Second Shift Theorem: Integral Explanation

[goldfish9776]

Homework Statement

for the alternative form of second shift property (4.8) , why he integral of (e^-sp) g(p+a) dp isn't equal to integral of (e^-sp) g(t) dp ? why it will become L{ g(t+a) } ?

Homework Equations

The Attempt at a Solution

 

[RUber]

Would you mind showing more work explaining your reasoning?
I imagine the answer is either in the endpoints of your integrals or the substitution.

Straight off, from your question, for $t = p+a$
$\int e^{-sp}g(p+a) dp = \int e^{-sp}g(t) dp$
but you can't do much with that...complete the substitution to change the whole integral into one with only dependence on t and no p's.

 

[goldfish9776]

Would you mind showing more work explaining your reasoning?
I imagine the answer is either in the endpoints of your integrals or the substitution.

Straight off, from your question, for $t = p+a$
$\int e^{-sp}g(p+a) dp = \int e^{-sp}g(t) dp$
but you can't do much with that...complete the substitution to change the whole integral into one with only dependence on t and no p's.

ya , why not the final ans not L {g(p+a)} ?

 

[RUber]

Write out the steps with the limits of integration. Do the substitutions where appropriate to make sure the new integrals are from zero to infinity.

 

[goldfish9776]

Write out the steps with the limits of integration. Do the substitutions where appropriate to make sure the new integrals are from zero to infinity.

is my working until here correct ?

 

[RUber]

It looks like you are treating g(p+a) as a multiplication. g(x) is a function.
Also, when you do substitutions within integrals, you have to change the limits as well.

For example:
Substituting $t = p+a\quad p = t-a\quad p=0 \implies t = a$
$\int_0^\infty e^{-sp}g(p+a) dp = \int_{a}^\infty e^{-s(t-a)}g(t) dt$
Does that make sense?

 

[goldfish9776]

It looks like you are treating g(p+a) as a multiplication. g(x) is a function.
Also, when you do substitutions within integrals, you have to change the limits as well.

For example:
Substituting $t = p+a\quad p = t-a\quad p=0 \implies t = a$
$\int_0^\infty e^{-sp}g(p+a) dp = \int_{a}^\infty e^{-s(t-a)}g(t) dt$
Does that make sense?

i gt stucked here , how to continue?

 

[RUber]

You are looking to show that $\mathcal{L}\{ g(t) \mathcal{H}(t-a) \} = e^{-as}\mathcal{L}\{ g(t+a)\}.$
Start by expanding the transform:
$\mathcal{L}\{ g(t) \mathcal{H}(t-a) \} =\int_0^\infty e^{-st}g(t) \mathcal{H}(t-a)dt$
Apply the Heaviside function:
$\qquad \qquad =\int_a^\infty e^{-st}g(t) dt$
Because this integral does not start at 0, we need to shift it with a substitution.
Substitute t = p+a ... p = t-a ... dt = dp ...and where t = a, p = 0 (for the limits of integration).
$\qquad \qquad =\int_0^\infty e^{-s(p+a)}g(p+a) dp=\int_0^\infty e^{-sp-sa}g(p+a) dp$
Since a and s are not variables inside the integral, you can move them to the outside:
$\qquad \qquad =e^{-as} \int_0^\infty e^{-sp}g(p+a) dp.$
In this form you can see that this is the same as
$e^{-as}\mathcal{L}\{ g(p+a)\}$ but since p could be any letter, you can name it t.
This will give you what you want to show:
$\mathcal{L}\{ g(t) \mathcal{H}(t-a) \} = e^{-as}\mathcal{L}\{ g(t+a)\}.$

 

[RUber]

i gt stucked here , how to continue?

Two things I noticed in your post #7...
1) There is no need to do integration by parts on this.
2) You did not distribute properly in your exponential: -s(t-a) = -st+sa.