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Thread: Limit of an integral

  1. #1
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    Limit of an integral

    Could someone please shed some light on how I might show that

    $\displaystyle \lim_{\epsilon\to 0}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\epsilon\over 2\pi [(x-x')^2+(y-y')^2+\epsilon^2]^{3\over2}}f(x,y)\;dx\,dy\,\,=f(x',y')$? I am not sure what conditions there is on $\displaystyle f(x,y)$, though I do know that $\displaystyle \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{z\over 2\pi [(x-x')^2+(y-y')^2+z^2]^{3\over2}}f(x,y)\;dx\,dy$ is well-defined (converges) for all $\displaystyle x,y\in R$ and $\displaystyle z>0$.

    Brainstorm: Perhaps we can show that $\displaystyle \lim_{\epsilon\to 0}{\epsilon\over 2\pi [(x-x')^2+(y-y')^2+\epsilon^2]^{3\over2}}$ is the Delta function $\displaystyle \delta(x-x')\delta(y-y')$? Or maybe changing variables?

    Thanks.
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  2. #2
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    Re: Limit of an integral

    Quote Originally Posted by wudup View Post
    Could someone please shed some light on how I might show that

    $\displaystyle \lim_{\epsilon\to 0}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\epsilon\over 2\pi [(x-x')^2+(y-y')^2+\epsilon^2]^{3\over2}}f(x,y)\;dx\,dy\,\,=f(x',y')$? I am not sure what conditions there is on $\displaystyle f(x,y)$, though I do know that $\displaystyle \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{z\over 2\pi [(x-x')^2+(y-y')^2+z^2]^{3\over2}}f(x,y)\;dx\,dy$ is well-defined (converges) for all $\displaystyle x,y\in R$ and $\displaystyle z>0$.

    Brainstorm: Perhaps we can show that $\displaystyle \lim_{\epsilon\to 0}{\epsilon\over 2\pi [(x-x')^2+(y-y')^2+\epsilon^2]^{3\over2}}$ is the Delta function $\displaystyle \delta(x-x')\delta(y-y')$? Or maybe changing variables?

    Thanks.
    For z>0, let $\displaystyle K_z(x,y) = \frac z{2\pi (x^2+y^2+z^2)^{3/2}}$, and check that it has the key properties of a kernel function (such as the FejÚr kernel). These are: (1) $\displaystyle K_z(x,y)\geqslant0$ (that's obviously true); (2) $\displaystyle \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}K_z(x,y)\,dx\,dy = 1$ (easy integral, if you express it in polar coordinates); and (3) given $\displaystyle \varepsilon>0$ and $\displaystyle r>0$ there exists $\displaystyle z_0>0$ such that $\displaystyle K_z(x,y)<\varepsilon$ whenever $\displaystyle z<z_0$ and $\displaystyle x^2+y^2\geqslant r^2.$ (That last condition essentially says that when z is small, $\displaystyle K_z(x,y)$ is also small except when (x,y) is close to the origin.)

    Once you have checked that those three conditions hold, you can calculate as follows:

    $\displaystyle \begin{aligned}\biggl| \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} &K_z(x-x',y-y')f(x,y)\,dx\,dy - f(x',y') \biggr| \\ &= \left| \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} K_z(x-x',y-y')\bigl(f(x,y)-f(x',y')\bigr)\,dx\,dy \right| \qquad\text{(by (2))} \\ &\leqslant \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}K_z(x-x',y-y')\left|f(x,y)-f(x',y')\right|\,dx\,dy .\end{aligned}$

    You now want to show that for small enough z, that last integral can be made arbitrarily small. To do that, split the plane into two regions $\displaystyle R_1 = \{(x,y) : |x-x'|^2 + |y-y'|^2 <r^2\}$ and $\displaystyle R_2 = \{(x,y) : |x-x'|^2 + |y-y'|^2 \geqslant r^2\}$, where r is chosen so that $\displaystyle |f(x,y)-f(x',y')|$ is small whenever $\displaystyle (x,y)\in R_1$ (that will require assuming that f is continuous). Then the integral over $\displaystyle R_1$ will be small by (2), and the integral over $\displaystyle R_2$ will be small by (3) provided we assume that |f| is integrable over $\displaystyle \mathbb{R}^2$ and provided that z is small enoungh.
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