Residue theorem — computing real integrals: Worked examples

4.10.6 · D3 · Maths › Advanced Topics (Elite Level) › Residue theorem — computing real integrals

Definition Is poori page par use hone wala notation

Yahan ke zyaadatar integrands rational functions hain: ek ratio $\frac{P ( x )}{Q ( x )}$ jahan == $P$ numerator polynomial hai aur $Q$ denominator polynomial== hai. Symbol $de g P$ ka matlab hai $P$ ki degree — usmein $x$ ki sabse badi power (jaise $de g (x^{2} + 1) = 2$ ). Gap ek single number hai: $gap = de g Q - de g P .$ Yeh measure karta hai ki denominator, numerator se door jaake kitni tezi se badhta hai — aur yahi decide karta hai ki hamare contour ka bada arc khatam hoga ya nahi. Gap $2$ ya usse zyada hone ka matlab hai $\frac{P}{Q}$ ki shrinking $\frac{1}{∣ z ∣ ^{2}}$ ya usse tez hoti hai.

Jab ek oscillating factor $e^{ia z}$ present ho, hum integrand ko $R (z) e^{ia z}$ mein split karte hain, jahan == $R$ wo rational part hai jo exponential ko peel karne ke baad bachti hai==. $R$ wahi hai jo Jordan's Lemma maangta hai ki infinity par vanish ho.

Scenario matrix

Is chapter ke har real integral ko inhi cells mein se ek mein rakha ja sakta hai. Neeche ka har worked example us cell ke saath labelled hai jo woh fill karta hai.

Cell	Kya special hai	Chhupi hui danger	Example
A Basic rational	gap $= de g Q - de g P \geq 2$ , simple poles	galat half-plane choose karna	Ex 1
B Higher-order pole	denominator mein repeated factor hai	residue ke liye differentiate karna padega	Ex 2
C Oscillatory (cos/sin)	integrand mein $cos a x$ ya $sin a x$ hai	$e^{ia z}$ use karo, $cos z$ nahi	Ex 3
D Odd integrand → sine	"imaginary part" branch	real part $0$ hai, sine answer hai	Ex 4
E Trig over $[0, 2 π]$	already ek loop hai → unit circle	mandatory $\frac{1}{i z}$ factor	Ex 5
F Degenerate: pole on the real axis	contour ek spike se guzarti hai	indent karo — principal value lo	Ex 6
G Gap $= 1$ with no oscillation	arc nahi marta	recipe fail — ise pehchano	Ex 7
H Word problem	physics/engineering dressing	ise ek jaane-pehchane cell mein translate karo	Ex 8

Ab hum sabhi aath par jaate hain. Labels dekhte rehna.

Ex 1 — Cell A: seedha-saadha rational integral

Worked example

$I = \int_{- \infty}^{\infty} \frac{x ^{2}}{x ^{4} + 1} d x$ Forecast: Integrand even aur positive hai, isliye answer ek positive real number hona chahiye, aur (aane wale algebra se) usme $2$ hoga. Aage padhne se pehle guess karo.

Step 1 — check karo ki arc marta hai ya nahi. Yeh step kyun? Boxed formula tabhi legal hai jab bada semicircle kuch contribute na kare. Yahan $P = x^{2}$ hai toh $de g P = 2$ , aur $Q = x^{4} + 1$ hai toh $de g Q = 4$ ; gap $= 4 - 2 = 2 \geq 2$ . ✓ Arc marta hai, machine legal hai.

Step 2 — poles dhundho. Kyun? Hume har wo spike chahiye jo contour trap kar sake. Solve karo $z^{4} + 1 = 0 \Rightarrow z^{4} = - 1 = e^{iπ}$ . Chaar fourth-roots hain: $z_{k} = e^{i (π + 2 π k) /4}, k = 0, 1, 2, 3,$ yaani angles $4 5^{\circ}, 13 5^{\circ}, 22 5^{\circ}, 31 5^{\circ}$ .

