi \alpha} \ln \frac{\sqrt{\pi}(2 \pi)^{\alpha}}{\Gamma\left(\frac{1}{2}-\alpha\right) \sqrt{\cos \pi \alpha}}


Mamlsten Integrals


\begin{aligned} I&=\int_0^1 \frac{\ln\left(\ln\left(\frac{1}{x}\right)\right)}{x^2+2x \cos(\theta)+1}\,dx\\ &=\int_0^1 \frac{\ln\left(\ln\left(\frac{1}{x}\right)\right)}{x^2+\left(e^{i \theta}+e^{-i \theta}\right)x+1}\,dx\\ &=\int_0^1 \frac{\ln\left(\ln\left(\frac{1}{x}\right)\right)}{(1+xe^{-i \theta})(1+xe^{i \theta})}\,dx\\ &=\int_0^1 \ln\left(\ln\left(\frac{1}{x}\right)\right) \left( \left(\sum_{k=0}^\infty (-1)^k e^{-i \theta n}x^k\right) \cdot \left(\sum_{k=0}^\infty (-1)^k e^{i \theta n}x^k\right)\right)\,dx\\ &=\int_0^1 \ln\left(\ln\left(\frac{1}{x}\right)\right)\left(\frac{1}{\sin\left(\theta \right)} \sum_{k=1}^\infty(-1)^{k-1} \sin(k \theta) x^{k-1}\right)\,dx\\ &=\frac{1}{\sin\left(\theta \right)} \sum_{k=1}^\infty(-1)^{k-1} \sin(k \theta)\int_0^1 \ln\left(\ln\left(\frac{1}{x}\right)\right)x^{k-1}\,dx\\ &=\frac{1}{\sin\left(\theta \right)} \sum_{k=1}^\infty(-1)^{k-1} \sin(k \theta)\left(-\frac{\gamma}{k} -\frac{\ln(k)}{k} \right)\\ &=-\frac{\gamma}{\sin\left(\theta \right)} \sum_{k=1}^\infty \frac{(-1)^{k-1} \sin(k \theta)}{k}-\frac{1}{\sin\left(\theta \right)} \sum_{k=1}^\infty \frac{(-1)^{k-1} \ln(k) \sin(k \theta)}{k}\\ &=-\frac{\gamma \theta}{2\sin\left(\theta \right)}-\frac{1}{\sin\left(\theta \right)} \left(\pi\ln\left(\Gamma\left(\frac12-\frac{\theta}{2 \pi}\right)\right)+\frac{\pi\ln\left(\sqr\cos\left(\frac{\theta}{2}\right)\right)}{2}-\left( \frac{\theta}{2 }\right)\left(\gamma+\ln(2)\right)-\left(\frac{\pi}{2}+\frac{\theta}{2 }\right)\ln(\pi) \right)\\ &=\frac{\pi}{\sin\left(\theta \right)} \left(-\ln\left(\Gamma\left(\frac12-\frac{\theta}{2 \pi}\right)\right)-\ln\left(\sqrt{\cos\left(\frac{\theta}{2 }\right)}\right)+\frac{\theta}{\pi}\ln(\sqrt{2\pi})+\ln(\sqrt{\pi}) \right)\\ &=\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) \qquad \blacksquare \end{aligned}


\boxed{\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) }(1)


If we let   \theta \to 2 \pi \alpha  in (1) we obtain the desired result:


\boxed{\int_{0}^{1} \frac{\ln \ln \frac{1}{x}}{1+2 x \cos 2 \pi \alpha+x^{2}} d x=\frac{\pi}{\sin 2 \pi \alpha} \ln \frac{\sqrt{\pi}(2 \pi)^{\alpha}}{\Gamma\left(\frac{1}{2}-\alpha\right) \sqrt{\cos \pi \alpha}}}
(2)

Appendix:


Cauchy Product


\begin{aligned} P&=\left(\sum_{n=0}^\infty a_nx^n\right) \cdot \left(\sum_{n=0}^\infty b_nx^n\right)\\ &=\left(a_0+a_1x+a_2x^2+\cdots\right) \cdot \left(b_0+b_1x+b_2x^2+\cdots\right)\\ &=a_0b_0+\left(a_0b_1x+a_1b_0x \right)+\left(a_0b_2x^2+a_1b_1x^2+a_2b_0x^2 \right)+\left(a_0b_3x^3+a_1b_2x^3+a_2b_1x^3+a_3b_0x^3 \right)+\cdots\\ &=a_0b_0+\left(a_0b_1+a_1b_0 \right)x+\left(a_0b_2+a_1b_1+a_2b_0 \right)x^2+\left(a_0b_3+a_1b_2+a_2b_1+a_3b_0 \right)x^3+\cdots\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n a_kb_{n-k} \right)x^n \end{aligned}


