Exercises — Laplace transform — definition, region of convergence

WHAT we do: plug $f(t)=1$ into the definition. $F(s)={\int }_0^{\infty }{e}^{-st}\cdot 1dt={\left[\frac{-1}{s}{e}^{-st}\right]}_0^{\infty }.$ WHY the upper limit needs a condition: the term $e^{-st}$ has size $|e^{-st}|=e^{-\sigma t}$ where $\sigma =\mathrm{Re}(s)$. It shrinks to $0$ only if $\sigma >0$. At $t=0$ we get $e^0=1$. $F(s)=0-\left(\frac{-1}{s}\right)=\boxed{\frac{1}{s}},\mathrm{Re}(s)>0.$

Combine the exponents first (cleaner integral): $F(s)={\int }_0^{\infty }{e}^{-(s-a)t}dt={\left[\frac{-1}{s-a}{e}^{-(s-a)t}\right]}_0^{\infty }.$ The upper limit dies iff $\mathrm{Re}(s-a)>0$, i.e. $\mathrm{Re}(s)>a$. $F(s)=\boxed{\frac{1}{s-a}},\mathrm{Re}(s)>a.$ The growth rate $a$ becomes the pole of $F(s)$ — the value of $s$ where the denominator $s-a$ is zero and $F$ blows up — and the boundary line of the ROC. Same number, two jobs.

False. The integrand is $e^{-st}{e}^{t^2}=e^{t^2-st}$. Fix any $s$; as $t\to \infty$ the exponent $t^2-st\to +\infty$ because $t^2$ eventually dominates any linear $st$. So the integrand grows without bound for every $s$ — the integral diverges everywhere. $f$ is not of exponential order (no fixed $a$ can bound $e^{t^2}$), hence no transform. See Exponential Order and Growth Rates.

WHY integration by parts (IBP): the integrand is a product of an algebraic factor $t$ and an exponential $e^{-st}$. IBP trades the awkward $t$ for a constant. Let $u=t,dv=e^{-st}dt\Rightarrow du=dt,v=-\frac{1}{s}{e}^{-st}.$ $F(s)={\left[-\frac{t}{s}{e}^{-st}\right]}_0^{\infty }+\frac{1}{s}{\int }_0^{\infty }{e}^{-st}dt.$ The boundary term vanishes for $\mathrm{Re}(s)>0$ (exponential decay beats the linear $t$). The remaining integral is $\mathcal{L}\{1\}=\frac{1}{s}$. $F(s)=\frac{1}{s}\cdot \frac{1}{s}=\boxed{\frac{1}{s^2}},\mathrm{Re}(s)>0.$

The picture below shows the IBP trade visually: the tall factor $t$ (cyan) gets swapped for the flat constant, while $e^{-st}$ (amber) crushes the $t$ down so the boundary term at $\infty$ is zero.

WHY the exponential form: we already own $\mathcal{L}\{e^{at}\}=\frac{1}{s-a}$. Splitting cosine into two exponentials lets us reuse it directly instead of doing IBP twice. $F(s)=\frac{1}{2}\left(\frac{1}{s-i\omega }+\frac{1}{s+i\omega }\right)=\frac{1}{2}\cdot \frac{(s+i\omega )+(s-i\omega )}{s^2+{\omega }^2}=\frac{1}{2}\cdot \frac{2s}{s^2+{\omega }^2}.$ $F(s)=\boxed{\frac{s}{s^2+{\omega }^2}},\mathrm{Re}(s)>0.$ ROC reasoning: $a=\pm i\omega$ are purely imaginary, so both need $\mathrm{Re}(s)>0$; cosine is bounded (growth rate $0$).

The integral is linear, so transform term by term: $\mathcal{L}\{3e^{2t}\}=\frac{3}{s-2}(\mathrm{Re}(s)>2),\mathcal{L}\{5\}=\frac{5}{s}(\mathrm{Re}(s)>0).$ WHY the ROC of a sum is the intersection: both integrals must individually converge for the whole thing to converge. The stricter condition wins: $\mathrm{Re}(s)>2$ implies $\mathrm{Re}(s)>0$, so $F(s)=\boxed{\frac{3}{s-2}+\frac{5}{s}},\mathrm{Re}(s)>2.$

$F(s)={\int }_0^{\infty }t{e}^{at}{e}^{-st}dt={\int }_0^{\infty }t{e}^{-(s-a)t}dt.$ This is exactly the $\mathcal{L}\{t\}$ computation but with $s$ replaced by $(s-a)$. From Exercise 2.1, ${\int }_0^{\infty }t{e}^{-pt}dt=\frac{1}{p^2}$ for $\mathrm{Re}(p)>0$. Set $p=s-a$: $F(s)=\boxed{\frac{1}{(s-a{)}^2}},\mathrm{Re}(s)>a.$ WHAT this reveals: multiplying $f(t)$ by $e^{at}$ shifts $F(s)\to F(s-a)$ and shifts the ROC boundary from $0$ to $a$. This is the first shifting theorem falling out of the definition.

