The Definition — Question 4

Question 4 – Solution

Strategy. An improper transform can fail at either endpoint. Rapid decay at infinity cannot repair a nonintegrable singularity at zero.

Step 1: Separate the endpoints. Write $a=s+1$. Near zero, $e^{-at}$ is bounded above and below by positive constants, and $\int_0^1t^{-1/2}\,dt=2$, so this endpoint is integrable for every fixed real $s$. At infinity, $a>0$ gives exponential decay. If $a=0$, the integral of $t^{-1/2}$ diverges; if $a<0$, the integrand eventually exceeds one. Hence convergence occurs exactly for $s>-1$, and is absolute.

Step 2: Evaluate the convergent integral. For $a>0$, use $u=\sqrt{at}$, so $t=u^2/a$ and $dt=2u\,du/a$. Then $$F(s)=\frac 2{\sqrt a}\int_0^\infty e^{-u^2}\,du =\boxed{\frac{\sqrt\pi}{\sqrt{s+1}},\qquad s>-1.}$$ Both the substitution and the square root require the already established condition $a>0$. An assigned value at the single point zero does not change the improper integral.

Step 3: Test the stronger singularity. For $h$, the integrand is $t^{-1}e^{-(s+1)t}$. On a sufficiently small interval $(0,\delta)$, its exponential factor is at least $1/2$. Thus $$\int_0^\delta t^{-1}e^{-(s+1)t}\,dt \ge\tfrac 12\int_0^\delta\frac{dt}{t}=\infty.$$ This holds for every real $s$. Consequently $h$ has no transform under this definition, even for parameters that give fast decay at infinity.

Step 4: Interpret the sufficient theorem. The first function is unbounded as $t\to 0^+$, so it does not satisfy the usual finite-limit requirement for piecewise continuity on $[0,A]$. Yet its endpoint singularity is integrable, and the direct calculation establishes existence for $s>-1$. A sufficient theorem guarantees existence when its hypotheses hold; failure of a hypothesis does not prove nonexistence. Here direct endpoint analysis gives the sharper conclusion.