Integral of a nascent delta function (Poisson kernel) times a generic function from 0 to $infty$












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I have a double integral of the form:
begin{equation} frac{1}{pi}int_0^infty domega frac{1}{pi} int_0^infty domega' frac{f(omega)}{f(omega)^2 + (omega-omega')^2} g(omega, omega')~, ~~~~~~~(1)end{equation}
where $g(omega,omega')$ is a generic function, smooth and evrything. It happens to be that $f(omega)$ is very small so that one could approximate $frac{f(omega)}{f(omega)^2 + (omega-omega')^2} sim pi delta(omega-omega')$ as the limit of the nascent delta function (Poisson kernel). After this approximation the integral becomes
$$ frac{1}{pi}int_0^infty domega int_0^infty domega' delta(omega-omega') g(omega, omega')~.$$
Computing this we end up with $left(int_0^infty dx delta(x-x_0)f(x)=theta(x_0)f(x_0)right)$
$$ frac{1}{pi}int_0^infty domega theta(omega)g(omega,omega)~,~~~~~~(2)$$
My question is if this procedure is correct or where it fails? Because checking the exact solution (1) and the approximate (2) I have a discrepance of a factor $frac{1}{2}$. (Missing a factor $frac{1}{2}$ in (2)). I have very little knowledge about delta functions.






calculus dirac-delta







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    I have a double integral of the form:
    begin{equation} frac{1}{pi}int_0^infty domega frac{1}{pi} int_0^infty domega' frac{f(omega)}{f(omega)^2 + (omega-omega')^2} g(omega, omega')~, ~~~~~~~(1)end{equation}
    where $g(omega,omega')$ is a generic function, smooth and evrything. It happens to be that $f(omega)$ is very small so that one could approximate $frac{f(omega)}{f(omega)^2 + (omega-omega')^2} sim pi delta(omega-omega')$ as the limit of the nascent delta function (Poisson kernel). After this approximation the integral becomes
    $$ frac{1}{pi}int_0^infty domega int_0^infty domega' delta(omega-omega') g(omega, omega')~.$$
    Computing this we end up with $left(int_0^infty dx delta(x-x_0)f(x)=theta(x_0)f(x_0)right)$
    $$ frac{1}{pi}int_0^infty domega theta(omega)g(omega,omega)~,~~~~~~(2)$$
    My question is if this procedure is correct or where it fails? Because checking the exact solution (1) and the approximate (2) I have a discrepance of a factor $frac{1}{2}$. (Missing a factor $frac{1}{2}$ in (2)). I have very little knowledge about delta functions.






    calculus dirac-delta







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      I have a double integral of the form:
      begin{equation} frac{1}{pi}int_0^infty domega frac{1}{pi} int_0^infty domega' frac{f(omega)}{f(omega)^2 + (omega-omega')^2} g(omega, omega')~, ~~~~~~~(1)end{equation}
      where $g(omega,omega')$ is a generic function, smooth and evrything. It happens to be that $f(omega)$ is very small so that one could approximate $frac{f(omega)}{f(omega)^2 + (omega-omega')^2} sim pi delta(omega-omega')$ as the limit of the nascent delta function (Poisson kernel). After this approximation the integral becomes
      $$ frac{1}{pi}int_0^infty domega int_0^infty domega' delta(omega-omega') g(omega, omega')~.$$
      Computing this we end up with $left(int_0^infty dx delta(x-x_0)f(x)=theta(x_0)f(x_0)right)$
      $$ frac{1}{pi}int_0^infty domega theta(omega)g(omega,omega)~,~~~~~~(2)$$
      My question is if this procedure is correct or where it fails? Because checking the exact solution (1) and the approximate (2) I have a discrepance of a factor $frac{1}{2}$. (Missing a factor $frac{1}{2}$ in (2)). I have very little knowledge about delta functions.






      calculus dirac-delta







      share|cite|improve this question













      I have a double integral of the form:
      begin{equation} frac{1}{pi}int_0^infty domega frac{1}{pi} int_0^infty domega' frac{f(omega)}{f(omega)^2 + (omega-omega')^2} g(omega, omega')~, ~~~~~~~(1)end{equation}
      where $g(omega,omega')$ is a generic function, smooth and evrything. It happens to be that $f(omega)$ is very small so that one could approximate $frac{f(omega)}{f(omega)^2 + (omega-omega')^2} sim pi delta(omega-omega')$ as the limit of the nascent delta function (Poisson kernel). After this approximation the integral becomes
      $$ frac{1}{pi}int_0^infty domega int_0^infty domega' delta(omega-omega') g(omega, omega')~.$$
      Computing this we end up with $left(int_0^infty dx delta(x-x_0)f(x)=theta(x_0)f(x_0)right)$
      $$ frac{1}{pi}int_0^infty domega theta(omega)g(omega,omega)~,~~~~~~(2)$$
      My question is if this procedure is correct or where it fails? Because checking the exact solution (1) and the approximate (2) I have a discrepance of a factor $frac{1}{2}$. (Missing a factor $frac{1}{2}$ in (2)). I have very little knowledge about delta functions.






      calculus dirac-delta




      calculus dirac-delta






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      asked Nov 26 at 10:20









      Andreu

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