I found, numerically, that $$\Im\Li(\sqrt i(\sqrt 2-1))=\frac34G+\frac18\pi\ln(\sqrt2-1).$$ How can we prove it?

My attempt of proving this equation: Using identity $$\Li(x)=\int_0^1\frac{x}{xt-1}\ln tdt,$$ we can deduce$$\begin{align}\Im\Li(\sqrt i(\sqrt 2-1))&=\frac1{2i}\int_0^1\left(\frac{\sqrt i(\sqrt2-1)}{\sqrt i(\sqrt2-1)t-1}-\frac{\sqrt {-i}(\sqrt2-1)}{\sqrt {-i}(\sqrt2-1)t-1}\right)\ln tdt\\ &=\int_0^1\frac{2-\sqrt{2}}{\left(4 \sqrt{2}-6\right) t^2-2 \left(\sqrt{2}-2\right) t-2}\ln tdt\\ &=\int_0^{2-\sqrt2}-\frac{1}{u^2-2u+2}\ln\frac u{2-\sqrt 2}du\\ &=\frac18\pi\ln(2-\sqrt2)-\int_{-1}^{1-\sqrt2}\frac{\ln(v+1)}{v^2+1}dv\\ &=\frac18\pi\ln(2-\sqrt2)-\int_{\pi/8}^{\pi/4}\ln(1-\tan x)dx\\ &=\frac18\pi\ln(\sqrt2-1)-\int_{\pi/8}^{\pi/4}\ln\sec x+\ln\sin\left(\frac\pi4-x\right)dx\\ \end{align}$$ I have no idea how to deal with the log-trig integral.

  • $\begingroup$ Try reducing the lower limit of your integral to zero and then use the identity ; $$G = \int_0^{π/4}\log \cot \theta d\theta $$ $\endgroup$ – Awe Kumar Jha Dec 7 '18 at 11:49
  • $\begingroup$ $\log\sin x$ and $\log\cos x$ have well-known Fourier series, and you just have to perform a termwise integration of them. $\endgroup$ – Jack D'Aurizio Dec 7 '18 at 17:06
  • $\begingroup$ What is $\sqrt{i}$? $\endgroup$ – FDP Dec 7 '18 at 18:14
  • $\begingroup$ @FDP $\sqrt i= e^{\pi i/4}=\frac{1+i}{\sqrt 2}$. $\endgroup$ – Kemono Chen Dec 8 '18 at 0:55
  • $\begingroup$ why not \begin{align}\sqrt i= -e^{\pi i/4}?\end{align} if $x_0^2=a$ then $(-x_0)^2=a$. $\endgroup$ – FDP Dec 8 '18 at 11:30

$\newcommand{\Cl}{\operatorname{Cl}}$In order to dodge the extensive usage of the Fourier Series Expansions I will use the Clausen Function $\Cl_2(z)$ to shorten things up; nevertheless the result will remain the same as one could get going all along the long way. Anyway, the only difficulty that remains after your attempt is the evaluation of the following integral

$$\mathfrak{I}~=~-\int_{\pi/8}^{\pi/4}\log(\sec x)+\log\left(\sin \frac\pi4-x\right)\mathrm{d}x\tag1$$

First of all I will do a bit of reshaping to actually apply useful formulae involving the Clausen Function. Therefore split up the first integral and enforce the substitution $x+\frac\pi4\mapsto x$ within the second integral to get

$$\begin{align*} \mathfrak{I}&=-\int_{\pi/8}^{\pi/4}\log(\sec x)+\log\left(\sin \frac\pi4-x\right)\mathrm{d}x\\ &=\int_0^{\pi/4}\log(\cos x)\mathrm{d}x-\int_0^{\pi/8}\log(\cos x)\mathrm{d}x-\int_{\pi/8}^{\pi/4}\log\left(\cos x+\frac \pi4\right)\mathrm{d}x\\ &=\int_0^{\pi/4}\log(\cos x)\mathrm{d}x-\int_0^{\pi/8}\log(\cos x)\mathrm{d}x-\int_{3\pi/8}^{\pi/2}\log(\cos x)\mathrm{d}x\\ &=\int_0^{\pi/4}\log(\cos x)\mathrm{d}x-\int_0^{\pi/8}\log(\cos x)\mathrm{d}x-\int_0^{\pi/2}\log(\cos x)\mathrm{d}x+\int_0^{3\pi/8}\log(\cos x)\mathrm{d}x \end{align*}$$

