Problem with a simple integration!

babaliaris

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 $$\displaystyle \tau \neq 1$$

$$\displaystyle \int_{1}^{2}e^{2 \tau} \delta ( \tau -1 )d \tau$$
can I say that $$\displaystyle \delta ( \tau -1 ) = 1$$ ?