# Limit of an integral

Printable View

• December 26th 2011, 03:19 AM
wudup
Limit of an integral
Could someone please shed some light on how I might show that

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

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

Thanks.
• December 27th 2011, 11:07 AM
Opalg
Re: Limit of an integral
Quote:

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

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

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

Thanks.

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

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

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 $R_1 = \{(x,y) : |x-x'|^2 + |y-y'|^2 <r^2\}$ and $R_2 = \{(x,y) : |x-x'|^2 + |y-y'|^2 \geqslant r^2\}$, where r is chosen so that $|f(x,y)-f(x',y')|$ is small whenever $(x,y)\in R_1$ (that will require assuming that f is continuous). Then the integral over $R_1$ will be small by (2), and the integral over $R_2$ will be small by (3) provided we assume that |f| is integrable over $\mathbb{R}^2$ and provided that z is small enoungh.
All times are GMT -8. The time now is 05:16 PM.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 32, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9455105066299438, "perplexity_flag": "head"}