# Dirac delta integration

1. Jul 11, 2011

### Jncik

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

find

$$\int_{-\infty}^{+\infty} x(t) \delta (\beta t - t_{0}) dt$$

for $$x(t) = e^{a t} u(t)$$

there is no information conserning a, β, or $$t_{0}$$...

3. The attempt at a solution

assuming that $$t_{0}$$ is a constant

$$\int_{-\infty}^{+\infty} x(t) \delta (\beta t - t_{0}) dt = \int_{-\infty}^{+\infty} e^{at}u(t) \delta (\beta t - t_{0}) dt = \int_{0}^{+\infty} e^{at} \delta (\beta t - t_{0}) dt = \int_{0}^{+\infty} e^{\frac{a}{b}t_{0}} dt = +\infty$$

2. Jul 11, 2011

### lanedance

when you integrate, the delta function is only non-zero at $t_0$ and
$$\int f(t) \delta (t-t_0) = f(t_0)$$

3. Jul 11, 2011

### lanedance

4. Jul 11, 2011

### Jncik

thanks

hence the result is just

$$e^{\frac{a}{b}t_{0}}$$

it would be 0 if we didn't integrate from 0 to infinity but from k>0 to infinity

correct?

edit: actually it would not be 0 if the integration was from k>t0/β

Last edited: Jul 11, 2011
5. Jul 11, 2011

### lanedance

you've got the right idea on the k>t0/b stuff, but i think you need to be careful with the scaling properties of the delta function

$$\int f(t) \delta (\beta t-t_0) dt$$

now try making the substitution
$$u = \beta t-t_0$$

6. Jul 11, 2011

### Jncik

$$u = \beta t - t_{0} => du = \beta dt$$

also

$$t = \frac{u + t_{0}}{\beta}$$

hence

$$\int_{-t_{0}}^{+\infty} f(\frac{u+t_{0}}{\beta}) \delta(u) \frac{du}{\beta}$$

which will give us

$$\frac{1}{\beta} f(\frac{t_{0}}{\beta}) = \frac{e^{\frac{a t_{0}}{\beta}}}{\beta}$$ $$t_{0}>0$$

:S, seems complicated.... is this correct? In my previous attempt I forgot to change the variables.. I thought that we just had to search when the function inside $$\delta$$ is 0, and we just put the $$t$$ for which δ is 0, in the function before δ..

7. Jul 11, 2011

### lanedance

that looks good to me - you could do it either way if you remember the scaling & translation properties of the delta function.

Whilst slightly longer I prefer the substitution as you don't need to remember any rules, and we got that extra factor of 1/beta correct without needing to remember it.

8. Jul 12, 2011

### Jncik

yes you're right, thanks for your help :)

9. Jul 12, 2011

no worries;)