The Definition — Question 7

Question 7 – Solution

Strategy. The integral depends on weighted areas, not just pointwise heights. Quantify the widths before invoking a growth criterion.

Step 1: Examine local regularity and height. Since $w_n<1$, the spike intervals are disjoint. Any bounded time interval meets only finitely many spikes, each with finite height and finite one-sided limits. Thus $f$ is piecewise continuous locally. If an exponential-order bound held, evaluating at $t=n$ for large integers would give $e^{n^2}\le Me^{an}$, or $n^2-an\le\ln M$. The left side tends to infinity, a contradiction.

Step 2: Estimate each weighted area. Fix real $s$. On the $n$th spike, $t\le n+1$ and $e^{-st}\le e^{|s|(n+1)}$. Its contribution is therefore at most $$e^{n^2}w_n e^{|s|(n+1)}=e^{-n^2+|s|(n+1)}.$$ For sufficiently large $n$, $|s|(n+1)\le n^2/2$, so these terms are at most $e^{-n^2/2}\le e^{-n/2}$. The comparison series converges. Since the integrand is nonnegative, the bound proves absolute convergence for every real $s$.

Step 3: Sum the exact contributions. Split finite integrals into their finitely many spike intervals, then pass to the limit using the convergent positive series. For $s\ne 0$, $$\boxed{F(s)=\sum_{n=1}^\infty e^{n^2-sn}\frac{1-e^{-sw_n}}s.}$$ The quotient is positive even for negative $s$. At $s=0$, each contribution is simply height times width, giving $$\boxed{F(0)=\sum_{n=1}^\infty e^{-n^2}.}$$ No closed form for this convergent series is required.

Step 4: Interpret the theorem correctly. The heights grow faster than every exponential, but the widths shrink fast enough that their areas are $e^{-n^2}$. Any fixed exponential weight still leaves a summable sequence of areas. The common existence theorem provides sufficient conditions; exponential order is not necessary for an individual function to have a Laplace transform. This example establishes that distinction by direct estimates, rather than by trying to apply a theorem whose hypothesis fails.