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Unformatted text preview: 1 Solutions 47 But this last integral does not exist. Since the Laplace transform does not
exist at , for any , the Laplace transform does not exist.
43. is of exponential type because it is continuous and bounded. On the
. Suppose there is a
and so that
. We need only show that there are some for
which this inequality does not hold. Since
and let’s focus on those for which
. This happens when
is a multiple of
. If the inequality
is valid for all
it is valid for
. We then get the
. Now divide by
, combine, complete the
square, and simplify to get the inequality
. Then this last inequality is not satisﬁed.
It follows that
is not of exponential type. Now consider the deﬁnite
and compute by parts: We get Since is bounded and it follows that Taking limits as
in the equation above gives
. The righthand side exists because
(a) Show that
(b) Show that satisﬁes the recursion formula
(Hint : Integrate by parts.)
(c) Show that
when is a nonnegative integer.
The second equality is obtained by integration by parts using
(c) Repeated use of (b) gives .
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.
- Fall '08