# Delta Function Integration

1. Sep 26, 2007

### EugP

1. The problem statement, all variables and given/known data

Evaluate:

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

2. Relevant equations

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

3. 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.

2. Sep 26, 2007

### 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

3. Sep 26, 2007

### user101

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

4. Sep 26, 2007

### Dick

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.