Integrating with a dirac delta function

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The integral of the Dirac delta function, \int_0^x \delta(x-y)f(y)dy, can be simplified using the property that \int_{a-ε}^{a+ε} f(y)\delta(y-a) dy = f(a). Instead of integration by parts, the integral can be manipulated to express it in terms of the function f. The result of the integration leads to f(x)(2H(x)-1), where H is the antiderivative of the Dirac delta function. Understanding the properties of the Dirac delta function is crucial for solving such integrals effectively. The discussion emphasizes the unique characteristics of the Dirac delta function in integration.
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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