Figure mein kya dekhna hai: chaar poles ek circle par evenly spaced baithe hain. Yellow path haara contour hai — real axis plus upper arc $Γ_{ρ}$ . Do pink dots ( $4 5^{\circ}$ aur $13 5^{\circ}$ ) loop ke andar hain (positive imaginary part), toh ye count hote hain; do blue crosses ( $22 5^{\circ}, 31 5^{\circ}$ ) axis ke neeche hain, loop ke bahar, toh inhe ignore karte hain. Jo do bachte hain: $z_{0} = e^{iπ /4} = \frac{1 + i}{2}, z_{1} = e^{i 3 π /4} = \frac{- 1 + i}{2} .$

Step 3 — residues (simple poles, $g / h^{'}$ use karo). $g / h^{'}$ kyun? Quartic ko factor karna ugly hai; formula $Res = g (z_{0}) / h^{'} (z_{0})$ use karne se bach jaate hain. Yahan $g = z^{2}$ , $h = z^{4} + 1$ , $h^{'} = 4 z^{3}$ , toh $Res_{z_{k}} = \frac{z _{k}^{2}}{4 z _{k}^{3}} = \frac{1}{4 z _{k}} .$ Step 4 — har UHP pole par evaluate karo. Kyunki $z_{k}$ unit circle par hai, $\frac{1}{z _{k}} = \overline{z_{k}} = e^{- i a r g z_{k}}$ . Toh $Res_{z_{0}} = \frac{1}{4} e^{- iπ /4}, Res_{z_{1}} = \frac{1}{4} e^{- i 3 π /4} .$ Step 5 — sum karo aur $2 π i$ se multiply karo. Kyun? Yahi residue theorem ka verdict hai. Har exponential ko rectangular form mein likho taaki add kar sakein: $e^{- iπ /4} = \frac{1 - i}{2}, e^{- i 3 π /4} = \frac{- 1 - i}{2} .$ Yeh kyun help karta hai: real parts $+ \frac{1}{2}$ aur $- \frac{1}{2}$ cancel ho jaate hain, aur imaginary parts $- \frac{i}{2}$ add ho jaate hain, ek purely imaginary sum bachta hai: $e^{- iπ /4} + e^{- i 3 π /4} = \frac{- 2 i}{2} = - i 2 .$ Isliye residue sum hai $\frac{1}{4} (- i 2) = - \frac{i 2}{4}$ , aur $I = 2 π i \cdot (- \frac{i 2}{4}) = \frac{2 π 2}{4} = \frac{π}{2} = \frac{π 2}{2} .$ (Note karo $2 π i \cdot (- i) = 2 π$ , kyunki $- i^{2} = 1$ — yahi imaginary sum ko real answer mein badalta hai.)

Verify: Answer positive hai aur $2$ carry karta hai — forecast se match karta hai. Numerically $π / 2 \approx 2.221$ , aur height $\leq \frac{1}{2}$ wale bump ka $\int$ kuch units ki width ke saath plausibly $\approx 2.2$ hai. ✓

Ex 2 — Cell B: ek repeated factor derivative force karta hai

Worked example

$I = \int_{- \infty}^{\infty} \frac{d x}{( x ^{2} + 4 ) ^{2}}$ Forecast: Even, positive, aur poles $\pm i$ ki jagah $\pm 2 i$ par hain — answer $\int \frac{d x}{( x ^{2} + 1 ) ^{2}} = π /2$ case se chota hona chahiye kyunki bump flatter hai. $π$ ka ek fraction guess karo.

Step 1 — arc check. $P = 1$ toh $de g P = 0$ ; $Q = (x^{2} + 4)^{2}$ toh $de g Q = 4$ ; gap $= 4 - 0 = 4 \geq 2$ . Arc marta hai. ✓

Step 2 — poles. $(z^{2} + 4)^{2} = (z - 2 i)^{2} (z + 2 i)^{2}$ . UHP pole hai $z = 2 i$ , order $m = 2$ . Order 2 kyun? Factor $(z - 2 i)$ squared appear karta hai.

