Residue theorem — computing real integrals: Question bank

4.10.6 · D5 · Maths › Advanced Topics (Elite Level) › Residue theorem — computing real integrals

Figure — Residue theorem — computing real integrals

True ya false — justify karo

Har line ek claim hai. True ya false decide karo, phir ek-sentence mein reason do.

Residue theorem ko chahiye ki

f

contour ke andar har jagah analytic ho.

False. Yeh andar analytic hota hai siwa isolated poles ke — wahi exceptions hain jo

2 π i \sum Res

term produce karte hain; ek fully analytic function to Cauchy's Integral Theorem & Formula se

0

deta.

Residue ek point ki property hai, us contour ki nahi jo tum uske around khinchte ho.

True.

Res_{z_{0}} f = c_{- 1}

z_{0}

par Laurent series se fix hota hai; contour sirf yeh decide karta hai ki woh residue count hoga ya nahi, uski value nahi.

Agar

de g Q = de g P + 1

, to semicircle formula

\int R d x = 2 π i \sum Res

phir bhi kaam karta hai.

False. Arc par height

\sim \frac{c}{ρ ^{1}}

hai aur arc ki length

π ρ

hai, to bound hai

\frac{c}{ρ} \cdot π ρ = π c

— ek nonzero constant jo vanish nahi karta, isliye boxed formula invalid hai.

\int_{- \infty}^{\infty} \frac{c o s x}{x ^{2} + 1} d x

ke liye tum equally lower half-plane mein bhi band kar sakte ho.

False.

e^{i z}

ke saath, decay

e^{- y}

sirf

y > 0

(UHP) ke liye hoti hai; LHP mein

e^{i z}

explode karta hai, isliye wahan Jordan's Lemma fail ho jaata hai.

Real integral ko UHP ya LHP mein band karne se same answer aata hai.

True (valid rational integrand ke liye) — value same hoti hai, lekin LHP mein

- 2 π i

(clockwise orientation) aur lower poles use hote hain, to arithmetic alag hoti hai jabki result same milta hai.

Removable singularity par residue hamesha zero hota hai.

True. Removable singularity mein negative-power terms nahi hote, isliye

c_{- 1} = 0

; woh point kisi bhi contour sum mein kuch contribute nahi karta.

Har function jiska

Res_{z_{0}} f = 0

hai, uska

z_{0}

ke paas antiderivative hota hai.

Locally True.

c_{- 1} = 0

ke saath har Laurent term

c_{n} (z - z_{0})^{n}

termwise integrate hokar single-valued function deta hai, isliye punctured disk mein local primitive exist karta hai.

Essential singularity ka koi residue nahi hota.

False. Uska bhi Laurent series hota hai aur isliye

c_{- 1}

bhi; residue theorem bilkul theek se apply hota hai — bas tidy pole formulas use nahi kar sakte.

\int_{0}^{2 π} F (cos θ, sin θ) d θ

mein tumhe infinity par decay check karni padti hai.

False.

z = e^{i θ}

substitution ise

∣ z ∣ = 1

par ek finite loop mein convert kar deta hai; wahan infinity ki taraf koi arc nahi jaata, to koi decay condition nahi — sirf yeh check karo ki "kaun se poles unit circle ke andar hain."

Why questions

Sirf fact nahi, reason do.

Sirf coefficient

c_{- 1}

ko residue kyun kehte hain, koi aur

c_{n}

ko nahi?

Kyunki

(z - z_{0})^{n}

ko ek chhote circle ke around integrate karne par

n \neq = - 1

ke liye

0

milta hai aur

n = - 1

ke liye

2 π i

—

c_{- 1}

woh akela term hai jo loop integral mein survive karta hai.

Hum by default rational integrands ke liye upper half-plane mein kyun band karte hain?

Sirf convention hai — rationals ke liye dono half work karte hain kyunki decay symmetric hai; hum UHP lete hain taaki same setup reuse kar sakein jab oscillatory integrals ke liye Jordan's Lemma (jo UHP chahiye) aata hai.

Jordan's Lemma sirf degree-1 gap mein kyun succeed karta hai jabki plain semicircle ko degree-2 gap chahiye?

Factor

∣ e^{ia z} ∣ = e^{- a ρ s i n θ}

arc par exponentially decay karta hai,

R

se aane wale mere

\frac{1}{ρ}

ko overcome kar leta hai, isliye

R

ko khud fast decay karne ki zaroorat nahi — dekho Jordan's Lemma.

Trig substitution

z = e^{i θ}

kabhi kabhi

z = 0

par pole kyun introduce karta hai?

Kyunki

d θ = \frac{d z}{i z}

ek factor

\frac{1}{z}

laata hai, aur

sin θ, cos θ

bhi

z

ki negative powers laate hain — ye origin par genuine singularity create kar sakte hain jo check karni zaroori hai.

