# Solve a Laplace transform puzzle

1. Jan 18, 2014

### davyyao

Hi~
I recently solve a Laplace transform problem as following
L[int{t,0}cosh(t'-1)U(t'-1)dt']=? U(t'-1) is the unit step function(=1 for t'>1, =zero otherwise)
According the standard Laplace formula :
(1)L[cosh(t-1)U(t-1)]=exp(-s)*s/(s^2-1);
(2)L(int{t,0}f(t')dt')=F(s)/s. where F(s)=L(f(t))

I conclude the answer to be:
exp(-s)/(s^2-1)

But from the direct integral formula:
L[int{t,0}cosh(t'-1)U(t'-1)dt']=L[sinh(t-1)]=(1/2)*(exp(-1)/(s-1)-exp(1)/(s+1)).

This two answers are not consistent!
I totally have no idea what's wrong with the two methods?

I believe the second method should be correct by basic definition of Laplace transformation.

Thanks very much!

2. Jan 18, 2014

### thelema418

How did you find the direct integral (the step at the first equal sign)?

3. Jan 19, 2014

### Ray Vickson

$$\int_0^t \cosh(t'-1) U(t'-1) \, dt' \neq \sinh(t-1)$$
because the integral vanishes if t < 1 but sinh(t-1) does not.

4. Jan 19, 2014

### HallsofIvy

$$\int_0^t cosh(t'- 1)U(t'-1)dt'= \int_1^t cos(t'- 1)dt'$$
and with the substitution u= t'- 1, du= dt', when t'= 1, u= 0 and when t'= t, u= t- 1 so that becomes
$$\int_0^{t-1} cosh(u)du$$.

5. Jan 20, 2014

### davyyao

Thanks Ray Vickson and Hallsoflvy for nice answering.
I think I make mistake to deal with the unit step function in the integral form.
It clear now for me!