I Could this function be approximated by Dirac delta function?

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The function N(k) defined as $$N(k)=\frac {-1} {|k|} e^{-c|k|}$$ approaches infinity as k approaches zero and decreases rapidly for non-zero k, but its integral is infinite, preventing it from being a true delta function. While N(k) meets two criteria of a delta function, it fails to satisfy the normalization condition, which requires the integral to equal one. A suggestion is made to modify N(k) and P(k) to $$\tilde{N}(k)= e^{-c|k|}/\sqrt{|k|}$$ and $$\tilde{P}(k)=P(k)/\sqrt{|k|}$$ to potentially achieve a better approximation. However, as c increases, the integral of the modified N(k) does not stabilize, indicating further adjustments are necessary. Ultimately, approximating N(k) as a delta function poses challenges in physics due to these normalization issues.
Haorong Wu
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Under what conditions, could a function be replaced by a delta function?
hi, there. I am doing some frequency analysis. Suppose I have a function defined in frequency space $$N(k)=\frac {-1} {|k|} e^{-c|k|}$$ where ##c## is some very large positive number, and another function in frequency space ##P(k)##. Now I need integrate them as $$ \int \frac {dk}{2 \pi} N(k) P(k).$$However, the integration wil be too complicated to be solved.

Meanwhile, I notice that, ##N(k)## is infinite when ##k\rightarrow 0##, and decreases to zero rapidly when ##k \ne 0##. However, the integral of $$ \int dk N(k)$$ is also infinite. I am not sure whether I could approximate ##N(k)## by ##\delta(k)## or not. Would it yield problems in physics?

Thanks!
 
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No. There are three facts that you want for a function that acts like a delta function are:

1. ##f(k) \rightarrow \text{very large}## when ##k \rightarrow 0##
2. ##f(k) \rightarrow 0## when ##k \rightarrow \infty##
3. ##\int_{-\infty}^{+\infty} f(k) dk = 1##

Your function has properties 1 and 2, but not 3.
 
You could shuffle some of N onto P. Like if you said ##\tilde{N}(k)= e^{-c|k|}/\sqrt{|k|}## and ##\tilde{P}(k)=P(k)/\sqrt{|k|}## you might be able to do something.

As c goes to infinity the integral of ##\tilde{N}## is not constant, so you would also need to shuffle around some ##c## stuff in order to get this to totally work.
 
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