The Definition — Question 1

Question 1

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

Let $f(t)=e^{2t}(1+t)$ for $t\ge 0$. A student obtains the expression $1/(s-2)+1/(s-2)^2$ and claims that it gives a transform value at every real $s\ne 2$.

Tasks

1. Evaluate the truncated defining integral from zero to $R$, treating $s=2$ separately.

2. Determine exactly which real $s$ give convergence, and find $F(s)$ on that set.

3. Evaluate the student’s expression at $s=1$ and explain why it cannot be the Laplace integral there.

4. Prove directly from the integral, without differentiating the final formula, that $F$ is positive and strictly decreasing throughout its convergence interval.

Question 1 – Solution

Strategy. Keep the upper endpoint finite until the parameter-dependent boundary terms have been examined.

Step 1: Evaluate the finite integral. Write $a=s-2$. Integration by parts gives, for $a\ne 0$, $$I_R=\int_0^R(1+t)e^{-at}\,dt =\frac 1a+\frac 1{a^2}-e^{-aR}[\frac{1+R}{a}+\frac 1{a^2}].$$ At $a=0$, direct integration gives $I_R=R+R^2/2$. The formula with denominators cannot be used at that parameter.

Step 2: Determine the convergence set. For $a>0$, both $e^{-aR}$ and $Re^{-aR}$ tend to zero. Therefore $$\boxed{F(s)=\frac 1{s-2}+\frac 1{(s-2)^2},\qquad s>2.}$$ For $a=0$, the displayed polynomial diverges. For $a<0$, the integrand $(1+t)e^{|a|t}$ is at least one, so $I_R\ge R\to\infty$. Thus $s>2$ is the entire real convergence set, and convergence there is absolute.

Step 3: Reject an algebraic continuation as an integral value. At $s=1$, the rational expression equals $-1+1=0$. But the actual integral is $$I_R=\int_0^R(1+t)e^t\,dt=Re^R\longrightarrow\infty.$$ The discarded boundary term grows instead of vanishing. A finite value of the algebraic expression outside the convergence interval does not assign a value to the defining improper integral.

Step 4: Prove order properties from the kernel. For $s>2$, the integrand is positive, so $F(s)>0$. If $s_2>s_1>2$, then for every $t>0$, $$0<e^{-s_2t}f(t)<e^{-s_1t}f(t).$$ Both integrals converge, and their difference is strictly positive on any fixed interval of positive length inside $(0,\infty)$. Hence $F(s_2)<F(s_1)$. This argument shows why positivity and monotonicity hold, independently of the rational simplification.