Step 3 — order-2 residue formula. Ek baar differentiate kyun? Order- $m$ pole ke liye hum $(z - 2 i)^{2} f$ ka $m - 1 = 1$ derivative lete hain: $Res_{2 i} = \frac{1}{1 !} lim_{z \to 2 i} \frac{d}{d z} [(z - 2 i)^{2} \cdot \frac{1}{( z - 2 i ) ^{2} ( z + 2 i ) ^{2}}] = \frac{d}{d z} (z + 2 i)^{- 2}_{2 i} .$ Step 4 — derivative karo. $\frac{d}{d z} (z + 2 i)^{- 2} = \frac{- 2}{( z + 2 i ) ^{3}} .$ $z = 2 i$ par: $(2 i + 2 i)^{3} = (4 i)^{3} = 64 i^{3} = - 64 i$ , toh $Res_{2 i} = \frac{- 2}{- 64 i} = \frac{1}{32 i}$ .

Step 5 — assemble karo. $I = 2 π i \cdot \frac{1}{32 i} = \frac{2 π}{32} = \frac{π}{16} .$

Verify: $π /16 \approx 0.196$ , jo $π /2 \approx 1.571$ se kaafi chota hai — "flatter bump" forecast se match karta hai. Sanity scaling: $\int \frac{d x}{( x ^{2} + 4 ) ^{2}}$ mein $x = 2 u$ substitute karne se $\frac{1}{2 ^{3}} \int \frac{d u}{( u ^{2} + 1 ) ^{2}} = \frac{1}{8} \cdot \frac{π}{2} = \frac{π}{16}$ milta hai. ✓

Ex 3 — Cell C: cosine → $e^{ia z}$ use karo, Jordan's Lemma apna kaam karta hai

Worked example

$I = \int_{- \infty}^{\infty} \frac{cos ( 3 x )}{x ^{2} + 4} d x$ Forecast: Positive (tail of $cos$ , $1/ x^{2}$ se damp hai aur $0$ ke paas peak dominate karta hai jahan $cos \approx 1$ ). Decay factor $e^{- a \cdot 2} = e^{- 6}$ ise bahut chota banata hai. Kuch $\frac{π}{2} e^{- 6}$ jaisa guess karo.

Step 1 — $cos 3 x$ ko $e^{i 3 z}$ se replace karo. $cos z$ kyun nahi? Kyunki $cos (3 z) = \frac{1}{2} (e^{i 3 z} + e^{- i 3 z})$ aur $e^{- i 3 z}$ UHP mein explode karta hai (uska modulus $e^{+ 3 y}$ hai). Sirf $e^{i 3 z}$ wahan decay karta hai (Jordan's Lemma). Toh set karo $f (z) = \frac{e ^{i 3 z}}{z ^{2} + 4}$ , $a = 3$ .

Step 2 — Jordan's Lemma legality. Yahan exponential ko peel karne ke baad jo rational leftover bachta hai woh hai $R (z) = \frac{1}{z ^{2} + 4}$ (notation box se yaad karo: $R$ woh non-oscillating part hai). Yeh $∣ z ∣ \to \infty$ par $\to 0$ hai, aur $a = 3 > 0$ . Jordan kehta hai arc integral $\to 0$ — chahe gap sirf $2$ ho, hume bonus milta hai ki yeh gap $1$ par bhi kaam karta. ✓

Step 3 — UHP pole & residue. $z = 2 i$ (simple). $g = e^{i 3 z}$ , $h^{'} = 2 z$ : $Res_{2 i} = \frac{e ^{i 3 (2 i)}}{2 ( 2 i )} = \frac{e ^{- 6}}{4 i} .$ $e^{- 6}$ kyun? $i \cdot 3 \cdot 2 i = 6 i^{2} = - 6$ .