The figure makes the shift literal: attaching $e^{at}$ to $f$ slides the whole $s$-picture (pole and ROC boundary together) to the right by $a$.

(i) $t^{10}$: any polynomial is beaten by every positive exponential, i.e. $t^{10}\le M{e}^{\epsilon t}$ for any $\epsilon >0$. So the smallest workable rate is $a=\boxed{0}$. (ROC $\mathrm{Re}(s)>0$.)

(ii) $e^{5t}\sin (3t)$: the sine is bounded by $1$, so $|f|\le e^{5t}$. The envelope grows like $e^{5t}$, so $a=\boxed{5}$.

(iii) $\cosh (4t)=\frac{1}{2}(e^{4t}+e^{-4t})$: the dominant piece is $e^{4t}$, so $a=\boxed{4}$. The $e^{-4t}$ piece only helps convergence.

Step 1 (pull the modulus inside): $|\int g|\le \int |g|$, and $|e^{-st}|=e^{-\sigma t}$: $|F(s)|\le {\int }_0^{\infty }{e}^{-\sigma t}|f(t)|dt.$ Step 2 (use exponential order): by the definition above, $|f(t)|\le M{e}^{at}$, so $|F(s)|\le M{\int }_0^{\infty }{e}^{-(\sigma -a)t}dt=\frac{M}{\sigma -a},\text{finite iff }\sigma >a.$ Step 3 (limit): as $\sigma \to \infty$, the bound $\frac{M}{\sigma -a}\to 0$, forcing $F(s)\to 0$. WHAT this proves: every genuine Laplace transform must decay to $0$ as $\mathrm{Re}(s)\to \infty$. So a "transform" like $F(s)=1$ or $F(s)=s$ can never come from a piecewise-continuous, exponential-order $f$.

The integrand is zero past $t=2$, so the infinite integral is finite: $F(s)={\int }_0^2{e}^{-st}dt={\left[\frac{-1}{s}{e}^{-st}\right]}_0^2=\frac{1}{s}\left(1-e^{-2s}\right).$ $F(s)=\boxed{\frac{1-e^{-2s}}{s}}.$ WHY the ROC is the entire $s$-plane: because the interval is finite, there is no $t\to \infty$ tail to tame — convergence never needed $\mathrm{Re}(s)>0$. The formula looks like it has a pole at $s=0$, but that pole is removable: expanding $1-e^{-2s}=2s-2s^2+\cdots$ gives $F(s)=\frac{2s-2s^2+\cdots }{s}=2-2s+\cdots$, so ${\lim }_{s\to 0}F(s)=2$ (which is just the area of the pulse, ${\int }_0^{2}1dt=2$). Since the apparent singularity cancels, $F(s)$ is finite for every $s$ — the ROC is the whole plane, with no point excluded. The picture below shows why: a signal that switches off can never produce an infinite integral.

Write $\cos 2t=\frac{1}{2}(e^{2it}+e^{-2it})$, then attach $e^{-t}$: $e^{-t}\cos 2t=\frac{1}{2}\left(e^{(-1+2i)t}+e^{(-1-2i)t}\right).$ Each term is $e^{at}$ with $a=-1\pm 2i$, giving $\frac{1}{s-a}$: $F(s)=\frac{1}{2}\left(\frac{1}{s+1-2i}+\frac{1}{s+1+2i}\right)=\frac{1}{2}\cdot \frac{2(s+1)}{(s+1{)}^2+4}.$ $F(s)=\boxed{\frac{s+1}{(s+1{)}^2+4}}.$ ROC: both poles have $\mathrm{Re}(a)=-1$, so we need $\mathrm{Re}(s)>-1$. The $e^{-t}$ factor pushed the ROC boundary left to $-1$ (it helps convergence).

