Find the value of the integral $$\int_0^{2\pi}\ln|a+b\sin x|dx$$ where $0\lt a\lt b$. What is the use of this inequality. I tried to integrate the integral by parts, but the integral of the 2nd term was quite messy.Please help.
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I looks like complex integral could be used here. – Nikita Evseev May 19 '14 at 05:04
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The answer is of the form m(3.141)ln(nb) where m,n are real numbers and mn=1. – User56192 May 19 '14 at 05:15
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1In both cases we have $I=\Re~\bigg{2\pi\cdot\ln\bigg[\dfrac a2\cdot\bigg(1+\sqrt{1-\dfrac{b^2}{a^2}}~\bigg)\bigg]\bigg}$. – Lucian May 19 '14 at 05:38
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Ha! AITS... integer type paper 1. – evil999man May 19 '14 at 07:42
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@Awesome: Which AITS was it? – Pranav Arora May 19 '14 at 07:55
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@PranavArora FIITJEE AITS Full Test XI – evil999man May 19 '14 at 07:59
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Ah, I guessed so, they really ask terrible problems. :D – Pranav Arora May 19 '14 at 08:01
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@PranavArora You have probably not seen the Open Test conducted by them. – evil999man May 19 '14 at 08:03
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I have seen most of the papers but not this one. I don't remember much about the Open test, was it too hard? :P – Pranav Arora May 19 '14 at 08:04
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@PranavArora It is just 4-5 times difficulty level of AITS. This is the paper they prepare for rest of the year. – evil999man May 19 '14 at 08:13
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@Awesome: Was a solution provided by FIITJEE to this problem? – Pranav Arora May 19 '14 at 09:22
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1@PranavArora No... They are too lazy. – evil999man May 19 '14 at 09:27
2 Answers
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This is just for $0<b < a $
\begin{align*} I &= \int_0^{2\pi}\log(a+\sqrt{a^2 - b^2}+be^{i\theta})d\theta + \int_0^{2\pi}\log(a+\sqrt{a^2 - b^2}+be^{-i\theta})d\theta\\ &= \int_0^{2\pi} \log \left(2a^2 +2a\sqrt{a^2 - b^2} + (a + \sqrt {a^2 - b^2}) b (e^{i\theta} + e^{-i\theta}) \right )d\theta\\ &= \int_0^{2\pi} \log (2 (a + \sqrt{a^2 - b^2}))d\theta + \int_0^{2\pi }\log(a + b \cos\theta)d\theta\\ \end{align*}
By Gauss MVT, the value of top two integral is $$4\pi\log(a + \sqrt{a^2 -b^2})$$ Hence the for $0<b<a$ is $$\int_0^{2\pi} \log(a+ b\cos\theta)d\theta = 2 \pi \log \left(a + \sqrt{a^2 - b^2}\over 2\right)$$
EDIT:: The integral for $\log (a + b \sin \theta )$ is equal to that of $\log (a + b \cos \theta )$ as $\theta$ goes around $2 \pi$. It can also be done by changing $be^{-i\theta}$ to $-b e^{-i\theta}$ is above derivation.
For $0<a < b$, the integral is equal to real part of the above solution as mentioned by @Lucian in comments.
S L
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{0 < a < b}$.
\begin{align} &\int_{0}^{2\pi}\ln\pars{\verts{a + b\sin\pars{x}}}\,\dd x= 2\pi\ln\pars{\verts{b}} +\color{#c00000}{\int_{0}^{2\pi}\ln\pars{\verts{\mu + \sin\pars{x}}}\,\dd x} \\[3mm]&\mbox{where}\quad \mu \equiv {a \over b}\,,\quad\mbox{Note than}\quad 0 < \mu < 1 \end{align}
\begin{align} &\color{#c00000}{\int_{0}^{2\pi}\ln\pars{\verts{\mu + \sin\pars{x}}}\,\dd x} =\int_{-\pi}^{\pi}\ln\pars{\verts{\mu - \sin\pars{x}}}\,\dd x =\sum_{\sigma = \pm}\int_{0}^{\pi}\ln\pars{\verts{\mu + \sigma\sin\pars{x}}}\,\dd x \\[3mm]&=\sum_{\sigma = \pm}\int_{-\pi/2}^{\pi/2} \ln\pars{\verts{\mu + \sigma\cos\pars{x}}}\,\dd x =2\sum_{\sigma = \pm}\int_{0}^{\pi/2} \ln\pars{\verts{\mu + \sigma\cos\pars{x}}}\,\dd x \\[3mm]&=2\int_{0}^{\pi/2}\ln\pars{\verts{\mu - \cos\pars{x}}}\,\dd x +2\int_{0}^{\pi/2}\ln\pars{\mu + \cos\pars{x}}\,\dd x \\[3mm]&=2\int_{0}^{\tilde{\mu}}\ln\pars{\cos\pars{x} - \mu}\,\dd x +2\int^{\pi/2}_{\tilde{\mu}}\ln\pars{\mu - \cos\pars{x}}\,\dd x +2\int_{0}^{\pi/2}\ln\pars{\mu + \cos\pars{x}}\,\dd x \end{align} where $\ds{\tilde{\mu} = \arccos\pars{\mu}}$.
$$ \color{#c00000}{\!\!\!\!\!\int_{0}^{2\pi}\ln\pars{\verts{\mu + \sin\pars{x}}}\,\dd x} = 2\int_{0}^{\tilde{\mu}}\ln\pars{\cos^{2}\pars{x} - \mu^{2}}\,\dd x +2\int^{\pi/2}_{\tilde{\mu}}\ln\pars{\mu^{2} - \cos^{2}\pars{x}}\,\dd x $$
Can you take it from here ?.
Felix Marin
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