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I have to resolve this integral:

$$\int e^{-2t}u(t)dt$$

Where $u(t)$ is heaviside's function. The result should be:


but I don't undestand why this result. Someone could explain me the step to get it?

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Is that Heaviside's function the one that equals 1 for positive argument, zero for negative one and whatever (many times, $\,0.5\,$) for zero? I doubt it as then I can't see how could you get what you say you do from the integral... – DonAntonio Jun 3 '12 at 10:18
Yes the heaviside function is that you said – Mazzy Jun 3 '12 at 10:19
up vote 4 down vote accepted

Since you give an indefinite integral, a constant of integration should be added. However, if we wish to compute $$ \int_0^xe^{-2t}u(t)\,\mathrm{d}t $$ We can proceed as follows. For $x<0$, since $u(t)=0$ when $t<x<0$, we get $$ \int_0^xe^{-2t}u(t)\,\mathrm{d}t=0\tag{1} $$ For $x\ge0$, $$ \begin{align} \int_0^xe^{-2t}u(t)\,\mathrm{d}t &=\int_0^xe^{-2t}\,\mathrm{d}t\\ &=-\tfrac{1}{2}\left[e^{-2x}-e^{0}\right]\\ &=\tfrac12-\tfrac12e^{-2x}\tag{2} \end{align} $$ Putting $(1)$ and $(2)$ together yields $$ \int_0^xe^{-2t}u(t)\,\mathrm{d}t=\left[\tfrac12-\tfrac12e^{-2x}\right]u(x)\tag{3} $$

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...but if the Heaviside function you mention is what I said, then the function $\,e^{-2t}u(t)\,$ is zero on the negative reals and thus $$\int e^{-2t}u(t)\,dt=\int e^{-2t}\,dt=-\frac{1}{2}e^{-2t}+C\,,\,\,C=\,\text{constant}$$ only when $\,t> 0\,$

I think that what you really had is something like $\,\,\displaystyle{\int_0^t e^{-2x}u(x)\,dx}\,\,$ , which gives exactly what you said, so that the integral isn't zero only if $\,t>0\,$ , but without the limits the integral is indefinite and you can't get what you say you did.

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Hint: The form of the answer should remind you of integration by parts: $\int_a^b f(t)g'(t)dt=f(t)g(t)|_a^b-\int_a^b f'(t)g(t)dt$ This will depend upon some properties of $u(t)$

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