- #1
Haorong Wu
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- TL;DR Summary
- How to calculate the following limits, when viewed as distributions?
Hi, there. I am reading this thesis. On page 146, it reads that
I do not know how to calculate the limits when they are viewed as distributions. I am trying to integrate a test function with the limits. So I try (##Q## is defined as ##Q>0##) $$\lim_ {r\rightarrow \infty} \int_{0}^\infty dQ \cos ((Q-Q')r )\frac {\sin ((Q-Q')r)}{Q-Q'}=\frac \pi 2,$$ while ##\int_{-\infty}^\infty dQ \cos ((Q-Q')r ) \delta (Q-Q')=1##. Then I only have ##\lim_{r\rightarrow \infty}\frac {\sin ((Q-Q')r)}{Q-Q'}=\pi \delta(Q-Q') /2##. Is this wrong? Thanks.
when viewed as distributions, one can show that the following limits holds:
$$\lim_{r\rightarrow \infty}\frac {\sin ((Q-Q')r)}{Q-Q'}=\pi \delta(Q-Q') ,$$
$$\lim_{r\rightarrow \infty}\frac {\cos ((Q+Q')r)}{Q+Q'}=0 .$$
I do not know how to calculate the limits when they are viewed as distributions. I am trying to integrate a test function with the limits. So I try (##Q## is defined as ##Q>0##) $$\lim_ {r\rightarrow \infty} \int_{0}^\infty dQ \cos ((Q-Q')r )\frac {\sin ((Q-Q')r)}{Q-Q'}=\frac \pi 2,$$ while ##\int_{-\infty}^\infty dQ \cos ((Q-Q')r ) \delta (Q-Q')=1##. Then I only have ##\lim_{r\rightarrow \infty}\frac {\sin ((Q-Q')r)}{Q-Q'}=\pi \delta(Q-Q') /2##. Is this wrong? Thanks.