luobo : i am looking forward to seeing your solution , actually , I have been shocked to see how you (or someone ) applied magic differentiation in the Integral Bee !! It is really a powerful tool (Clapping)
i think the solution is two straight lines passing throguh the origin (at,bt) and (-at,bt) (Happy)
Here's another solution to my proposed problem using a powered tool:
Now since and is an integrable function on besides on we get and on we have hence the Dominated Convergence Theorem applies and
1) therefore applying integration by parts twice gives us:
2) we know that where is the sign function.
3) using some routine trigonometry identities we get: .
4) applying 1) to 3) and then using 2) will finally give us:
The problem would be easier to solve if it changes this way:
Solve for real values of
Step (1): Show for a fixed , the following integral is proportional to , i.e.
Step (2): Solve for real values of such that
For Step (1), can we use the "differentiation"? hahaha...