Question bank — Transforms of standard functions — proofs

False — only for $==s>0==$. The antiderivative gives ${\left[\frac{e^{-st}}{-s}\right]}_0^{\infty }$, and $e^{-st}\to 0$ at infinity only when $s>0$; for $s\le 0$ the area runs off to infinity.

True and consistent — at $a=0$ the formula gives $\frac{0}{s^2}=0$, which matches $\mathcal{L}\{0\}=0$. A good sanity check that the formula behaves at the degenerate input.

True — $\frac{s}{s^2+0}=\frac{1}{s}$, exactly $\mathcal{L}\{1\}$. Because $\cos (0\cdot t)=1$, the cosine formula must collapse to the constant one, and it does.

False — it is $==s>|a|==$. $\cosh at$ contains both $e^{at}$ and $e^{-at}$; we need $s>a$ and $s>-a$ simultaneously, and the tighter of the two is $s>|a|$.

True — the integral becomes ${\int }_0^{\infty }{e}^{-(s-a)t}dt$, structurally identical to the constant case but with $s-a$ in place of $s$, hence ROC $s-a>0$, i.e. $s>a$.

False — that only kills the lower limit. The upper limit vanishes for a different reason: $e^{-st}$ decays faster than any polynomial $t^n$ grows, and only when $s>0$.

True — $s^2+(ia{)}^2=s^2-a^2$ and the numerator $ia$ over $i$ cancels the $i$ (since $\sinh at$ relates to $\sin$ by $\sin (iat)=i\sinh at$), landing on $\frac{a}{s^2-a^2}$. See Euler's Formula $e^{i\theta}$.

False — linearity is about sums and scalar multiples: $\mathcal{L}\{\alpha f+\beta g\}=\alpha \mathcal{L}\{f\}+\beta \mathcal{L}\{g\}$. Products do NOT transform to products (that would be convolution in reverse). See Linearity and First Shifting Theorem.

The missing $==n!==$. The recursion $I_n=\frac{n}{s}{I}_{n-1}$ multiplies by a fresh integer factor at every step, so unrolling gives $n(n-1)\cdots 1=n!$, not $1$. Correct: $\frac{n!}{s^{n+1}}$.

Sign error in the denominator. Hyperbolic functions use $\frac{1}{2}(e^{at}\pm e^{-at})$, giving real poles at $s=\pm a$: the denominator is $s^2==-==a^2$. So $\mathcal{L}\{\sinh at\}=\frac{a}{s^2-a^2}$.

The ROC $s>a$. Without it the exponential $e^{-(s-a)t}$ would grow instead of decay and the integral diverges — the condition is part of the answer, not an afterthought.

One IBP turns $\sin$ into $\cos$, not into a number. You need two IBPs to get $\sin$ back and form two linear equations in $S$ and $C$, then solve. See Integration by Parts.

You cannot read off the real part before rationalising. Multiply by the conjugate: $\frac{1}{s-ia}=\frac{s+ia}{s^2+a^2}$. Now the real part is $\frac{s}{s^2+a^2}=\mathcal{L}\{\cos at\}$, not $\frac{1}{s}$.

Missing the factorial: $\mathcal{L}\{t^3\}=\frac{3!}{s^4}=\frac{6}{s^4}$, so the answer is $\frac{30}{s^4}$. Linearity handles the $5$, but the $3!=6$ is intrinsic to $t^3$.

Because $s$ is a tunable knob. Making $s$ larger increases the decay so that even fast-growing $f(t)$ eventually gets crushed to finite area — and it is exactly this multiplication-by-$s$ machinery that later turns $\frac{d}{dt}$ into "multiply by $s$".

$\cosh at=\frac{1}{2}(e^{at}+e^{-at})$ is a sum of transformable pieces, so linearity applies. But $\cos (at)\cdot t$ is a product, which linearity cannot split; it needs the derivative-of-transform rule instead.

It gets both transforms in one shot: $\mathcal{L}\{e^{iat}\}=\frac{s+ia}{s^2+a^2}$ has $\cos$ as its real part and $\sin$ as its imaginary part. IBP would need the two-equation dance twice. See Euler's Formula $e^{i\theta}$.

Because the defining integral is improper — an infinite upper limit. Such an integral only names a finite number for certain $s$; outside the ROC the "formula" is a symbol with no value. See Improper Integrals.

A polynomial doesn't fight the decay — any positive $s$ eventually wins, so $s>0$. But $e^{at}$ grows, so the decay $e^{-st}$ must out-race it, needing $s-a>0$.

Real roots of the denominator are poles = places where the growth of the time-function would blow up the transform. Hyperbolic functions grow like $e^{|a|t}$, so poles sit at $s=\pm a$; oscillating functions never grow, so no real poles.

$t^0=1$, and the formula gives $\frac{0!}{s^1}=\frac{1}{s}$ since $0!=1$. Fully consistent — this is the base case $I_0$ of the recursion.

Yes, gently — for $n=0$ the lower boundary is $t^0=1\ne 0$, which is why $\mathcal{L}\{1\}$ is computed directly (no IBP) and serves as the seed the recursion builds on.

Both $\to 0$: $\frac{a}{s^2\pm a^2}\to \frac{0}{s^2}=0$. Makes sense — $\sin 0=\sinh 0=0$ as functions of $t$, so their transforms must vanish.

$\cosh 0=1$, so it reduces to $\mathcal{L}\{1\}$ with ROC $s>0$. And indeed $s>|0|$ is $s>0$ — the general condition $s>|a|$ specialises correctly.

$s>a$ where $a$ is negative, so e.g. $a=-3$ gives ROC $s>-3$ — the transform even exists for some negative $s$, because a decaying $f(t)$ needs less help from the weight.

Yes — the "decay" condition is on the real part: $\mathrm{Re}(s-ia)=s>0$. Since $|e^{iat}|=1$ is bounded, $s>0$ suffices for convergence.