May 7th, 2019, 09:00 AM  #1 
Senior Member Joined: Oct 2015 From: Greece Posts: 120 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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integration, problem, simple 
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