Is the sin function dependant on t & x or is the x a typo?
Use the infinite geometric series for to get that for t>0.
Thus
And you can find either via integration by parts or via a table of integrals. Integrating each term thus gives you an infinite series for your answer.
--Kevin C.
Besides complex variables , i have a nice solution for this question .
I'd like to make use of these two lemmas :
and
Since
the series becomes :
Use the lemma given , replace by
We have
The integral
Although you may think it is a stupid method , i think it is very interesting indeed !
OK, you asked for it.
Let's integrate the function around the rectangular contour with vertices at , , and in the complex plane , indented at and at with quarter circles of radius . is analytic inside and on and so . This leads to
where represents the integral around the indentation at . Simplify, simplify...
(*) where .
It's easy to show that as irrespective of the value of , and by the indentation theorem,
and .
Obviously we have a problem letting tend to zero in the first two terms of (*). The key observation here is the identity
. So we change to in (*) to get
.
Now changing to in the second integral yields
, and dividing by gives
(**) .
Are you still watching this? Good. Now add (*) and (**) to see what happens. Don't forget the identity!
Before rigour mortis sets in we should let and to arrive at these equations:
The second term on the left is , so finally we have full view of the answer:
.
Isn't Cauchy wonderful?
You can see how much easier the series method is. I'm off for a nice cup of tea.