Now it is time to apply the first useful formula of the Clausen Function, namely

$$\int_0^t \log(\cos x)\mathrm{d}x~=~\frac12\Cl_2(\pi-2t)-t\log(2)\tag2$$

Formula $(2)$ can be shown rather easy be utilizing the well-known Fourier Series Expansion of $\log(\cos x)$ combined with the series representation of the $\Cl_2(z)$ function. However, with this knowledge we can rewrite the integrals from above in terms of the Clausen Function to get

$$\begin{align*} \mathfrak{I}&=\frac12\Cl_2\left(\pi-2\frac\pi4\right)-\frac12\Cl_2\left(\pi-2\frac\pi8\right)-\frac12\Cl_2\left(\pi-2\frac\pi2\right)+\frac12\Cl_2\left(\pi-2\frac{3\pi}8\right)\\ &~~~\underbrace{-\frac\pi4\log(2)+\frac\pi8\log(2)+\frac\pi2\log(2)-\frac{3\pi}8\log(2)}_{=0}\\ &=\frac12\left[\Cl_2\left(\frac\pi2\right)-\Cl_2\left(0\right)+\Cl_2\left(\frac\pi4\right)-\Cl_2\left(\frac{3\pi}4\right)\right] \end{align*}$$

We are almost done! It is time to throw some more important formulae in: firstly the already mentioned series representation and secondly the so-called Duplication Formula

$$\begin{align*} \Cl_2(z)~&=~\sum_{n=1}^\infty \frac{\sin(nz)}{n^2}\tag3\\ \Cl_2(2z)~&=~2\Cl_2(z)-2\Cl_2(\pi-z)\tag4 \end{align*}$$

From $(3)$ we can direcetly conclude that $\Cl_2\left(\frac\pi2\right)=G$ and that $\Cl_2(0)=0$ where $G$ denotes Catalan's Constant. Using the $(4)$ with $z=\frac\pi4$ we get a representation for the other terms from above. Putting this all together gives us the final value

$$\begin{align*} \mathfrak{I}&=\frac12\left[\underbrace{\Cl_2\left(\frac\pi2\right)}_{=G}-\underbrace{\Cl_2\left(0\right)}_{=0}+\underbrace{\Cl_2\left(\frac\pi4\right)-\Cl_2\left(\frac{3\pi}4\right)}_{=\frac G2}\right]\\ &=\frac12\left[G+\frac G2\right] \end{align*}$$

$$\therefore~\mathfrak{I}~=~-\int_{\pi/8}^{\pi/4}\log(\sec x)+\log\left(\sin \frac\pi4-x\right)\mathrm{d}x~=~\frac34 G$$

I recommend to study the Clausen Function hence it reduces the number of caculations needed for linear logarithmo-trigonometric integrals tremendously. If you are feeling uncomfortable with a part of the proof let me know and I will try to clear your doubts.


\begin{align} I&=-\int_{\pi/8}^{\pi/4}\log(\sec x)+\log\sin \left(\frac\pi4-x\right)\ dx\\ &=\int_{\pi/8}^{\pi/4}\log(\cos x)\ dx-\int_{\pi/8}^{\pi/4}\log\sin \left(\frac\pi4-x\right)\ dx\\ &\{\text{break the interval for the first integral and let } \frac{\pi}{4}-x\mapsto x \text{ for the second integral}\}\\ &=\int_{0}^{\pi/4}\log(\cos x)\ dx-\int_{0}^{\pi/8}\log(\cos x)\ dx-\int_{0}^{\pi/8}\log(\sin x)\ dx\\ &=\int_{0}^{\pi/4}\log(\cos x)\ dx-\int_{0}^{\pi/8}\log(\cos x\ \sin x)\ dx\\ &=\int_{0}^{\pi/4}\log(\cos x)\ dx-\int_{0}^{\pi/8}\log\left(\frac{\sin(2x)}{2}\right)\ dx\\ &=\int_{0}^{\pi/4}\log(\cos x)\ dx-\frac12\int_{0}^{\pi/4}\log\left(\frac{\sin x}{2}\right)\ dx,\quad \{\ln(\cos x)=\ln(\sin x)-\ln(\tan x)\}\\ &=\int_{0}^{\pi/4}\log(\sin x)\ dx-\int_{0}^{\pi/4}\log(\tan x)\ dx-\frac12\int_{0}^{\pi/4}\log(\sin x)\ dx+\frac12\int_{0}^{\pi/4}\ln2\ dx\\ &=\frac12\int_{0}^{\pi/4}\log(\sin x)\ dx-\int_{0}^{\pi/4}\log(\tan x)\ dx+\frac12\int_{0}^{\pi/4}\ln2\ dx\\ &=\frac12\left(-\frac{G}{2}-\frac{\pi}{4}\ln2\right)-(-G)+\frac12\ln2\left(\frac{\pi}{4}\right)\\ &=\frac34G \end{align}


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