Step 4 — full complex integral, phir real part lo. $\int_{- \infty}^{\infty} \frac{e ^{i 3 x}}{x ^{2} + 4} d x = 2 π i \cdot \frac{e ^{- 6}}{4 i} = \frac{π}{2} e^{- 6} .$ Yeh already real hai, toh Real part $= \frac{π}{2} e^{- 6}$ aur (imag part $0$ ) $\int \frac{sin 3 x}{x ^{2} + 4} d x = 0$ — consistent hai kyunki woh integrand odd hai.

Verify: $\frac{π}{2} e^{- 6} \approx 1.5708 \times 0.00248 \approx 0.00389$ — positive aur tiny, exactly as forecast. ✓

Ex 4 — Cell D: odd integrand, answer sine mein rehta hai

Worked example

$I = \int_{- \infty}^{\infty} \frac{x sin x}{x ^{2} + 1} d x$ Forecast: Note karo ki $x sin x$ even hai (odd $\times$ odd), toh integral nonzero aur positive hai. Yeh woh case hai jahan sine (imaginary part) answer carry karta hai. Kuch $π / e$ -jaisa guess karo.

Step 1 — complex integrand banao. Kyun? $sin x = Im (e^{i x})$ , toh use karo $f (z) = \frac{z e ^{i z}}{z ^{2} + 1}$ , $a = 1$ .

Step 2 — arc / Jordan check. Exponential peel karne ke baad rational leftover hai $R (z) = \frac{z}{z ^{2} + 1} \to 0$ jab $∣ z ∣ \to \infty$ (gap $= 2 - 1 = 1$ ), aur $a = 1 > 0$ . Jordan's Lemma apply hota hai — arc exactly isliye marta hai kyunki exponential usse kill karta hai chahe gap $1$ hi ho. Issi liye hume Jordan chahiye tha, naive gap- $2$ rule nahi.

Step 3 — UHP pole & residue. $z^{2} + 1 = 0 \Rightarrow z = i$ (UHP). Simple pole, $g = z e^{i z}$ , $h^{'} = 2 z$ : $Res_{i} = \frac{i e ^{i \cdot i}}{2 i} = \frac{e ^{- 1}}{2} .$ Kyun: $e^{i \cdot i} = e^{i^{2}} = e^{- 1}$ , aur $g$ mein $i$ upar-neeche $2 i$ ke saath cancel hokar $\frac{1}{2}$ banta hai.

Step 4 — complex integral assemble karo. $\int_{- \infty}^{\infty} \frac{x e ^{i x}}{x ^{2} + 1} d x = 2 π i \cdot \frac{e ^{- 1}}{2} = \frac{π i}{e} .$ Step 5 — imaginary part lo. Imaginary kyun? Kyunki hume $sin$ chahiye, aur $sin x = Im (e^{i x})$ , toh $I = Im (\frac{π i}{e}) = \frac{π}{e} .$ Real part $0$ hai — yeh match karta hai ki $\frac{x cos x}{x ^{2} + 1}$ odd hai, toh uska integral zero hota hai.

Verify: $π / e \approx 1.156$ , forecast ke anusaar positive. Cross-check: pairing "real part $\to$ cos $\to$ 0 (odd), imag part $\to$ sin $\to$ answer" exactly wahi parity bookkeeping hai. ✓

Ex 5 — Cell E: $[0, 2 π]$ par ek trig integral, $\frac{1}{i z}$ mandatory hai

Worked example

$I = \int_{0}^{2 π} \frac{d θ}{5 + 4 cos θ}$ Forecast: Denominator $5 \pm 4 = [1, 9]$ ke beech ghoomta hai, hamesha positive, toh $I > 0$ . Kyunki $\frac{1}{5 + 4 c o s θ}$ ka average $2 π$ length par $1$ se kam hai, expect karo $I < 2 π \approx 6.28$ . Kuch $2 π /3$ ke aas-paas guess karo.

