# Integrate by parts because of two functions

1. Oct 17, 2008

### Cyrus

I found this interesting little problem when thinking about convolution:

$$\int x( \tau) \delta(t-\tau) d\tau$$

Normally to solve something like this you would have to integrate by parts because of two functions in $$\tau$$

Using the fact that:

$$\int u *dv = u*v - \int v*du$$

Where

$$u=x(\tau)$$

$$dv= \delta(t-\tau) d\tau$$

Then:

$$du=x'(\tau) d\tau$$

$$v= 1$$

If you plug this back in you get:

$$x(\tau) - x(\tau) = 0$$

Total nonsense!

2. Oct 17, 2008

### D H

Staff Emeritus
Re: Convolution

When you do your math wrong you get total nonsense. The integral of the delta distribution is the Heaviside function, not 1.

Using definite integrals (which is the only thing that makes sense when using the delta distribution), the goal is to evaluate the definite integral

$$\int_a^b x(\tau)\delta(t-\tau)\,d\tau$$

where $a<b$ and x(t) is well-behaved over (a,b). There are three cases to investigate:
1. $t<a$, which should yield zero,
2. $t>b$, which should also yield zero, and
3. $t\in(a,b)$, which should yield $x(t)$.

Integrating by parts,

$$\int_a^b x(\tau)\delta(t-\tau)\,d\tau = -\,x(\tau)H(t-\tau)\Bigl|_{\tau=a}^b + \int_a^b x^{\prime}(\tau)H(t-\tau)\,d\tau$$

In case (1), the first two terms vanish because $H(t-a)=H(t-b)=0$ since both $t-a$ and $t-b$ are negative. Similarly, the integral also vanishes, so the result is zero.

In case (2), the Heaviside function $H(t-\tau)$ is identically one for all $\tau\in(a,b)$. The results of integrating by parts reduces to
$$x(a)-x(b)+ \int_a^b x^{\prime}(\tau)\,d\tau = x(a)-x(b)+x(b)-x(a) = 0$$

In case (3), the Heaviside function changes from 1 to 0 at $\tau=t$. The results of integrating by parts reduces to
$$x(a)+ \int_a^t x^{\prime}(\tau)\,d\tau = x(a)+x(t)-x(a) = x(t)$$

Which is exactly what was expected in all three cases.

3. Oct 17, 2008

### Cyrus

Re: Convolution

I gotta run right now, but for now, THANKS DH!

4. Oct 18, 2008

### Cyrus

Re: Convolution

Hi DH, I think you might have an error in your signs.

How come the its = - ... +

I think it should be + .... -

No?

5. Oct 18, 2008

### D H

Staff Emeritus
Re: Convolution

No.

The goal is to integrate $\int x(\tau) \delta(t-\tau)\,d\tau$ by parts. Setting $u=x(\tau)$ and $dv=\delta(t-\tau)d\tau$, then $du=x'(\tau)d\tau$ and $v=-\,H(t-\tau)$. The sign change results from the $t-\tau$ argument to the delta distribution.

6. Oct 18, 2008

### Cyrus

Re: Convolution

I see. I want the more general case, where the upper and lower limits are +/- infinity. So then, the last case holds just as you stated but a is -inft instead of a finite value.

When it comes to the last integral, it would have to be split up as:

$$\int^\infty_{t+} + \int^t_{-\infty}$$

Where the second integral is *just after* t, right?

Otherwise, if it were split at the value of time t, you would get 2x(t) when you did the integral piecewise.

Last edited: Oct 18, 2008