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

[Rome_Leader]

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!

 

[Dick]

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.

 

[Rome_Leader]

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.

 

[Dick]

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.

 

[Rome_Leader]

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!

 

[Ray Vickson]

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.