# Second shift theorem

1. Feb 2, 2016

### goldfish9776

1. The problem statement, all variables and given/known data
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) } ?

2. Relevant equations

3. The attempt at a solution

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2. Feb 2, 2016

### 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.

3. Feb 2, 2016

### goldfish9776

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

4. Feb 2, 2016

### 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.

5. Feb 2, 2016

### goldfish9776

is my working until here correct ?

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6. Feb 3, 2016

### 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?

7. Feb 3, 2016

### goldfish9776

i gt stucked here , how to continue?

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8. Feb 4, 2016

### 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:
$\mathcal{L}\{ g(t) \mathcal{H}(t-a) \} = e^{-as}\mathcal{L}\{ g(t+a)\}.$

9. Feb 4, 2016

### RUber

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.