$G(s)=\frac{1}{2}\left(\frac{1}{s-3}-\frac{1}{s+3}\right)=\frac{1}{2}\cdot \frac{(s+3)-(s-3)}{s^2-9}=\frac{1}{2}\cdot \frac{6}{s^2-9}=\boxed{\frac{3}{s^2-9}}.$ ROC reasoning (the key insight): the term $e^{3t}$ needs $\mathrm{Re}(s)>3$; the term $-e^{-3t}$ needs $\mathrm{Re}(s)>-3$. Both must hold, so intersection gives the stricter one, $\mathrm{Re}(s)>3$. The fastest-growing piece always sets the boundary — that's why $+3$ wins over $-3$.

Let $I_n={\int }_0^{\infty }{t}^n{e}^{-st}dt$. Use IBP with $u=t^n,dv=e^{-st}dt$: $I_n={\left[-\frac{t^n}{s}{e}^{-st}\right]}_0^{\infty }+\frac{n}{s}{\int }_0^{\infty }{t}^{n-1}{e}^{-st}dt.$ The boundary term vanishes for $\mathrm{Re}(s)>0$ (exponential beats any power). So $I_n=\frac{n}{s}{I}_{n-1}.$ Unwind the recurrence starting from $I_0=\mathcal{L}\{1\}=\frac{1}{s}$: $I_n=\frac{n}{s}\cdot \frac{n-1}{s}\cdots \frac{1}{s}\cdot \frac{1}{s}=\frac{n!}{s^{n+1}}.$ $\boxed{\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}},\mathrm{Re}(s)>0.$ Check $n=1$: $\frac{1!}{s^2}=\frac{1}{s^2}$ ✓, matches Exercise 2.1.

Start from the known result with $a=i\omega$: $\mathcal{L}\{e^{i\omega t}\}=\frac{1}{s-i\omega },\mathrm{Re}(s)>\mathrm{Re}(i\omega )=0.$ Rationalise by multiplying top and bottom by $s+i\omega$: $\frac{1}{s-i\omega }=\frac{s+i\omega }{s^2+{\omega }^2}=\underset{\text{real}}{\underbrace{\frac{s}{s^2+{\omega }^2}}}+i\underset{\text{imag}}{\underbrace{\frac{\omega }{s^2+{\omega }^2}}}.$ But $e^{i\omega t}=\cos \omega t+i\sin \omega t$, so by linearity the real part is $\mathcal{L}\{\cos \omega t\}$ and the imaginary part is $\mathcal{L}\{\sin \omega t\}$: $\boxed{\mathcal{L}\{\cos \omega t\}=\frac{s}{s^2+{\omega }^2}},\boxed{\mathcal{L}\{\sin \omega t\}=\frac{\omega }{s^2+{\omega }^2}},\mathrm{Re}(s)>0.$ WHY one calculation gives two answers: matching real and imaginary parts of a single complex identity is twice as efficient as two real integrals — the same trick underlies the Fourier Transform.

The two convergence checks (WHY both matter):

1. Near $t=0$: $t^{-1/2}$ is unbounded, but the integral ${\int }_0{t}^{-1/2}dt$ still converges because the exponent $-\frac{1}{2}$ is greater than $-1$ (recall ${\int }_0{t}^{-p}dt$ converges exactly when $p<1$). So the singularity is integrable — fine.
2. Near $t=\infty$: the factor $e^{-st}$ tames the tail as long as $\mathrm{Re}(s)>0$.

The computation, step by step. Substitute $u=st$ (so $t=u/s$ and $dt=du/s$), valid for $\mathrm{Re}(s)>0$: $F(s)={\int }_0^{\infty }{t}^{-1/2}{e}^{-st}dt={\int }_0^{\infty }{\left(\frac{u}{s}\right)}^{-1/2}{e}^{-u}\frac{du}{s}.$ Pull the constant powers of $s$ out. Note ${\left(\frac{u}{s}\right)}^{-1/2}=u^{-1/2}{s}^{1/2}$, and there is a further $\frac{1}{s}$ from $dt$, so the combined $s$-factor is $s^{1/2}\cdot s^{-1}=s^{-1/2}$: $F(s)=s^{-1/2}{\int }_0^{\infty }{u}^{-1/2}{e}^{-u}du.$ Recognise the gamma function. By definition $\Gamma (z)={\int }_0^{\infty }{u}^{z-1}{e}^{-u}du$; here $z-1=-\frac{1}{2}$ so $z=\frac{1}{2}$, giving ${\int }_0^{\infty }{u}^{-1/2}{e}^{-u}du=\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi }$. Therefore $F(s)=s^{-1/2}\cdot \sqrt{\pi }=\frac{\sqrt{\pi }}{\sqrt{s}}=\boxed{\sqrt{\frac{\pi }{s}}},\mathrm{Re}(s)>0.$