Question bank — Transforms of standard functions — proofs

False — sirf $==s>0==$ ke liye. Antiderivative deta hai ${\left[\frac{e^{-st}}{-s}\right]}_0^{\infty }$, aur $e^{-st}\to 0$ infinity par tabhi jaata hai jab $s>0$ ho; $s\le 0$ ke liye area infinity tak bhaag jaata hai.

True aur consistent — $a=0$ par formula deta hai $\frac{0}{s^2}=0$, jo $\mathcal{L}\{0\}=0$ se match karta hai. Yeh ek achha sanity check hai ki formula degenerate input par bhi theek behave karta hai.

True — $\frac{s}{s^2+0}=\frac{1}{s}$, exactly $\mathcal{L}\{1\}$. Kyunki $\cos (0\cdot t)=1$ hai, cosine formula ko constant one mein collapse hona chahiye, aur hota bhi hai.

False — yeh $==s>|a|==$ hai. $\cosh at$ mein $e^{at}$ aur $e^{-at}$ dono hain; hume $s>a$ aur $s>-a$ ek saath chahiye, aur dono mein tighter condition $s>|a|$ hai.

True — integral ban jaata hai ${\int }_0^{\infty }{e}^{-(s-a)t}dt$, structurally bilkul constant case jaisa lekin $s$ ki jagah $s-a$ hai, isliye ROC $s-a>0$, yaani $s>a$.

False — yeh sirf lower limit ko kill karta hai. Upper limit ek alag wajah se vanish hota hai: $e^{-st}$ kisi bhi polynomial $t^n$ ki growth se zyada fast decay karta hai, aur yeh sirf tab hota hai jab $s>0$ ho.

True — $s^2+(ia{)}^2=s^2-a^2$ aur numerator $ia$ ke upar se $i$ cancel ho jaata hai (kyunki $\sinh at$ ka $\sin$ se relation $\sin (iat)=i\sinh at$ hai), aur result aata hai $\frac{a}{s^2-a^2}$. Dekho Euler's Formula $e^{i\theta}$.

False — linearity sums aur scalar multiples ke baare mein hai: $\mathcal{L}\{\alpha f+\beta g\}=\alpha \mathcal{L}\{f\}+\beta \mathcal{L}\{g\}$. Products, products mein transform NAHI hote (yeh toh convolution ka ulta hoga). Dekho Linearity and First Shifting Theorem.

Missing $==n!==$ hai. Recursion $I_n=\frac{n}{s}{I}_{n-1}$ har step par ek naya integer factor multiply karta hai, toh unrolling karne par $n(n-1)\cdots 1=n!$ milta hai, $1$ nahi. Sahi answer: $\frac{n!}{s^{n+1}}$.

Denominator mein sign error hai. Hyperbolic functions $\frac{1}{2}(e^{at}\pm e^{-at})$ use karte hain, jisse $s=\pm a$ par real poles milte hain: denominator $s^2==-==a^2$ hai. Toh $\mathcal{L}\{\sinh at\}=\frac{a}{s^2-a^2}$.

ROC $s>a$. Iske bina exponential $e^{-(s-a)t}$ decay karne ki jagah grow karega aur integral diverge ho jaayega — yeh condition answer ka hissa hai, afterthought nahi.

Ek IBP $\sin$ ko $\cos$ mein badalta hai, kisi number mein nahi. $\sin$ ko wapas laane aur $S$ aur $C$ mein do linear equations form karne ke liye two IBPs chahiye, phir solve karo. Dekho Integration by Parts.

Real part rationalize karne se pehle nahi nikaala ja sakta. Conjugate se multiply karo: $\frac{1}{s-ia}=\frac{s+ia}{s^2+a^2}$. Ab real part hai $\frac{s}{s^2+a^2}=\mathcal{L}\{\cos at\}$, $\frac{1}{s}$ nahi.

Factorial missing hai: $\mathcal{L}\{t^3\}=\frac{3!}{s^4}=\frac{6}{s^4}$, toh answer hai $\frac{30}{s^4}$. Linearity $5$ ko handle karti hai, lekin $3!=6$ toh $t^3$ ka intrinsic part hai.

Kyunki $s$ ek tunable knob hai. $s$ ko bada karne par decay zyada ho jaati hai, toh fast-growing $f(t)$ bhi eventually finite area mein crush ho jaata hai — aur exactly yahi multiplication-by-$s$ machinery baad mein $\frac{d}{dt}$ ko "$s$ se multiply karo" mein convert karti hai.

$\cosh at=\frac{1}{2}(e^{at}+e^{-at})$ transformable pieces ka sum hai, isliye linearity apply hoti hai. Lekin $\cos (at)\cdot t$ ek product hai, jise linearity split nahi kar sakti; iske liye derivative-of-transform rule chahiye.

Isse ek hi shot mein dono transforms mil jaate hain: $\mathcal{L}\{e^{iat}\}=\frac{s+ia}{s^2+a^2}$ ka real part $\cos$ hai aur imaginary part $\sin$. IBP ko do baar woh two-equation dance karni padti. Dekho Euler's Formula $e^{i\theta}$.

Kyunki defining integral improper hai — infinite upper limit ke saath. Aisa integral sirf certain $s$ ke liye ek finite number name karta hai; ROC se bahar woh "formula" ek aisa symbol hai jiska koi value nahi. Dekho Improper Integrals.

Polynomial decay se fight nahi karta — koi bhi positive $s$ eventually jeet jaata hai, toh $s>0$. Lekin $e^{at}$ grow karta hai, toh decay $e^{-st}$ ko usse aage nikalana hoga, jiske liye $s-a>0$ chahiye.

Denominator ke real roots poles hain = woh jagah jahan time-function ki growth transform ko blow up kar de. Hyperbolic functions $e^{|a|t}$ jaisi grow karte hain, toh poles $s=\pm a$ par hote hain; oscillating functions kabhi grow nahi karte, toh koi real poles nahi hote.

$t^0=1$ hai, aur formula deta hai $\frac{0!}{s^1}=\frac{1}{s}$ kyunki $0!=1$. Fully consistent — yeh recursion ka base case $I_0$ hai.

Haan, thoda sa — $n=0$ ke liye lower boundary $t^0=1\ne 0$ hai, isliye $\mathcal{L}\{1\}$ directly compute kiya jaata hai (koi IBP nahi) aur yeh woh seed hai jis par recursion build hoti hai.

Dono $\to 0$ ho jaate hain: $\frac{a}{s^2\pm a^2}\to \frac{0}{s^2}=0$. Sense banta hai — $\sin 0=\sinh 0=0$ $t$ ke functions ki tarah hain, toh unke transforms bhi vanish hone chahiye.

$\cosh 0=1$, toh yeh $\mathcal{L}\{1\}$ mein reduce ho jaata hai jiska ROC $s>0$ hai. Aur wakai $s>|0|$ hi $s>0$ hai — general condition $s>|a|$ sahi se specialise hoti hai.

$s>a$ jahan $a$ negative hai, toh e.g. $a=-3$ deta hai ROC $s>-3$ — transform kuch negative $s$ ke liye bhi exist karta hai, kyunki decaying $f(t)$ ko weight se kam help chahiye.

Haan — "decay" condition real part par hai: $\mathrm{Re}(s-ia)=s>0$. Kyunki $|e^{iat}|=1$ bounded hai, $s>0$ convergence ke liye kaafi hai.