Don't understand this application of the transform of an integral

[Turion]

Homework Statement

$$L\{ 1*{ t }^{ 3 }\}$$

Homework Equations

$$L\{ \int _{ 0 }^{ t }{ f(\tau )d\tau } \} =\frac { L\{ f(t)\} }{ s }$$

The Attempt at a Solution

$$L\{ 1*{ t }^{ 3 }\} \\ =L\{ \int _{ 0 }^{ t }{ { (t-\tau ) }^{ 3 }d\tau } \} \\ =\frac { L\{ { t }^{ 3 }\} }{ s } \\ =\frac { 6 }{ { s }^{ 5 } }$$

The function in the integral is a function of t and tau so how was the identity under "relevant equations" applied?

 

[szynkasz]

Yes, you have $\int_0^tf(\tau,t)\,d\tau$ in the integral instead of $\int_0^tf(\tau)\,d\tau$ and the formula for integral doesn't work. This is convolution so:

$L\{1*t^3\}=L\{1\}\cdot L\{t^3\}$

 

[vela]

Homework Statement

$$L\{ 1*{ t }^{ 3 }\}$$

Homework Equations

$$L\{ \int _{ 0 }^{ t }{ f(\tau )d\tau } \} =\frac { L\{ f(t)\} }{ s }$$

The Attempt at a Solution

$$L\{ 1*{ t }^{ 3 }\} \\ =L\{ \int _{ 0 }^{ t }{ { (t-\tau ) }^{ 3 }d\tau } \} \\ =\frac { L\{ { t }^{ 3 }\} }{ s } \\ =\frac { 6 }{ { s }^{ 5 } }$$

The function in the integral is a function of t and tau so how was the identity under "relevant equations" applied?

Use the substitution $u=t-\tau$ on the integral to convert it to a more familiar form.

 

[Turion]

@szynkasz: Thanks. I solved it.

Use the substitution $u=t-\tau$ on the integral to convert it to a more familiar form.

Any suggestions?

 

[vela]

Remember that $u$ is a dummy variable.

 

[Turion]

Remember that $u$ is a dummy variable.

Yeah, I substituted it back in the last step so there is no more u. I'm not sure how that would help me.

 

[vela]

I don't think you understood my point. In the integral $\int_0^t u^3\,du$, it doesn't matter what letter we use for the variable of integration. It's a dummy variable.

http://mathworld.wolfram.com/DummyVariable.html

If you evaluated the integral, what variable is it a function of?

 

[Turion]

I don't think you understood my point. In the integral $\int_0^t u^3\,du$, it doesn't matter what letter we use for the variable of integration. It's a dummy variable.

http://mathworld.wolfram.com/DummyVariable.html

If you evaluated the integral, what variable is it a function of?

It's a function of u which is the same thing as saying it's a function of t - tau.

 

[vela]

Really? Did you actually calculate $\int_0^t u^3\,du$?

 

[Turion]

Really? Did you actually calculate $\int_0^t u^3\,du$?

No, I didn't. The textbook says to apply the identity under "2. Homework Equations " first.

 

[vela]

You wrote

$$L\{ \int _{ 0 }^{ t }{ f(\tau )d\tau } \} =\frac { L\{ f(t)\} }{ s }$$

What variable is $f$ a function of on the left-hand side? What variable is $f$ a function of on the right-hand side? Are they the same? Why not?

 

[Turion]

You wrote
What variable is $f$ a function of on the left-hand side? What variable is $f$ a function of on the right-hand side? Are they the same? Why not?

On the left hand side, f is a function of tau. On the right hand side, f is a function of t. They are not the same. I'm not sure why they're not the same but I can dig up the proof for it I guess. I think they're not the same because we're doing a transformation.

 

[Turion]

I found the proof:

Honestly, that proof (as most proofs) went right over my head.