Exercises — Transforms of standard functions — proofs

• $\mathcal{L}\{1\}=\frac{1}{s}$, for $s>0$.
• $\mathcal{L}\{t\}=\frac{1!}{s^2}=\frac{1}{s^2}$, for $s>0$.
• $\mathcal{L}\{t^4\}=\frac{4!}{s^5}=\frac{24}{s^5}$, for $s>0$.

Why the factorial? The recursion $I_n=\frac{n}{s}{I}_{n-1}$ tacks on one new factor each step, so $t^4$ carries $4!=24$, not just $1$.

Here $a=3$, and $a^2=9$: $\mathcal{L}\{\cos 3t\}=\frac{s}{s^2+9},\mathcal{L}\{\sin 3t\}=\frac{3}{s^2+9},s>0.$ The numerator of $\cos$ is $s$ (it equals $1$ at $t=0$); the numerator of $\sin$ is $a=3$.

• $\mathcal{L}\{e^{5t}\}=\frac{1}{s-5}$, for $s>5$ (need $s-5>0$ so $e^{-(s-5)t}$ decays).
• $\mathcal{L}\{\cosh 2t\}=\frac{s}{s^2-4}$, for $s>2$ (i.e. $s>|a|$ with $a=2$; there are real poles at $\pm 2$).

Term by term (linearity lets us transform each piece and add): $\mathcal{L}\{3t^2\}=3\cdot \frac{2!}{s^3}=\frac{6}{s^3},\mathcal{L}\{2\sin 4t\}=2\cdot \frac{4}{s^2+16}=\frac{8}{s^2+16},\mathcal{L}\{5\}=\frac{5}{s}.$ $\boxed{\mathcal{L}\{3t^2-2\sin 4t+5\}=\frac{6}{s^3}-\frac{8}{s^2+16}+\frac{5}{s}}s>0.$

Expand $(t+1{)}^2=t^2+2t+1$ so every piece is a table entry: $\mathcal{L}\{t^2\}=\frac{2}{s^3},\mathcal{L}\{2t\}=\frac{2}{s^2},\mathcal{L}\{1\}=\frac{1}{s}.$ $\boxed{\mathcal{L}\{(t+1{)}^2\}=\frac{2}{s^3}+\frac{2}{s^2}+\frac{1}{s}}s>0.$ Why expand? $\mathcal{L}$ has no "product rule" — you cannot transform $(t+1{)}^2$ as a product. Expanding turns it into a sum, where linearity works.

${\cos }^{2}t=\frac{1}{2}+\frac{1}{2}\cos 2t$. With $a=2$: $\mathcal{L}\{{\cos }^{2}t\}=\frac{1}{2}\cdot \frac{1}{s}+\frac{1}{2}\cdot \frac{s}{s^2+4}=\frac{1}{2s}+\frac{s}{2(s^2+4)}s>0.$ Combine over a common denominator if desired: $\frac{(s^2+4)+s^2}{2s(s^2+4)}=\frac{2s^2+4}{2s(s^2+4)}=\frac{s^2+2}{s(s^2+4)}$.

Split the fraction to match table shapes (numerators $s$ and constant separately): $\frac{2s+5}{s^2+9}=2\cdot \frac{s}{s^2+9}+\frac{5}{3}\cdot \frac{3}{s^2+9}.$ The first is $\cos 3t$'s transform, the second is $\sin 3t$'s (with $a=3$): $\boxed{{\mathcal{L}}^{-1}\left\{\frac{2s+5}{s^2+9}\right\}=2\cos 3t+\frac{5}{3}\sin 3t}$ Why rewrite the $5$ as $\frac{5}{3}\cdot 3$? The sine entry needs a bare $a=3$ on top; we manufacture it and rebalance with $\frac{5}{3}$.

This matches $\frac{a}{s^2-a^2}$ with $a=2$, but that entry has an $a=2$ on top. We have $1$, so pull out $\frac{1}{2}$: $\frac{1}{s^2-4}=\frac{1}{2}\cdot \frac{2}{s^2-4}\Rightarrow \boxed{{\mathcal{L}}^{-1}\left\{\frac{1}{s^2-4}\right\}=\frac{1}{2}\sinh 2t}$ The minus sign in $s^2-4$ signals hyperbolic, not trig. (See the Inverse Laplace Transform — Partial Fractions note for the general splitting method.)

Use the first shifting theorem: multiplying by $e^{3t}$ shifts $s\to s-3$. We know $\mathcal{L}\{\cosh 2t\}=\frac{s}{s^2-4}$ for $s>2$. Therefore $\mathcal{L}\{e^{3t}\cosh 2t\}=\frac{(s-3)}{(s-3{)}^2-4}.$ Convergence shifts too: the original needed $s>2$; after the shift we need $s-3>2$, i.e. $\boxed{s>5}$.

