Integrating with a dirac delta function

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SUMMARY

The discussion centers on integrating the Dirac delta function, specifically the integral \int_0^x \delta(x-y)f(y)dy. Participants clarify that the Dirac delta function is zero everywhere except at zero, where it is infinite. The integral can be simplified using the property \int_{a-ε}^{a+ε} f(y)\delta(y-a) \ dy = f(a), leading to the conclusion that the result is f(x)(2H(x)-1), where H is the antiderivative of the Dirac delta function.

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  • Understanding of the Dirac delta function and its properties
  • Knowledge of integration techniques, including integration by parts
  • Familiarity with antiderivatives and their applications
  • Basic calculus concepts, particularly limits and continuity
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fred_91
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Homework Statement



I have to integrate:

\int_0^x \delta(x-y)f(y)dy

Homework Equations





The Attempt at a Solution



I know that the dirac delta function is zero everywhere except at 0 it is equal to infinity:
\delta(0)=\infty
I have to express the integral in terms of function f only and i am unsure how to do this.
Do I have to use integration by parts?

Thank you very much in advance
 
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fred_91 said:

Homework Statement



I have to integrate:

\int_0^x \delta(x-y)f(y)dy

Homework Equations





The Attempt at a Solution



I know that the dirac delta function is zero everywhere except at 0 it is equal to infinity:
\delta(0)=\infty
I have to express the integral in terms of function f only and i am unsure how to do this.
Do I have to use integration by parts?

Thank you very much in advance
Dirac's Delta is...weird. You have to integrate \int_0^x \delta(x-y)f(y)dy?

Integration by parts sounds...okay? However, one of the properties of the Dirac Delta "function" is that \displaystyle \forall ε > 0, \int_{a-ε}^{a+ε} f(y)\delta(y-a) \ dy = f(a). Can you manipulate your integral into that form? :wink:

By my calculations, you should end up with f(x)(2H(x)-1), where H is the antiderivative of δ.
 

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