Integral of dirac delta function at x=0

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SUMMARY

The integral of the Dirac delta function, represented as \delta(x), over the entire real line is defined as \int_{-\infty}^{+\infty}\delta(x)dx=1. However, evaluating \delta(0) directly is incorrect, as the Dirac delta function is not a conventional function and cannot be evaluated at a specific point. The proper interpretation is that the delta function serves as a distribution, and its integral over its entire domain yields the value of 1.

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vsravani
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Hi

Can somebody help me with this...
Is is correct to say that, Integral(delta(0)) = 1 (limits are from -infinity to +infinity)
I don't know latex and sorry for the inconvenience in readability.

Thanks,
VS
 
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Yes.
 
vsravani said:
Hi

Can somebody help me with this...
Is is correct to say that, Integral(delta(0)) = 1 (limits are from -infinity to +infinity)
I don't know latex and sorry for the inconvenience in readability.

Thanks,
VS
Examine the definition of the delta function. You will see that yes is the obvious answer.
 
vsravani said:
Hi

Can somebody help me with this...
Is is correct to say that, Integral(delta(0)) = 1 (limits are from -infinity to +infinity)
I don't know latex and sorry for the inconvenience in readability.

Thanks,
VS

The \delta(x) is not a function, so you can't evaluate it at x=0 and write \delta(0). Maybe you meant

\int_{-\infty}^{+\infty}\delta(x)dx=1

this is correct. Instead

\int_{-\infty}^{+\infty}\delta(0)dx

is meaningless, so in particular it's not 1.
 
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