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everyone,here I have a question about how to calculate $$\int e^t H(t) dt$$ where $H(t)$ is Heaviside step function thank you for your answering!!

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Try splitting the integral into two parts, either side of zero. – Mark Pattison Jun 26 '14 at 13:24

Hint: The Heaviside function is defined as \begin{aligned} \text{H}(t) & = 0, \quad t<0,\\ \text{H}(t) & = 1, \quad t\geq0. \end{aligned} That means your integrand will be zero for all $t<0$, and $e^{t}$ for $t\geq1$. Can you figure out the rest?

Note: If you use the half-maximum convention though, you will find it to be: \begin{aligned} \text{H}(t) & = 0, \quad t<0,\\ \text{H}(t) & = \frac{1}{2}, \quad t=0,\\ \text{H}(t) & = 1, \quad t>0. \end{aligned} If so, you will have the integrand is $\frac{1}{2}e^{t}$ in $t=0$, and the rest is similar to the above.

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use computer to solve,I find answer:(e^t-1)*H(t) – user113644 Jun 26 '14 at 13:38
However,use integration by part,int e^tH(t) dt=int H(t) d(e^t)=e^tH(t)-int e^t d(H(t))= e^t*H(t)-delta(t)*e^t – user113644 Jun 26 '14 at 13:41
where delta(t) is dirac-delta function,so I still can't understand why there is a subtle difference between computer and my answer from Integrantion By Part – user113644 Jun 26 '14 at 13:44*H%28t%29+dt – user113644 Jun 26 '14 at 13:46
Don't forget that you are integrating over the Dirac-delta function (it doesn't make much sense otherwise), which means that you evaluate the integrand in the point where the delta function has a zero argument. Just like you wrote: $I = \int e^{t} \text{H}(t) dt = e^{t}\text{H}(t) - \int e^{t} \frac{d}{dt}\left(\text{H}(t)\right)dt$ where $\frac{d}{dt}\text{H} = \delta(t)$, so $I = (e^{t}-1)\text{H}(t)$. The last integral turns into $\text{H}(t)e^{0}$ instead of just $e^{0}$, since the original expression is zero for all $t<0$. – fromGiants Jun 26 '14 at 13:56

$$ \int f(t) H(t), dt = \int_0^{+\infty} f(t) $$

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