Step 1 — $z = e^{i θ}$ substitute karo. Kyun? Integral already ek loop hai; unit circle par yeh ek contour integral ban jaata hai jise hum residues se attack kar sakte hain. Use karo (parent): $cos θ = \frac{z + z ^{- 1}}{2}, d θ = \frac{d z}{i z} .$ $\frac{1}{i z}$ kyun? Kyunki $d z = i e^{i θ} d θ = i z d θ$ . Ise bhoolna Cell-E ka classic trap hai.

Step 2 — rewrite karo. $I = \oint_{∣ z ∣ = 1} \frac{1}{5 + 2 ( z + z ^{- 1} )} \cdot \frac{d z}{i z} = \oint \frac{d z}{i z ( 5 + 2 z + 2 z ^{- 1} )} = \oint \frac{d z}{i ( 2 z ^{2} + 5 z + 2 )} .$ $z^{- 1}$ kyun vanish ho gaya? Numerator aur denominator ko $z$ se multiply karne se woh clear ho gaya; note karo isse $z = 0$ par ek pole aa sakta hai — check karo! Yahan $z = 0$ pe $2 z^{2} + 5 z + 2 = 2 \neq = 0$ , toh origin par koi extra pole nahi.

Step 3 — $∣ z ∣ = 1$ ke andar poles. Solve karo $2 z^{2} + 5 z + 2 = 0$ : $z = \frac{- 5 \pm 25 - 16}{4} = \frac{- 5 \pm 3}{4}$ , jisse milte hain $z = - \frac{1}{2}$ aur $z = - 2$ .

Figure mein kya dekhna hai: yellow unit circle haara naya contour hai. $z = - \frac{1}{2}$ par pink dot andar hai (distance $0.5 < 1$ ), toh yeh count hota hai; $z = - 2$ par blue cross bahar hai (distance $2 > 1$ ), toh isse discard karte hain. Sirf ek residue matter karta hai.

Step 4 — $z_{0} = - \frac{1}{2}$ par residue (simple). $g = \frac{1}{i}$ , $h = 2 z^{2} + 5 z + 2$ , $h^{'} = 4 z + 5$ ke saath: $Res_{- 1/2} = \frac{1/ i}{4 ( - \frac{1}{2} ) + 5} = \frac{1/ i}{3} = \frac{1}{3 i} .$ Step 5 — assemble karo. $I = 2 π i \cdot \frac{1}{3 i} = \frac{2 π}{3} .$

Verify: $\frac{2 π}{3} \approx 2.094$ , positive aur $2 π$ se kam — forecast se match. General formula check: $\int_{0}^{2 π} \frac{d θ}{a + b c o s θ} = \frac{2 π}{a ^{2} - b ^{2}}$ se milta hai $\frac{2 π}{25 - 16} = \frac{2 π}{3}$ . ✓

Ex 6 — Cell F: real axis par baitha ek pole (degenerate → principal value)

Worked example

$I = P.V. \int_{- \infty}^{\infty} \frac{sin x}{x} d x$ Forecast: Yeh famous "sinc" integral hai; integrand even hai, average-par-positive hai, aur answer $π$ ka ek clean multiple hai. $π$ guess karo.

Step 1 — degeneracy. Yeh special kyun hai? $f (z) = \frac{e ^{i z}}{z}$ ka pole $z = 0$ par hai, jo haare contour par andar nahi, par hi hai. Seedha recipe illegal hai; hum contour ko ek choti semicircle $γ_{ε}$ of radius $ε$ ke saath indent karte hain jo origin ke upar se detour karti hai (dekho Principal Value Integrals).

Figure mein kya dekhna hai: yellow path real axis ke saath chalti hai lekin origin ke aas-paas ek chota pink half-loop $γ_{ε}$ banati hai, phir bade upper arc $Γ_{ρ}$ se close hoti hai. Kyunki detour $z = 0$ ke around step karta hai, koi pole enclosed nahi — poora loop zero integrate ho jaata hai. Answer tiny arc se aayega, enclosed residue se nahi.

