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May 7th, 2019, 09:00 AM   #1
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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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