Time-shifted Integral with Unit Impulse Function

[Les talons]

Homework Statement

Evaluate the integral.
$$\int_{-\infty}^\infty f(2 -t) \delta(3 -t) \, dt$$

Homework Equations

The Unit Impulse function is defined
$$\hspace{18mm} \delta(0) = 0 \hspace{10mm} t \neq 0$$
$$\hspace{-15mm} \int_{-\infty}^\infty \delta(t) \, dt = 1$$

The Attempt at a Solution

Using a u-substitution $u = 3 -t$, $dt = -du$, $t = 3 -u$
$$\int_{-\infty}^\infty f(2 -t) \delta(3 -t) \, dt$$
$$= \int_{-\infty}^\infty f(2 -(3 -u)) \delta(u) (-1) \, du$$
$$= -\int_{-\infty}^\infty f(u -1) \delta(u) \, du$$
$$= -f(0 -1)$$
$$= -f(-1)$$

Please let me know if the above is unclear. My question is 1) Is this the right solution? and 2) Trusted sources tell me that the answer is just $f(-1)$. Which is the correct answer? Is there some mathematical detail I am missing? Hope this is clear, and hope you can help. $\mathcal{Thanks}$

 

[DEvens]

You did a substitution of u = 3-t. And so you got a du instead of a dt. Good. What happens to the limits on the integral?

 

[Orodruin]

Is there some mathematical detail I am missing?

Yes, you have changed the integration limits when you did the variable substitution. $u\to -\infty$ when $t\to +\infty$

 

[Les talons]

For the lower limit of the integral, since $t \rightarrow -\infty$, $u \rightarrow 3 -(-\infty) \rightarrow 3 +\infty \rightarrow \infty$, and in the upper limit of the integral, $t \rightarrow \infty$, implying $u \rightarrow 3 -\infty \rightarrow -\infty$. The net effect is to annihilate the $-1$ in front of the integral, giving the evaluation $f(-1)$. Thank you both for the insight!

 

[DEvens]

Alternatively, you could use the fact that the Dirac delta is even. That means $\delta(x) = \delta(-x)$. So you could just get rid of the annoying minus sign.

https://en.wikipedia.org/wiki/Dirac_delta_function