- #1

- 143

- 2

Please ignore this thread.

Last edited:

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- Thread starter Turion
- Start date

- #1

- 143

- 2

Please ignore this thread.

Last edited:

- #2

- 143

- 2

The correct solution:

- #3

- 143

- 2

Actually, I think the solutions manual is wrong. Please ignore this thread.

- #4

- 329

- 34

##Lf = \int_0^{\infty} e^{-st}f(t)dt ##

which you can split up into the sum of integrals from (0,a) (a,2a) ... ([n-1]a,na) (na,t), where na < t < (n+1)a.

Each of these is easy enough to evaluate, but be careful to separate out the case where n is even and n is odd (why?). I'll bet you can factor some stuff out of the resulting sum.

Why don't you try this out? And why don't you quote theorem 4.3.3 -- maybe it's an easier way to do things.

- #5

- 143

- 2

##Lf = \int_0^{\infty} e^{-st}f(t)dt ##

which you can split up into the sum of integrals from (0,a) (a,2a) ... ([n-1]a,na) (na,t), where na < t < (n+1)a.

Each of these is easy enough to evaluate, but be careful to separate out the case where n is even and n is odd (why?). I'll bet you can factor some stuff out of the resulting sum.

Why don't you try this out? And why don't you quote theorem 4.3.3 -- maybe it's an easier way to do things.

Sorry for wasting your time, but I already posted:

Actually, I think the solutions manual is wrong. Please ignore this thread.

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