Graduate Integral of Dirac function from 0 to a.... value

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The discussion revolves around integrating the Dirac delta function within the context of Laplace transforms and convolution formulas. The user expresses confusion about how to properly evaluate the first integral involving the Dirac delta function, seeking clarification on its integration. Participants clarify that the integral of the Dirac delta function yields the value of the integrand at the point where the delta function is centered, specifically at t' = 0. There is mention of discrepancies in numerical results when using algebra software for partial fraction expansion. The conversation emphasizes the importance of understanding the placement of the delta function within the Laplace transform for accurate integration.
maistral
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Hi. So I'm trying to use Laplace transforms in inverting a particular s-function via the convolution formula.

I ended up with this terrifying-looking thing:

convo.png


So distributing, I ended up with:
convo2.png


Evaluating the second integral poses no problem for me (although I think the integration will definitely be 'hairy'). I have a problem with the first integral though. How on Earth do I integrate the dirac delta? Help! I am totally at a loss here. Or am I doing something wrong?
 

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What is your problem? The integral should be the value of the integrand (without the delta function) at t' = 0.
 
mathman said:
What is your problem? The integral should be the value of the integrand (without the delta function) at t' = 0.
sin(wt) right?

But I did partial fraction expansion using algebra software (mathcad) and the answer was different numerically (I mean, I numerically integrated that function, taking into account that the first integral is sin(w*t)
 
You might want half that value if your integration border is right where the delta is.
It depends on how the delta appears in the Laplace transform.
 
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Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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