Example

\begin{aligned} P&= \left(\sum_{n=0}^\infty (-1)^ne^{-i \theta n}x^n\right) \cdot \left(\sum_{n=0}^\infty (-1)^ne^{i \theta n}x^n\right)\\ &=\left(\sum_{n=0}^\infty (-e^{-i \theta })^nx^n\right) \cdot \left(\sum_{n=0}^\infty (-e^{i \theta })^nx^n\right) \\ \end{aligned}


We have that a_k=(-e^{-i \theta })^k and b_k=(-e^{i \theta })^k , we therefore obtain


\begin{aligned} P&= \left(\sum_{n=0}^\infty (-1)^ne^{-i \theta n}x^n\right) \cdot \left(\sum_{n=0}^\infty (-1)^ne^{i \theta n}x^n\right)\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n (-e^{-i \theta })^k(-e^{i \theta })^{n-k} \right)x^n\\ &=\sum_{n=0}^\infty(-1)^n\left(\sum_{k=0}^n e^{i n \theta }e^{-2i k \theta } \right)x^n\\ &=\sum_{n=0}^\infty(-1)^n e^{i n \theta }\left(\sum_{k=0}^n e^{-2i k \theta } \right)x^n\\ &=\sum_{n=0}^\infty(-1)^n e^{i n \theta }\left(\frac{1-e^{-2i (n+1) \theta }}{1-e^{-2i \theta }} \right)x^n \qquad \left(\text{geometric sum} \right)\\ &=\sum_{n=0}^\infty(-1)^n e^{i n \theta }\frac{e^{i \theta}}{e^{i \theta}}\left(\frac{1-e^{-2i (n+1) \theta }}{1-e^{-2i \theta }} \right)x^n \\ &=\sum_{n=0}^\infty(-1)^n\left(\frac{e^{i (n+1)\theta}-e^{-i (n+1) \theta }}{e^{i \theta}-e^{-i \theta }} \right)x^n \\ &=\sum_{n=0}^\infty(-1)^n\left(\frac{\sin\left((n+1)\theta \right)}{\sin\left(\theta \right)} \right)x^n \\ &=\sum_{n=1}^\infty(-1)^{n-1}\left(\frac{\sin\left(n\theta \right)}{\sin\left(\theta \right)} \right)x^{n-1}\\ &=\frac{1}{\sin\left(\theta \right)} \sum_{n=1}^\infty(-1)^{n-1} \sin(n \theta) x^{n-1}\qquad \blacksquare\\ \end{aligned}

Evaluation of the integral:


\begin{aligned} I&=\int_0^1 x^{n-1}\ln\left(\ln\left( \frac{1}{x}\right) \right)\,dx\\ &=\int_0^\infty \left(e^{-x}\right)^{n-1}\ln\left(x\right) e^{-x}\,dx \qquad \left(\frac{1}{x} \to x\right)\\ &=\int_0^\infty e^{-nx} \ln\left(x\right) \,dx \\ &=\frac{1}{n}\int_0^\infty e^{-x} \ln\left(\frac{x}{n}\right) \,dx \qquad (nx \to x)\\ &=\frac{1}{n}\int_0^\infty e^{-x} \ln\left(x \right) \,dx -\frac{\ln(n)}{n}\int_0^\infty e^{-x} \,dx \\ &=-\frac{\gamma}{n} -\frac{\ln(n)}{n} \qquad \blacksquare\\ \end{aligned}

Recall the sine of a difference formula


\sin(x-y)=\sin(x)\cos(y)-\sin(y)\cos(x)


And The Fourier series for the LogGamma function proved here


\ln(\Gamma(x))= \Big(\gamma+\log(2)\Big)\Big( \frac{1}{2}- x\Big)+(1-x)\log(\pi)-\frac{\ln(\sin(\pi x))}{2}+\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log( k)\sin(2 \pi k x)}{ k}


If we let      2 \pi k x =\pi k- \theta k \,\Rightarrow x=\frac12-\frac{\theta}{2 \pi}


We obtain


\ln\left(\Gamma\left(\frac12-\frac{\theta}{2 \pi}\right)\right)= \left( \frac{\theta}{2 \pi}\right)\left(\gamma+\log(2)\right)+\left(\frac12+\frac{\theta}{2 \pi}\right)\log(\pi)-\frac{\log\left(\cos\left(\frac{\theta}{2}\right)\right)}{2}+\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}\log( k)\sin\left(k\theta \right)}{ k}


Or


\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}\log( k)\sin\left(k\theta \right)}{ k}=\ln\left(\Gamma\left(\frac12-\frac{\theta}{2 \pi}\right)\right)+\frac{\log\left(\cos\left(\frac{\theta}{2}\right)\right)}{2}-\left( \frac{\theta}{2 \pi}\right)\left(\gamma+\log(2)\right)-\left(\frac12+\frac{\theta}{2 \pi}\right)\log(\pi)


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