Subsections

2-4

$\displaystyle \Int dx\, f(x)\delta(x^2-1)$ (10)

2-4解答

$\displaystyle \Int dx\, f(x)\delta(x^2-1)$ $\displaystyle =\left(\int_{-\infty}^{-1-\epsilon} +\int_{-1-\epsilon}^{-1+\epsi...
...+\int_{1+\epsilon}^{+\infty}\right) dx\, f(x)\delta(x^2-1)\,; \qquad \epsilon>0$    
  $\displaystyle =\int_{-1-\epsilon}^{-1+\epsilon} dx\, f(x)\delta(x^2-1)+ \int_{+1-\epsilon}^{+1+\epsilon} dx\, f(x)\delta(x^2-1)$    
  $\displaystyle =\int_{\epsilon^2 -2\epsilon}^{\epsilon^2 +2\epsilon} dX\,\frac{f...
...on \\ \hline X & \epsilon^2 +2\epsilon & \to & \epsilon^2-2\epsilon \end{array}$    
  $\displaystyle \,\,\,\,\, + \int_{\epsilon^2 -2\epsilon}^{\epsilon^2 +2\epsilon}...
...on \\ \hline X & \epsilon^2 -2\epsilon & \to & \epsilon^2+2\epsilon \end{array}$    
  $\displaystyle =\frac{1}{2}\left\{ f(-1)+f(+1)\right\}$    

著者: 茅根裕司 chinone_at_astr.tohoku.ac.jp