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Trying to find another way to answer this question, I arrived at a point where I cannot simplify anymore.

The solution being $$I_1=\int_0^{\frac \pi 2}\csc (x) \log \left(\frac{1+b \sin (x)}{1-b \sin (x)}\right)=\pi \sin ^{-1}(b)$$

I am stuck with $$I_2=\frac 12\Bigg[a_+ \Phi \left(a_+^2,2,\frac{1}{2}\right)+a_- \Phi \left(a_-^2,2,\frac{1}{2}\right)\Bigg]\quad \text{where}\quad a_\pm=b\pm\sqrt{b^2-1}$$ $\Phi(.)$ being the Lerch transcendent function.

How to prove that, for $\color{red}{-1 \le b \le 1}$, $\color{red}{I_2=I_1}$ ?

Claude Leibovici
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1 Answers1

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As $-1\le b\le 1$, we define \begin{equation} b=\cos\beta \end{equation} with $0\le\beta\le\pi$. Then \begin{align} a_{\pm}&=b\pm\sqrt{b^2-1}\\ &=e^{\pm i\beta} \end{align} From the definition of the Lerch transcendent function, \begin{equation} \Phi(z,s,a)=\sum_{k=0}^\infty\frac{z^k}{\left[(a+k)^2\right]^{s/2}} \end{equation} for $|z|<1 \bigvee \left(|z|=1\bigwedge \Re s>1\right)\bigwedge -a\notin \mathbb{N}$, we have \begin{align} I_2&=\frac 12\left[a_+ \Phi \left(a_+^2,2,\frac{1}{2}\right)+a_- \Phi \left(a_-^2,2,\frac{1}{2}\right)\right]\\ &=\frac12\left[e^{i\beta}\Phi\left(e^{2i\beta},2,\frac12\right)+e^{-i\beta}\Phi\left(e^{-2i\beta},2,\frac12\right)\right]\\ &=4\sum_{k=0}^\infty\frac{\cos\left((2k+1)\beta\right)}{(2k+1)^2} \end{align} Using the classical Fourier series, \begin{equation} |\beta|=\frac\pi2-\frac{4}{\pi}\sum_{k=0}^\infty\frac{\cos\left((2k+1)\beta\right)}{(2k+1)^2} \end{equation} we deduce \begin{align} I_2&=\pi\left( \frac\pi2-\beta \right)\\ &=\pi\sin^{-1}(b) \end{align} as expected.

Paul Enta
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  • Hi Paul ! Nice way to do it. I did not think about Fourier series (which I did not use a single time in the last sixty years or more). $\to +1$ for sure (at least for the time being). Cheers :-) – Claude Leibovici Jan 15 '22 at 09:54
  • Hi Claude! In 2022, Fourier series are back! :-) Cheers. – Paul Enta Jan 15 '22 at 10:01
  • Paul : just out of curiosity, what is $I_2$ for $|b|>1$. Its real part is $\frac{\pi ^2}{2}$ (OK with that) but what about the imaginary part ? – Claude Leibovici Jan 15 '22 at 15:50
  • One can express the Lerch function in terms of polylog here $$\Phi(z,2,1/2)=\frac2{\sqrt z}\left[\operatorname{Li}2(\sqrt z)-\operatorname{Li}_2(\sqrt z)\right]$$ and with the identity $$\operatorname{Li}_2(z)+\operatorname{Li}_2(\frac1z)=-\frac{\pi^2}6-\frac12\log^2(-z)$$ as $a+a_-=1$, we obtain $$I_2=\frac{\pi^2}{2}-i\pi\log (a_+)$$ which seems numerically correct. – Paul Enta Jan 15 '22 at 22:33