Problem with a simple integration!

Oct 2015
141
8
Greece
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?