Integral Calculus (9hrs) · Engineering Mathematics - I

Elliptic integral

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Elliptic integral

1. What an Elliptic Integral Is

An elliptic integral arises when the integrand contains a square root of a cubic or quartic polynomial, e.g. arc length of ellipse, pendulum period — it cannot be expressed in elementary functions, so it defines a new special function.

Normal (Legendre) forms for 0≤k2<10≤k²<1 (modulus kk, parameter m=k2m=k²), angle φφ:

  • Incomplete elliptic integral of the first kind:
F(φ,k)=∫0φdθ/√[1−k2sin2θ]F(φ,k)= ∫_{0}^{φ} dθ / √[1 - k² sin²θ]
  • Second kind:
E(φ,k)=∫0φ√[1−k2sin2θ]dθE(φ,k)= ∫_{0}^{φ} √[1 - k² sin²θ] dθ
  • Third kind:
Π(φ,n,k)=∫0φdθ/[(1−nsin2θ)√[1−k2sin2θ]]Π(φ,n,k)= ∫_{0}^{φ} dθ / [(1 - n sin²θ)√[1 - k² sin²θ]]

Complete versions are φ=π/2φ=π/2:

K(k)=F(π/2,k),E(k)=E(π/2,k)K(k)=F(π/2,k), E(k)=E(π/2,k)

etc. These occur pervasively.

2. Where They Come From (for intuition)

  • Arc length of ellipse x2/a2+y2/b2=1x²/a²+ y²/b²=1:
L=a∫0π/2√[1−e2sin2θ]dθ=aE(e)L = a ∫_{0}^{π/2} √[1 - e² sin²θ] dθ = a E(e)

where ee = eccentricity.

  • Period of simple pendulum amplitude φ0φ_0:
T=4√(l/g)⋅K(sin(φ0/2))T = 4√(l/g)· K(sin(φ_0/2))

So any "ellipse perimeter" or "large pendulum" problem is elliptic.

3. Reduction and Evaluation (what exams expect)

General ∫dx/√[(1−x2)(1−k2x2)]∫ dx/√[(1-x²)(1-k²x²)] via x=sinθx= sinθ becomes F(φ,k)F(φ,k). Standard trick: put x=sinθx= sinθ to reach Legendre form.

(a) I = ∫_{0}^{1} dx/√[(1-x²)(1 -k² x²)] : set x= sinθ ⇒ dx= cosθ dθ, √(1-x²)=cosθ ⇒ I=∫_{0}^{π/2} dθ/√(1 -k² sin²θ)=K(k) ✓

(b) Ellipse perimeter for a=5,b=3: e=√(1-b²/a²)=0.8 ⇒ L=4a E(e)=20·E(0.8) ; table gives E(0.8)≈1.276 ⇒ L≈25.5 vs 2π√((a²+b²)/2) approx 25.5 close ✓

(c) ∫_{0}^{π/2} dθ/√(1 -½ sin²θ)= K(1/√2) — leave as K, numerical via series K= π/2[1+ (1/4)k²+9k⁴/64+...] ✓

4. Relation to Beta/Gamma?

For k=0k=0, F(φ,0)=φ,E(φ,0)=φF(φ,0)=φ, E(φ,0)=φ — elementary. For general kk, no reduction to Beta/Gamma; series expansion uses Beta coefficients. Complete integrals have series: K(k)=π/2⋅[1+(1⋅1/2⋅2)2k2+...]K(k)= π/2·[1+ (1·1/2·2)² k²+...].

5. Exam Points

  • Define incomplete/complete elliptic integrals of first/second kind — 3-5 marks, frequent.
  • Show arc length of ellipse / pendulum period reduces to EE or KK — 5 marks, occasional (often as example of non-elementary integral).
  • Evaluate ∫01dx/√[(1−x2)(1−k2x2)]∫_{0}^{1} dx/√[(1-x²)(1-k² x²)] as K(k)K(k) — 5 marks, frequent.
  • Mention that ∫dx/√(quartic)∫ dx/√(quartic) is elliptic — 3 marks.
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