The Definition — Question 6

Question 6

For real $s$, use $F(s)=\int_0^\infty e^{-st}f(t)\,dt$ whenever this ordinary improper integral converges.

Compare $f_+(t)=e^{t^2}$ and $f_-(t)=e^{-t^2}$ for $t\ge 0$, and write their transforms as $F_+$ and $F_-$.

Tasks

1. Determine the real convergence set of each transform directly from the defining integral.

2. Complete the square to express $F_-(s)$ through a Gaussian tail for every real $s$. No special-function notation is required.

3. For a truncation at $T$ with $T+s/2>0$, prove an explicit upper bound for the omitted tail of $F_-(s)$.

4. At $s=-2$, find the smallest integer $T>1$ for which your bound certifies an error at most $10^{-4}$. Explain why negative $s$ does not destroy convergence for $f_-$.

Question 6 – Solution

Strategy. Compare the quadratic growth or decay with the linear term supplied by the Laplace weight, and use the completed square to bound numerical truncation.

Step 1: Classify both integrals. For any real $s$, $t^2-st\ge 0$ when $t\ge\max\{s,0\}$, so $e^{t^2-st}\ge 1$ on a half-line. Hence $F_+$ diverges for every real $s$. For $F_-$, $$-t^2-st=-(t+s/2)^2+s^2/4.$$ A shifted Gaussian has an integrable tail and no finite-endpoint singularity, so $F_-$ converges absolutely for every real $s$.

Step 2: Express the exact transform. Substitution $u=t+s/2$ gives $$\boxed{F_-(s)=e^{s^2/4}\int_{s/2}^\infty e^{-u^2}\,du,\qquad s\in\mathbb R.}$$ The lower endpoint is allowed to be negative; the finite portion of the integral then remains harmless. This definite integral is an exact answer.

Step 3: Bound the omitted tail. Let $a=T+s/2>0$. Since $1\le u/a$ for $u\ge a$, $$\int_a^\infty e^{-u^2}\,du\le\frac 1a\int_a^\infty u e^{-u^2}\,du =\frac{e^{-a^2}}{2a}.$$ Consequently the positive truncation error satisfies $$\boxed{\int_T^\infty e^{-t^2-st}\,dt \le\frac{e^{-T^2-sT}}{2T+s}.}$$ The condition $2T+s>0$ is part of the bound and cannot be dropped.

Step 4: Certify a cutoff. For $s=-2$, the bound is $e^{-T^2+2T}/(2T-2)$, decreasing for $T>1$. At $T=3$ it is $e^{-3}/4\approx 0.01245$, while at $T=4$ it is $e^{-8}/6\approx 5.59\times 10^{-5}$. Thus the smallest integer cutoff certified by this bound is $\boxed{T=4}$. Negative $s$ makes the weight grow exponentially, but the quadratic decay still dominates. The graph shows the weighted integrands, whose areas give the transforms.