My Math Forum Problem with a simple integration!

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 May 7th, 2019, 09:00 AM #1 Senior Member   Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 Problem with a simple integration! Things I know: $\displaystyle \delta ( \tau -1 ) = 1, \tau=1$ else $\displaystyle \delta ( \tau -1 ) = 0, \tau \neq 1$ and $\displaystyle \int \delta (\tau -1) = u(\tau -1)$ where $\displaystyle u ( \tau -1 ) = 1, \tau>1$ else $\displaystyle u ( \tau -1 ) = 0, \tau<1$ Integrate to Solve: $\displaystyle e^{-2t} \int_{1}^{t}e^{\tau} + e^{2 \tau} \delta ( \tau -1 )d \tau$ Solution: $\displaystyle e^{-2t} \cdot [ \int_{1}^{t}e^{\tau}d \tau + \int_{1}^{t}e^{2 \tau} \delta ( \tau -1 )d \tau ] = e^{-2t} \cdot [ e^{\tau}|_{1}^{t} + \int_{1}^{2}e^{2 \tau} \delta ( \tau -1 )d \tau + \int_{2}^{t}e^{2 \tau} \delta ( \tau -1 )d \tau]$ I know that $\displaystyle \int_{2}^{t}e^{2 \tau} \delta ( \tau -1 )d \tau = 0$ because $\displaystyle \delta ( \tau -1 ) = 0$ for $\displaystyle \tau \neq 1$ but what about $\displaystyle \int_{1}^{2}e^{2 \tau} \delta ( \tau -1 )d \tau$ can I say that $\displaystyle \delta ( \tau -1 ) = 1$ ? Or that 2 on the upper limit of the integral is messing this up?

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