The whole method rests on one computation. Integrating by parts, the transform of a derivative is s times the transform of the function, minus the value of the function at zero. Differentiation in time becomes multiplication by s, and the initial value appears on its own, without being asked for. Applied twice, the rule gives the transform of a second derivative, which carries both the initial value and the initial slope.

A currency exchange that charges a fixed fee. Every derivative, changed into the world of s, becomes a multiplication by s, less a fee equal to the starting value.

If $f$ is continuous and of exponential order and $f'$ is piecewise continuous, then $\mathcal{L}\{f'\}=sF(s)-f(0)$. Applying the rule to $f'$ gives the second derivative, and in general each derivative brings one more factor of $s$ and one more initial value. The same computation, run in $s$ instead, gives the transform of $tf(t)$: it is $-F'(s)$.

• $f=\sin t$: $\mathcal{L}\{\cos t\}=s\cdot\tfrac{1}{s^{2}+1}-0=\tfrac{s}{s^{2}+1}$, as the table says.

Integrate by parts, putting the derivative on the exponential. The boundary term at the far end dies, because the function grows more slowly than the weight decays, and at zero it leaves minus the initial value. The integral that remains is s times the transform of the function itself.

Integrate by parts, differentiating the exponential.

Evaluate the boundary term.

Exponential order makes the far end vanish.

Let T grow: the integral on the right becomes the transform of f.

The transform of a derivative is s times the transform of the function, minus the starting value.

Great work! You now know how to:

• transform first and second derivatives
• prove the rule by integration by parts
• transform an initial value problem into algebra
• say why the rule needs a continuous function