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in below link, (formula (34)-(40)) there are some definition of Dirac delta function in terms of other functions such as Airy function, Bessel function of the first kind, Laguerre polynomial,....

http://mathworld.wolfram.com/DeltaFunction.html

Is there any definition of the Dirac delta function in terms of the sech (secant hyperbolic) or cosh (cosine hyperbolic) functions?

Please say your references. Thanks

pluton
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Mr. Nobody
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2 Answers2

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Whenever $f$ is a nonnegative function with integral 1, one has approximation by convolution. Specifically, define $f_c(x) = f(x/c)/c$. Then $\lim_{c \to 0^+} g * f_c = g$ for mildly restricted $g$. (For example, $L^p$ for $1 \leq p < \infty$ is enough.) In particular you can take $f(x)=\text{sech}(x)/\pi$.

Ian
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The delta "function" $\delta(x)$ is supposed to be zero as $x$ gets large in either direction, so basing one on $\cosh(x)$ is hard because $\cosh(x) \to +\infty$ as $x$ gets large in either direction. That makes $\operatorname{sech} (x)$ a good candidate. Since $\int_{-\infty}^{+\infty} \operatorname{sech} x dx=\pi$, you can use $\lim_{k \to \infty}\frac k \pi \operatorname{sech} (kx)$ as a delta "function".

Ross Millikan
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  • thanks for your helpful answer. is "(Sqrt[k]/Pi).Sech[kx] " is a delta function?(for k>infinity ,it's integral becomes zero).thank you a lot. – Mr. Nobody Nov 15 '14 at 05:37
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    You need the integral to stay $1$, as that is part of the delta function definition. That is why I made it proportional to $k$. The $k$ in the integral narrows it up, and the $k$ outside makes the peak tall. – Ross Millikan Nov 15 '14 at 05:44