1
$\begingroup$

I am solving this question -

$\int_0^{\frac{\pi}{2}} log(\sin x)dx$

But I am unable to solve. Then I see this link.

In George Lowther answer. I understand how he reached to

$\frac12\int_0^{\pi/2}\log(\sin x\cos x)\,dx$

But I am unable to understand next step. And how this leads to an answer.

  • 1
    $$\sin(x)\cos(x)=\frac{1}{2}\sin(2x)$$ – 2017-02-12
  • 0
    I have a detailed answer here http://math.stackexchange.com/questions/1992462/prove-int-01-frac-sin-1xx-dx-frac-pi2-ln2/1992580#1992580 – 2017-02-12
  • 0
    @Jack D'Aurizio and how last step leads to the answer? – 2017-02-12
  • 0
    @Amar: after that you perform a change of variable and get something like $$ I = \frac{1}{2}I+K $$ from which $I=2K$ follows. – 2017-02-12
  • 0
    Can you solve it please. – 2017-02-12
  • 0
    @Amar Please see my answer I linked you. – 2017-02-12
  • 0
    @Ahmed S.Attaalla any simple method? – 2017-02-12

1 Answers 1

2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\quad\ln\pars{z} = \ln\pars{\verts{z}} + \,\mrm{arg}\pars{z}\ic\,;\quad -\pi < \,\mrm{arg}\pars{z} < \pi\,,\quad z \not= 0}$ and the following contour $\ds{\pars{~\mbox{where}\ \bbox[#dfd,5px,border:1px groove navy]{R = 1}~}}$

enter image description here

we'll have \begin{align} &\int_{0}^{\pi/2}\ln\pars{\sin\pars{x}}\,\dd x = \left.\Re\int_{0}^{\pi/2}\ln\pars{z - 1/z \over 2\ic z}\,{\dd z \over \ic z}\, \right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ \left.\Im\int_{0}^{\pi/2}\ln\pars{{1 - z^{2} \over 2z}\,\ic}\,{\dd z \over z}\, \right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim} &\,\,\, \overbrace{-\Im\int_{1}^{\epsilon}\ln\pars{1 + y^{2} \over 2y}\,{\dd y \over y}} ^{\ds{=\ 0}}\ -\ \Im\int_{\pi/2}^{0}\ln\pars{{1 \over 2\epsilon}\, \exp\pars{\ic\bracks{{\pi \over 2} - \theta}}}\,\ic\,\dd\theta \\[5mm] - &\ \Im\int_{\epsilon}^{1}\bracks{% \ln\pars{1 - x^{2} \over 2x} + {\pi \over 2}\,\ic}\,{\dd x \over x} \\[1cm] = &\ -\,{1 \over 2}\,\pi\ln\pars{2\epsilon} + {1 \over 2}\,\pi\ln\pars{\epsilon} = \bbx{\ds{-\,{1 \over 2}\,\pi\ln\pars{2}}} \end{align}

  • 0
    creative approach (+1) – 2017-02-13
  • 0
    @tired Thanks. It's useful in several situations. – 2017-02-13