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Hello I am trying to solve an incredible integral given by $$ \int_0^{\pi/2}\ln\big[1-\cos^2 x(\sin^2\alpha-\sin^2\beta \sin^2 x)\big]dx=\pi \ln\bigg[\frac{1}{2}\left(\cos^2\alpha +\sqrt{\cos^4 \alpha +\cos^2\frac{\beta}{2} \sin^2 \frac{\beta}{2}}\right)\bigg],\qquad \alpha > \beta >0. $$ I defined $$ I\equiv \int_0^{\pi/2}\ln\big[1-\cos^2 x(\sin^2\alpha-\sin^2\beta \sin^2 x)\big]dx $$ and using $\cos^2 x=1-\sin^2 x$ but obtained a more complicated expression. Usually it is easier to work with a closed form of Log Sine so I was trying this. I am not really sure how else to approach this at all.

The result looks like very nice and similar to something we all know :)

I am looking for real or complex methods to solve this problem. I am not sure of what substitutions to make but maybe we could work in hyperbolic space

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  • $\begingroup$ Note that $sin^2 \frac {\beta} {2} cos^2 \frac {\beta} {2} = 1/4 \sin^2 \beta$. Hence you can simply abbreviate $a = \cos^2 \alpha, b = \sin^2 \beta$ everywhere. Next, you could differentiate with respect to $a$ to obtain a rational function of $\sin x$ and $\cos x$. $\endgroup$ – user111187 Apr 30 '14 at 13:00
  • $\begingroup$ Have you checked this result numerically? I tried a few values and it didn't match. Perhaps I made a mistake. $\endgroup$ – user111187 Apr 30 '14 at 17:03
  • $\begingroup$ @user111187 I have checked it yes $\endgroup$ – Jeff Faraci Apr 30 '14 at 17:05
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\pi/2} \ln\pars{1-\cos^{2}\pars{x}\bracks{\sin^{2}\pars{\alpha} - \sin^{2}\pars{\beta} \sin^{2}\pars{x}}}\,\dd x:\ {\large ?}}$

\begin{align} &\color{#c00000}{\int_{0}^{\pi/2} \ln\pars{1-\cos^{2}\pars{x}\bracks{\sin^{2}\pars{\alpha} - \sin^{2}\pars{\beta} \sin^{2}\pars{x}}}\,\dd x} \\[3mm]&=\half\int_{-\pi/2}^{\pi/2} \ln\pars{1-\cos^{2}\pars{x}\bracks{\sin^{2}\pars{\alpha} - \sin^{2}\pars{\beta} \sin^{2}\pars{x}}}\,\dd x \\[3mm]&=\half\int_{-\pi/2}^{\pi/2} \ln\pars{1-{1 + \cos\pars{2x} \over 2}\bracks{\sin^{2}\pars{\alpha} - \sin^{2}\pars{\beta}\,{1 - \cos\pars{2x} \over 2}}}\,\dd x \\[3mm]&={1 \over 4}\int_{-\pi}^{\pi} \ln\pars{1 - \half\,\sin^{2}\pars{\alpha} + {1 \over 4}\,\sin^{2}\pars{\beta} -\bracks{\half\,\sin^{2}\pars{\alpha} + {1 \over 4}\,\sin^{2}\pars{\beta}}\cos\pars{x}}\,\dd x \end{align}

Let's $$ \verts{\sin\pars{\alpha}} = {\root{2}\sin\pars{\theta/2} \over a}\,,\quad \verts{\sin\pars{\beta}} = {2\cos\pars{\theta/2} \over a}\,,\qquad 0 \leq \theta < \pi $$ such that \begin{align} &-\,\half\,\sin^{2}\pars{\alpha} + {1 \over 4}\,\sin^{2}\pars{\beta} ={\cos\pars{\theta} \over a^{2}}\,,\qquad \half\,\sin^{2}\pars{\alpha} + {1 \over 4}\,\sin^{2}\pars{\beta} ={1 \over a^{2}} \\[3mm]&\mbox{and}\quad\theta =2\arctan\pars{\root{2}\,{\verts{\sin\pars{\alpha}} \over \verts{\sin\pars{\beta}}}} \,,\qquad \verts{a} = {2 \over \root{2\sin^{2}\pars{\alpha} + \sin^{2}\pars{\beta}}} \end{align} Note that $\ds{\verts{a} \geq {2\root{3} \over 3} \approx 1.1547 > 1}$.