Residue theorem apply hone ke liye poles isolated kyun hone chahiye?

Non-isolated singularity (singularities ka limit point, ya branch cut) ka koi single Laurent series nahi hota, isliye koi well-defined

c_{- 1}

nahi — aise cases ke liye Contour Integration & Branch Cuts chahiye.

Odd integrand jaise

\frac{s i n x}{x ^{2} + 1}

imaginary part ka answer bina extra kaam ke kyun confirm karta hai?

Odd function ko

(- \infty, \infty)

par symmetrically integrate karne par

0

milta hai, jo imaginary part se match karta hai jab wo vanish honi chahiye — residue arithmetic par ek free consistency check.

Jab koi pole real axis par ho, to residue theorem directly kyun use nahi kar sakte?

Contour singularity se gujari hogi, to integral ordinary sense mein defined bhi nahi hai; tumhe uske around indent karna hoga aur Principal Value Integrals use karna hoga.

Residue sum ko

2 π i

se multiply karne par (sirf

π i

se nahi) closed-contour value kyun milti hai?

Full circle

n = - 1

par

\int_{0}^{2 π} e^{i (n + 1) θ} d θ = 2 π

sweep karta hai; half-circle

π

deta, jo exactly wahi hai kyun indented semicircular detours real poles ke around half-residues contribute karte hain.

Edge cases

Boundary aur degenerate scenarios — precisely batao kya hota hai.

\int_{- \infty}^{\infty} R (x) d x

kya hoga agar

R

ke koi UHP poles nahi hain aur

de g Q \geq de g P + 2

Decay hypothesis ke saath arc vanish ho jaata hai, to empty residue sum exactly

0

deta hai; bina degree gap ke arc survive karta hai aur yeh conclusion fail ho jaata hai — hypothesis optional nahi, essential hai.

Kya hota hai formula mein agar

ρ \to \infty

hone par ek pole exactly semicircular arc par aa jaata hai?

Fixed finite set of poles ke liye yeh large

ρ

par kabhi nahi hota (poles bounded hote hain), isliye yeh koi issue nahi — bas

ρ

ko sabse bade

∣ z_{j} ∣

se bada lo.

Agar do poles coincide ho jaayein (

Q

ka repeated root), to theorem toot jaata hai kya?

Nahi — wo merge hokar ek higher-order pole banaate hain, aur tum order-

m

derivative formula use karte ho; theorem same rehta hai, sirf residue computation mushkil ho jaata hai.

Trig-substitution case mein

∣ z ∣ = 1

par aane wale simple pole of $R$ par residue kya hoga?

Iska matlab hai original real integral mein ek

θ

tha jahan

2 + cos θ = 0

-type blow-up tha; integrand path par singular hai, isliye principal-value treatment chahiye — plain formula fail ho jaata hai.

\int_{- \infty}^{\infty} \frac{e ^{ia x}}{x ^{2} + 1} d x

mein

a \to 0^{+}

hone par kya hota hai?

Yeh smoothly

\int \frac{d x}{x ^{2} + 1} = π

ki taraf jaata hai, jo

\frac{π}{e ^{a}} \to π

se match karta hai; residue

\frac{e ^{- a}}{2 i}

se aane wali exponential decay

e^{- a}

1

ban jaati hai.

\int R (x) e^{ia x} d x

mein agar

a < 0

ho to kya?

Tab

e^{ia z}

lower half-plane mein decay karta hai, isliye tumhe neeche close karna hoga (clockwise, factor

- 2 π i

) aur LHP poles use karne honge — upar close karne par arc blow up ho jaata.

Agar degree gap exactly 2 hai lekin leading coefficients

R

ko sirf

\frac{1}{ρ ^{2}}

jaisa decay karaate hain, to kya arc phir bhi die karta hai?

Haan —

\frac{c}{ρ ^{2}} \cdot π ρ = \frac{π c}{ρ} \to 0

, isliye degree-2 sharp threshold hai: yeh kaam karta hai, degree-1 nahi.

Recall Ek-line self-test

Answers chhupo aur "Spot the error" section mein race karo; agar tum har ek ka fix das second mein naam le sako, to tumne traps internalize kar liye hain.

Connections

Residue Theorem — Computing Real Integrals (parent)
Jordan's Lemma (exponential-decay edge cases)
Cauchy's Integral Theorem & Formula ("analytic ⇒ zero" baseline)
Laurent Series & Classification of Singularities (kyun $c_{- 1}$ , essential vs removable)
Principal Value Integrals (real axis par poles)
Contour Integration & Branch Cuts (non-isolated singularities)

Question bank — Residue theorem — computing real integrals