How Do You Correctly Integrate the Delta Function in This Equation?

[EugP]

Homework Statement

Evaluate:

\int_{-3}^{5} e^{-2t} sin(t-3) \delta(t-5) dt

Homework Equations

\int_{-\infty}^{\infty} f(t) \delta(at-t_0) dt = \frac{1}{|a|}f(\frac{t_0}{a})

The Attempt at a Solution

e^{-2(5)} sin (5-3) = e^{-10} sin (2)

The solution given by the professor was:

\frac{1}{2} e^{-10} sin (2)

I don't understand where he got the \frac{1}{2} from.

If anyone could help me it would be greatly appreciated.

 

[genneth]

It's a bit cheeky, and technically your professor's answer is contestable. Because your integration limits are only up to 5, and the delta function is delta(t-5), you've only integrated "half" of the delta spike. Now, if that sounds absurd -- "how the hell can you only integrate half of an infinitesimally thin spike?!" -- just remember that the delta function itself is not really very sensible as a function anyway. Personally, I treat that sort of integral as badly defined; more specifically, since I'm a physicist, I arrange my physics so that the maths never requires me to contemplate these soul-searching issues

 

[user101]

Hm, interesting... I never knew that you integrate only half the delta spike.

 

[Dick]

Hm, interesting... I never knew that you integrate only half the delta spike.

You shouldn't. As genneth was pointing out, there are perfectly fine representations of the delta function in which you can get a completely different answer. The correct answer is 'undefined', with all due respect to the composer of the solutions.