Integrate by parts because of two functions

[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!

 

[D H]

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

Then:

v= 1

Total nonsense!

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 =<br /> -\,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.

 

[Cyrus]

I got to run right now, but for now, THANKS DH!

 

[Cyrus]

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

How come the its = - ... +

I think it should be + ... -

No?

 

[D H]

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&#039;(\tau)d\tau and v=-\,H(t-\tau). The sign change results from the t-\tau argument to the delta distribution.

 

[Cyrus]

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.