Start with $\mathcal{L}\{\sin 3t\}=\frac{3}{s^2+9}$ ($s>0$). Multiplying by $e^{-2t}$ means shift $s\to s-(-2)=s+2$: $\boxed{\mathcal{L}\{e^{-2t}\sin 3t\}=\frac{3}{(s+2{)}^2+9}}s>-2.$ The abscissa moved from $0$ to $-2$ (shift by $a=-2$).

$\mathcal{L}\{t\}=\frac{1}{s^2}$ ($s>0$). Multiply the time function by $e^{2t}$ → shift $s\to s-2$: $\boxed{\mathcal{L}\{t{e}^{2t}\}=\frac{1}{(s-2{)}^2}}s>2.$ Both readings ("shift the $t$ transform" and "$t^1$ times $e^{2t}$") are the same theorem; they agree.

First, $\mathcal{L}\{t{e}^{iat}\}$: this is $t^1$ multiplied by $e^{iat}$, so shift $\mathcal{L}\{t\}=\frac{1}{s^2}$ by $s\to s-ia$: $\mathcal{L}\{t{e}^{iat}\}=\frac{1}{(s-ia{)}^2}.$ Rationalise by multiplying top and bottom by $(s+ia{)}^2$: $\frac{1}{(s-ia{)}^2}=\frac{(s+ia{)}^2}{((s-ia)(s+ia){)}^2}=\frac{(s+ia{)}^2}{(s^2+a^2{)}^2}.$ Expand the top: $(s+ia{)}^2=s^2-a^2+2ias$. So $\mathcal{L}\{t{e}^{iat}\}=\frac{(s^2-a^2)+i(2as)}{(s^2+a^2{)}^2}.$ Since $t{e}^{iat}=t\cos at+it\sin at$, the real part gives $\mathcal{L}\{t\cos at\}$: $\boxed{\mathcal{L}\{t\cos at\}=\frac{s^2-a^2}{(s^2+a^2{)}^2}}(\text{and }\mathcal{L}\{t\sin at\}=\frac{2as}{(s^2+a^2{)}^2}).$

Take $I={\int }_0^{\infty }{e}^{-st}tdt$. Choose $u=t$, $dv=e^{-st}dt$, so $du=dt$, $v=\frac{e^{-st}}{-s}$: $I={\left[t\cdot \frac{e^{-st}}{-s}\right]}_0^{\infty }-{\int }_0^{\infty }\frac{e^{-st}}{-s}dt.$ Boundary term. At $t=0$ it is $0$. At $t\to \infty$, $t{e}^{-st}\to 0$ because $s>0$ — the exponential beats the linear factor. (If $s\le 0$ this blows up: this is where the condition is born.) Remaining integral. $-{\int }_0^{\infty }\frac{e^{-st}}{-s}dt=\frac{1}{s}{\int }_0^{\infty }{e}^{-st}dt=\frac{1}{s}\cdot \frac{1}{s}=\frac{1}{s^2}$. $\boxed{\mathcal{L}\{t\}=\frac{1}{s^2},s>0.}$

$\mathcal{L}\{e^{at}\}={\int }_0^{\infty }{e}^{-st}{e}^{at}dt={\int }_0^{\infty }{e}^{-(s-a)t}dt={\left[\frac{e^{-(s-a)t}}{-(s-a)}\right]}_0^{\infty }.$ At $t\to \infty$ the exponential $\to 0$ iff the exponent is negative, i.e. $s-a>0$. Then the upper limit contributes $0$ and the lower gives $-\frac{1}{-(s-a)}=\frac{1}{s-a}$: $\boxed{\mathcal{L}\{e^{at}\}=\frac{1}{s-a},s>a.}$

Substitute the second equation into the first: $S=\frac{a}{s}\left(\frac{1}{s}-\frac{a}{s}S\right)=\frac{a}{s^2}-\frac{a^2}{s^2}S.$ Collect $S$: $S\left(1+\frac{a^2}{s^2}\right)=\frac{a}{s^2}\Rightarrow S\cdot \frac{s^2+a^2}{s^2}=\frac{a}{s^2}$, so $S=\frac{a}{s^2+a^2}.$ Back-substitute into $C=\frac{1}{s}-\frac{a}{s}S$: $C=\frac{1}{s}-\frac{a}{s}\cdot \frac{a}{s^2+a^2}=\frac{(s^2+a^2)-a^2}{s(s^2+a^2)}=\frac{s^2}{s(s^2+a^2)}=\frac{s}{s^2+a^2}.$ $\boxed{\mathcal{L}\{\sin at\}=\frac{a}{s^2+a^2},\mathcal{L}\{\cos at\}=\frac{s}{s^2+a^2}.}$

Which level am I comfortable with? ::: L1 read table · L2 linearity on sums · L3 invert + track convergence · L4 shifting + complex exponentials · L5 prove from the integral. The one theorem that dominates L3–L4 ::: First shifting: $\mathcal{L}\{e^{at}f(t)\}=F(s-a)$, and the abscissa shifts by $+a$ too.