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I am trying to figure out properties of the following integral:

$$p(t)=\int_{0}^{t} e^{\alpha(t-t')} f(t')dt', \hspace{1 cm} t>t'$$

I would google and read more info about this integral but I do not know a proper specific name for it. It seems like an exponential smoothing, filter, Laplace transform?

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  • $\begingroup$ Since $\int^t_0e^{\alpha(t-t^\prime)}f(t^\prime)dt^\prime=e^{\alpha t}\int^t_0e^{\alpha t^\prime}f(t^\prime)dt^\prime$, see the last entry in the table here. $\endgroup$
    – user122283
    Apr 27, 2014 at 22:20

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It is a convolution integral

$$p(t)=(f*g)(t)$$

with $f(t)$ equal to zero for $t<0$ and $g(t)=e^{\alpha t}u(t)$ ($u(t)$ is the step function).

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