If a function ??(??) meets two technical conditions, then its Laplace transform ??(??) exists. To simplify this discussion, we will label functions ??(??) that satisfy these two conditions "reasonable" in this activity.
What was not mentioned is that under the same conditions, not only does ??(??) exist, but its limit as s goes to infinity is always zero:
lim ??(??) = 0. ??→∞
A non-technical explanation for that is simple. According to the definition,
∞
??(??) = ∫ ??-??????(??)????. 0
As ?? goes to infinity, the factor ??-???? in the integral goes to zero very fast, which causes the whole integral to go to zero.
Knowing this gives us a way of determining that certain functions of ?? cannot be Laplace transforms of reasonable functions: if lim ??(??) =? 0 or if that limit does not exist, we know that ??→∞???(??) cannot be the Laplace transform of a reasonable function ??(??).
1. Explain why ??(??) = arctan ?? cannot occur as a Laplace transform of a reasonable function ??(??).
2. There is exactly one constant C for which ??(??) = arctan ?? + ?? satisfies lim ??(??) = 0. ??→∞ ?Find this constant. No explanation is necessary. Then find the inverse Laplace transform of ??(??) = arctan ?? + ?? for this constant. Hint: use the formula L{(-??)??(??)} = ??′(??).
3. Is the unit impulse aka Dirac Delta function ??(??) a reasonable function or not? Explain by using what we learned about its Laplace transform.
4. Combine what you learned in the previous two parts to find an unreasonable function whose Laplace transform is arctan ??.