Table Of Laplace Transforms — Question 3

Question 3 – Solution

Strategy. Generate missing table rows through operations whose normalization can be checked independently.

Step 1: Differentiate with respect to $s$. Since $-\frac{d}{ds}[a/(s^2+a^2)]=2as/(s^2+a^2)^2$, $$\boxed{\mathcal L^{-1}\{A_a\}=\frac{t\sin(at)}{2a}.}$$ The factor $2a$ must be divided out.

Step 2: Differentiate with respect to $a$. The parameter derivative gives $\mathcal L\{t\cos(at)\}=(s^2-a^2)/(s^2+a^2)^2$. For fixed $s>0$, $t e^{-st}$ dominates the differentiated integrand uniformly in $a$, justifying differentiation under the integral. Subtracting this row from $\mathcal L\{\sin(at)/a\}$ yields $$\frac 1{s^2+a^2}-\frac{s^2-a^2}{(s^2+a^2)^2} =\frac{2a^2}{(s^2+a^2)^2}.$$ Thus $$\boxed{\mathcal L^{-1}\{B_a\}= \frac{\sin(at)-at\cos(at)}{2a^3}.}$$

Step 3: Check the onset powers. The inverses start as $t^2/2+O(t^4)$ and $t^3/6+O(t^5)$. Their leading transforms are respectively $1/s^3$ and $1/s^4$, matching $A_a$ and $B_a$. This checks both factorials and the cancellation of the linear term in the second numerator.

Step 4: Remove the apparent parameter singularities. At each fixed $t$, the Taylor expansions give limits $t^2/2$ and $t^3/6$. Their transforms are $1/s^3$ and $1/s^4$, agreeing with the limits of the rational functions for $s>0$. Directly inserting $a=0$ into the inverse quotients produces $0/0$; the limits require cancellation first. The figure uses $a=1$ and shows that the repeated denominator can produce oscillations with increasing size, unlike a single sine entry.