# Convolution theorem for Laplace Transform proof

1. May 5, 2015

### marksman95

Hi all.

Sorry about creating this new threat despite existing some others on the same topic.

I have a problem in understanding a very specific step in the mentioned proof.

Let me take the proof given in this link as our guide.

My problem is just at the ending. When it says:

"The region of integration for this last iterated integral is the wedge-shaped region in the (t, τ) plane shown in Figure 12.28. We reverse the order of integration in the integral to get:"

$$F(s)G(s) = \int_0^\infty \left[ \int_0^t f(t)·g(\tau-t)·e^{-st}\; \mathrm{d}\tau\right]\; \mathrm{d}t$$

I don't understand how this reversion gives these limits of integration. How do we get from

$$F(s)G(s) = \int_0^\infty \left[ \int_\tau^\infty f(t)·g(\tau-t)·e^{-st}\; \mathrm{d}t\right]\; \mathrm{d}\tau$$

to there?

Thanks.

2. May 5, 2015

### Orodruin

Staff Emeritus
It is simply two different ways of defining the integration domain, which is shown in figure 12.28.

Since both parameters can be 0 to $\infty$, the outer integral necessarily needs to be from 0 to $\infty$. The inner integral must be such that you integrate over the shaded domain and nothing else. If you have fixed $t$ what values can $\tau$ take within this domain? If you have fixed $\tau$, what values can $t$ take in this domain?

3. May 5, 2015

### marksman95

Thanks Orodruin!

But now I think my problem is with this integration domain which is shown in the figure...

Could you (or someone else) explain how this integration domain is found and understood?

Thanks again!

4. May 5, 2015

### Orodruin

Staff Emeritus
The integration domain looks the way it does because of how the integral was constructed in the first place. This comes about when they change variables from $\sigma$ to $t$ in the inner integral.

5. May 8, 2015

### Clack_Attack

That's right, the visual picture on the 2nd page really helps to see this.
The original inside integral (which you have posted 2nd marskman) is integrating first from left to right or, from t= τ (tau) to ∞, then the outside covers the bounds (τ=0 to τ=∞) But then by reversing the order you'll go from τ=0 to τ=t as the new inside function then cover the remaining bounds, again, from t=0 to t=∞.

Manipulating the order of integration made many problems easier in multivariable calc.

6. May 8, 2015

### certainly

I'm scared.........................

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