Then, \begin{align} &\color{#c00000}{\int_{0}^{\pi/2} \ln\pars{1-\cos^{2}\pars{x}\bracks{\sin^{2}\pars{\alpha} - \sin^{2}\pars{\beta} \sin^{2}\pars{x}}}\,\dd x} \\[3mm]&={1 \over 4}\int_{-\pi}^{\pi} \ln\pars{1 + {\cos\pars{\theta} \over a^{2}} - {\cos\pars{x} \over a^{2}}}\,\dd x \\[3mm]&=\color{#c00000}{-\pi\ln\pars{\verts{a}}+{1 \over 4}\int_{-\pi}^{\pi} \ln\pars{\mu - \cos\pars{x}}\,\dd x}\,,\qquad \mu \equiv a^{2} + \cos\pars{\theta} \\[3mm]&\mbox{with}\quad \mu \geq {2\root{3} \over 3} + \cos\pars{\theta} \geq {2\root{3} - 3 \over 3} \approx 0.1547 \end{align}

The answer is \begin{align} &\color{#00f}{\large\int_{0}^{\pi/2} \ln\pars{1-\cos^{2}\pars{x}\bracks{\sin^{2}\pars{\alpha} - \sin^{2}\pars{\beta} \sin^{2}\pars{x}}}\,\dd x} \\[3mm]&=\color{#00f}{\large% -\pi\ln\pars{\verts{a}} +\half\,\pi\ln\pars{\mu + \root{\mu^{2} - 1} \over 2}}\,,\qquad\mu > 1 \end{align}

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Here is a sketchy calculation. It matches numerically for all values of $a,b$ that I tried.

Abbreviate $a = \cos^2 \alpha$ and $b = \sin^2 \beta$. Then $0 \leq a,b \leq 1$. We calculate $$ I(a,b)= \int _{0} ^{\pi/2} \ln(1-\cos^2 x(1-a-b \sin^2 x)) \,dx $$ We have $$ \partial_aI(a,b)= \int _{0} ^{\pi/2} \frac {\cos^2 x} {1-\cos^2 x(1-a-b \sin^2 x)} \,dx \\ = \int _{0} ^{\pi/2} \frac {1} {\tan^2 x+a+b \sin^2 x} \,dx \\ =\int _{0} ^{\infty} \frac {1} {(1+t^2)(t^2+a+b \frac {t^2} {1+t^2})} \,dt \\ $$ (by substituting $\tan x = t$ and using some trig identities) $$ = \frac{1}{2} \int _{-\infty} ^{\infty} \frac {1} {t^4+(a+b+1)t^2+a} \,dt \\ = \frac{1}{2} \frac {\pi} {\sqrt{a(2 \sqrt{a}+a+b+1)}} $$ (for example using residues; here it is crucial that $(a+b+1)^2 - 4a^2 \geq 0 $, which is the case because $0 \leq a,b \leq 1$)

By integrating with respect to $a$ (for example by setting $u = \sqrt{a}$), we hence obtain $$ I(a,b) = \pi \ln{ \left(1 + \sqrt{a} + \sqrt{1 + 2 \sqrt{a} + a + b} \right) } + C(b) $$ Here $C(b)$ is an integration constant depending on $b$ only. To fix it, we substitute $a = 1$ in the integral. We obtain, using some trig identities,

$$ I(1,b) = \int _{0} ^{\pi/2} \ln(1+b \sin^2 x \cos^2 x) \,dx \\ = \frac{1}{2} \int _{0} ^{\pi} \ln\left(1+ \frac{1}{4} b \sin^2 x\right) \,dx \\ = \int _{0} ^{\pi/2} \ln\left(1+ \frac{1}{4} b \sin^2 x\right) \,dx \\ = \pi \ln \left(\frac{1+\sqrt{1+b/4}}{2} \right) \\ $$ Here I used a previously calculated integral. On the other hand, we have $I(1,b) = C(b) + \pi \ln(2(1+\sqrt{1+b/4}))$. Comparing yields the value of $C(b)$. Plugging this back in, we get: $$ I(a,b)=\pi \ln\left( \frac{1+ \sqrt{a} + \sqrt{2\sqrt{a} + a+b + 1}}{4} \right) $$

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  • $\begingroup$ I am not sure this is equivalent to what I have. You do always have clever solutions, so I am curious I will go through this. I have a russian source or american source where this came from, which do you prefer? It is a quite standard integral, and I have many variations of it too $\endgroup$ – Jeff Faraci Apr 30 '14 at 21:48
  • $\begingroup$ Can you give a link or mention the book and page number of your sources? Also please double-check that you copied it correctly into your question. I am quite confident that this solution is correct, because Mathematica agrees. $\endgroup$ – user111187 Apr 30 '14 at 21:54
  • $\begingroup$ I have just verified again what I posted in the question, it is correct.,.. Ok pg 561 in Gradshteyn, Solomonovich ; Ryzhik, Iosif Moiseevich (Russian version) $\endgroup$ – Jeff Faraci Apr 30 '14 at 21:59
  • $\begingroup$ also would you like me to e-mail you anything in regards to these integrals? $\endgroup$ – Jeff Faraci Apr 30 '14 at 22:05

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