Step 2 — closed contour koi pole enclose nahi karta. Kyun? Sirf pole $z = 0$ tiny detour se exclude hai, aur koi strictly inside nahi hai. Toh Cauchy's Integral Theorem & Formula se: $\int_{real, indented} \frac{e ^{i z}}{z} d z + \int_{γ_{ε}} \frac{e ^{i z}}{z} d z + \int_{Γ_{ρ}} \frac{e ^{i z}}{z} d z = 0.$ Step 3 — bada arc marta hai (Jordan). $R (z) = 1/ z \to 0$ , $a = 1 > 0$ ⟹ $\int_{Γ_{ρ}} \to 0$ .

Step 4 — chota arc ek half residue contribute karta hai. Half kyun? Tiny semicircle simple pole $z = 0$ ke around $π$ angle (half a full loop) clockwise sweep karta hai. Simple pole ke liye, $ϕ$ angle ka chota arc $i ϕ \cdot Res$ contribute karta hai; yahan $ϕ = - π$ (clockwise). $Res_{0} \frac{e ^{i z}}{z} = e^{0} = 1$ , toh $\int_{γ_{ε}} \frac{e ^{i z}}{z} d z \to - iπ \cdot 1 = - iπ .$ Step 5 — collect karo. Indented real integral principal value $P.V. \int \frac{e ^{i x}}{x} d x$ ki taraf tend karta hai: $P.V. \int_{- \infty}^{\infty} \frac{e ^{i x}}{x} d x - iπ + 0 = 0 \Rightarrow P.V. \int \frac{e ^{i x}}{x} d x = iπ .$ Step 6 — imaginary part lo. $sin x = Im (e^{i x})$ : $\int_{- \infty}^{\infty} \frac{s i n x}{x} d x = Im (iπ) = π .$ (Real part deta hai $P.V. \int \frac{c o s x}{x} d x = 0$ — sahi hai, woh integrand odd hai.)

Verify: $π \approx 3.1416$ , forecast se match karta hai. Standard result: $\int_{0}^{\infty} \frac{s i n x}{x} d x = \frac{π}{2}$ , evenness se double karke $= π$ . ✓

Ex 7 — Cell G: arc marta nahi — trap pehchano

Worked example

$I = \int_{- \infty}^{\infty} \frac{x}{x ^{2} + 1} d x$ — kya machine legal hai? Forecast: Integrand odd hai, toh principal value $0$ hai, lekin ordinary integral diverge karta hai (large $∣ x ∣$ par $\int \frac{d x}{x}$ jaisa). Naive residue recipe tumhe ek galat finite number degi — fraud pakdo.

Step 1 — pehle degree/decay check. Pehle kyun? Kyunki yeh guardrail hai. $P = x$ toh $de g P = 1$ ; $Q = x^{2} + 1$ toh $de g Q = 2$ ; gap $= 2 - 1 = 1$ . Koi oscillating $e^{ia z}$ nahi hai, toh Jordan apply nahi karta. Toh arc integral vanish nahi hota — boxed formula yahan illegal hai.

Step 2 — dikhao ki arc sach mein survive karta hai. $Γ_{ρ}$ par, $z = ρ e^{i θ}$ : $\frac{z}{z ^{2} + 1} \sim \frac{1}{ρ}, arc length = π ρ \Rightarrow bound \sim \frac{1}{ρ} \cdot π ρ = π \neq \to 0.$ Estimate ek constant par ruk jaata hai — arc contribute karta hai aur drop nahi ho sakta.

Step 3 — jo IS sach hai. Principal value oddness se exist karta hai: $P.V. \int_{- \infty}^{\infty} \frac{x}{x ^{2} + 1} d x = lim_{R \to \infty} \int_{- R}^{R} \frac{x d x}{x ^{2} + 1} = lim_{R \to \infty} [\frac{1}{2} ln (x^{2} + 1)]_{- R}^{R} = 0,$ kyunki integrand odd hai aur limits symmetric hain. Lekin ordinary two-sided limit diverge karta hai.

