Integrating e^t*H(t): Solving for Integral Limits of -∞ to ∞

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Homework Help Overview

The problem involves evaluating the integral of the product of a step function H(t) and an exponential function e^(-2t) over the limits from negative infinity to positive infinity. Participants are exploring the implications of the step function on the convergence of the integral and the behavior of the exponential function at the specified limits.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the nature of the integral, questioning how the multiplication of H(t) and e^(-2t) affects convergence. There is consideration of whether integration by parts is necessary and how the limits of integration influence the result.

Discussion Status

Some participants have provided insights into the behavior of the functions involved, noting that H(t) effectively changes the limits of integration. There is an acknowledgment of the calculator's output and a discussion on why it might suggest a specific value, prompting further exploration of the integral's evaluation.

Contextual Notes

Participants note that H(t) is a step function, which is 1 for t >= 0 and 0 for t < 0, leading to a focus on the integral from 0 to infinity rather than from negative infinity. There is an ongoing examination of the implications of this setup on the integral's convergence.

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Homework Statement



-∞ H(t)*e-2tdt

Homework Equations



See above.

The Attempt at a Solution



I know if I were just integrating H(t) by itself, I would get a ramp function. I also know e^-2t by itself will not converge for the given limits of +/- infinity. I just want to know what is going on when the two multiply together, and if integration by parts would serve me here (if it is even necessary?)?

I have tried to evaluate it with my graphing calculator, and it seems to say the answer is 1/2, though I'm not sure how it has arrived there.

Any help would be greatly appreciated!
 
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Rome_Leader said:

Homework Statement



-∞ H(t)*e-2tdt

Homework Equations



See above.

The Attempt at a Solution



I know if I were just integrating H(t) by itself, I would get a ramp function. I also know e^-2t by itself will not converge for the given limits of +/- infinity. I just want to know what is going on when the two multiply together, and if integration by parts would serve me here (if it is even necessary?)?

I have tried to evaluate it with my graphing calculator, and it seems to say the answer is 1/2, though I'm not sure how it has arrived there.

Any help would be greatly appreciated!

H(t) is a step function. It's 1 if t>=0 and 0 if t<0. So H(t)f(t)=f(t) if t>=0 and 0 if t<0.
 
Why does my calculator give 1/2, then? If I follow your logic, then I should plug in my limits of inf and -inf to e^(-2t), and this gives me 0 -0.
 
Rome_Leader said:
Why does my calculator give 1/2, then? If I follow your logic, then I should plug in my limits of inf and -inf to e^(-2t), and this gives me 0 -0.

e^(-2t) for the limits inf and -inf diverges, like you said before. I mean the function you are integrating vanishes where t is in (-inf,0) so you can ignore that part. Just use 0 as the lower bound instead of -inf.
 
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Ahh, I totally understand now! Thanks for that last bit, it helped give my mind a kick. Essentially it's the same as integrating from 0 to inf of just e^-2t given that H(t) is 1 starting from 0 and going on to infinity! Thanks!
 
Rome_Leader said:
Why does my calculator give 1/2, then? If I follow your logic, then I should plug in my limits of inf and -inf to e^(-2t), and this gives me 0 -0.

You are just evaluating the elementary integral
\int_0^{\infty} e^{-2t} \, dt.
That's what H does: it zeros-out the part from t = -∞ to t = 0.
 

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