1

I have a pboblem as: $\int_0^{2\pi } {\delta \left( {{\rm{cos}}x} \right)dx} $.

I have done this:

$\begin{array}{l} g\left( x \right) = {\rm{cos}}x = 0 \Rightarrow \left[ \begin{array}{l} x = \frac{\pi }{2}\\ x = - \frac{\pi }{2} \end{array} \right. \Rightarrow \left[ \begin{array}{l} g'\left( {\frac{\pi }{2}} \right) = - \sin \left( {\frac{\pi }{2}} \right) = - 1\\ g'\left( { - \frac{\pi }{2}} \right) = - \sin \left( { - \frac{\pi }{2}} \right) = 1 \end{array} \right.\\ \Rightarrow \int_0^{2\pi } {\delta \left( {{\rm{cos}}x} \right)dx} = \int_0^{2\pi } {\delta \left( {x - \frac{\pi }{2}} \right)dx} + \int_0^{2\pi } {\delta \left( {x + \frac{\pi }{2}} \right)dx = 2 } \end{array}$

Howerver, my pro said that I did it wrong, and require me rethouht!!! Where is my false? Thanks

MacArthur Nguyen
  • 273
  • 1
    Your set of roots of $\cos x$ is incomplete and contains roots outside of the interval in question. Domain of integration is $[0,2\pi]$, the second root in this interval is ... – Lutz Lehmann Jan 05 '14 at 13:06
  • Indeed, what are the roots of $\cos x$ inside the interval $[0,2\pi]$? – Branimir Ćaćić Jan 05 '14 at 13:07
  • 1
    Thanks! I have done as: $\begin{array}{l} g\left( x \right) = {\rm{cos}}x = 0 \Rightarrow \left[ \begin{array}{l} x = \frac{\pi }{2}\ x = \frac{{3\pi }}{2} \end{array} \right. \Rightarrow \left[ \begin{array}{l} g'\left( {\frac{\pi }{2}} \right) = - \sin \left( {\frac{\pi }{2}} \right) = - 1\ g'\left( {\frac{{3\pi }}{2}} \right) = - \sin \left( {\frac{{3\pi }}{2}} \right) = 1 \end{array} \right.\ = \int_0^{2\pi } {\delta \left( {x - \frac{\pi }{2}} \right)dx} + \int_0^{2\pi } {\delta \left( {x - \frac{{3\pi }}{2}} \right)dx =2 } \end{array}$ Right???? – MacArthur Nguyen Jan 05 '14 at 13:41

1 Answers1

2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \color{#00f}{\large% \int_{0}^{2\pi}\delta\pars{\vphantom{\LARGE A}\cos\pars{x}}\,\dd x} = \int_{0}^{2\pi} \bracks{{\delta\pars{x - \pi/2} \over \verts{-\sin\pars{\pi/2}}} + {\delta\pars{x - 3\pi/2} \over \verts{-\sin\pars{3\pi/2}}}}\,\dd x = \color{#00f}{\large 2} $$

Felix Marin
  • 94,079