Verify: Antiderivative $\frac{1}{2} ln (x^{2} + 1)$ even hai, toh symmetrically evaluate karne par $0$ deta hai; upper aur lower limits ko independently infinity par bhejo toh blow up ho jaata hai. Lesson: residues sum karne se pehle hamesha arc/decay check karo. ✓ (P.V. $= 0$ .)

Ex 8 — Cell H: disguise mein ek word problem

Worked example Ek damped resonance ki signal energy

Ek electrical filter ka response hai $H (ω) = \frac{1}{ω ^{2} + ω _{0}^{2}}$ resonance parameter $ω_{0} > 0$ ke saath. Total spectral weight hai $I = \int_{- \infty}^{\infty} H (ω) d ω = \int_{- \infty}^{\infty} \frac{d ω}{ω ^{2} + ω _{0}^{2}}$ . Ise dhundho, aur $ω_{0} = 2$ ke liye evaluate karo. Forecast: Zyaada damping (bada $ω_{0}$ ) curve ko spread aur lower karna chahiye, toh $I$ ko decrease karna chahiye jab $ω_{0}$ badhta hai — expect karo $I \propto 1/ ω_{0}$ .

Step 1 — cell pehchano. Kyun? Rename karo $ω \to x$ : yeh Cell A hai, basic rational integral. Yahan $P = 1$ ( $de g P = 0$ ) aur $Q = ω^{2} + ω_{0}^{2}$ ( $de g Q = 2$ ), toh gap $= 2 \geq 2$ — machine legal hai. Poles $\pm i ω_{0}$ par hain. Step 2 — arc check. Gap $= 2$ . ✓ Step 3 — UHP pole & residue. $z = i ω_{0}$ (simple), $h^{'} = 2 z$ : $Res_{i ω_{0}} = \frac{1}{2 i ω _{0}} .$ Step 4 — assemble karo. $I = 2 π i \cdot \frac{1}{2 i ω _{0}} = \frac{π}{ω _{0}} .$ Step 5 — $ω_{0} = 2$ plug karo. $I = \frac{π}{2}$ .

Verify: $I = π / ω_{0}$ sach mein shrink karta hai jab $ω_{0}$ badhta hai — "more damping → lower weight" forecast se match. Units: $π / ω_{0}$ mein $1/ ω_{0}$ scaling hai, ek broader, lower Lorentzian ke saath consistent. Numerically $π /2 \approx 1.571$ , aur elementary check $\int \frac{d ω}{ω ^{2} + 4} = \frac{1}{2} arctan \frac{ω}{2}_{- \infty}^{\infty} = \frac{1}{2} \cdot π = \frac{π}{2}$ . ✓

Kaun sa cell skip kiya, aur kyun nahi?

Recall Humne matrix ki har cell cover ki

A basic rational ::: Ex 1 ( $π / 2$ ) B higher-order pole ::: Ex 2 ( $π /16$ ) C cosine via $e^{ia z}$ ::: Ex 3 ( $\frac{π}{2} e^{- 6}$ ) D odd $\to$ sine, imaginary part ::: Ex 4 ( $π / e$ ) E trig over $[0, 2 π]$ ::: Ex 5 ( $2 π /3$ ) F pole on the axis, principal value ::: Ex 6 ( $π$ ) G arc survives — recipe illegal ::: Ex 7 (P.V. $= 0$ ) H word problem ::: Ex 8 ( $π / ω_{0}$ )

Connections

Residue theorem — computing real integrals (parent — woh machine jinhe ye examples drive karte hain)
Jordan's Lemma (Cells C, D, F: kyun oscillatory arcs marte hain)
Principal Value Integrals (Cell F: real-axis pole ke around indenting)
Laurent Series & Classification of Singularities (kyun $c_{- 1}$ Ex 2 mein residue hai)
Cauchy's Integral Theorem & Formula (Ex 6: koi enclosed pole nahi ⟹ loop $0$ hai)
Contour Integration & Branch Cuts (agla terrain: fractional powers, keyholes)
Argument Principle & Rouché's Theorem (locate karna ki kaun se poles contour ke andar padte hain)