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!!
$\endgroup$ 14 Answers
$\begingroup$$$ \int f(t) H(t), dt = \int_0^{+\infty} f(t) $$
$\endgroup$ $\begingroup$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.
$\endgroup$ 6 $\begingroup$The derivative that you need is zero in the negatives and an exponential in the positives. Hence take a constant in the negatives and an exponential in the positives.
$$H(-x)+e^xH(x)$$ will be continuous. No solution can be differentiable at $x=0$.
$\endgroup$ $\begingroup$In fact the result is just trivially
$\int e^tH(t)~dt=e^tH(t)+C$
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