- #1
jostpuur
- 2,116
- 19
What do we have to assume of the function f so that following limit is correct,
[tex]
\lim_{L\to\infty} \int\limits_{-L}^L f(\frac{x}{L}) \frac{\sin(x)}{x} dx = f(0)\int\limits_{-\infty}^{\infty} \frac{\sin(x)}{x} dx = \pi f(0)
[/tex]
If we first fix the integration domain like this
[tex]
\lim_{L\to\infty} \int\limits_{-\infty}^{\infty} f(\frac{x}{L})\frac{\sin(x)}{x} \chi_{[-L,L]}(x)dx
[/tex]
the problem is that the limit of the integrand is not Lebesgue integrable over [itex]\mathbb{R}[/itex], so the standard convergence results do not settle this immediately.
[tex]
\lim_{L\to\infty} \int\limits_{-L}^L f(\frac{x}{L}) \frac{\sin(x)}{x} dx = f(0)\int\limits_{-\infty}^{\infty} \frac{\sin(x)}{x} dx = \pi f(0)
[/tex]
If we first fix the integration domain like this
[tex]
\lim_{L\to\infty} \int\limits_{-\infty}^{\infty} f(\frac{x}{L})\frac{\sin(x)}{x} \chi_{[-L,L]}(x)dx
[/tex]
the problem is that the limit of the integrand is not Lebesgue integrable over [itex]\mathbb{R}[/itex], so the standard convergence results do not settle this immediately.