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Evaluating Real Integrals Using the Residue Theorem | Complex Analysis

Added: 2026-08-07

Deep Dive

Euler's Formula and Complex Exponentials: Understanding how to represent trigonometric functions like cosine and sine in terms of complex exponentials (e.g., cos(θ) = (e^{iθ} + e^{-iθ})/2).

Nombres Complexes - Cours sur les Nombres Complexes - 2 Bac SM S.ex- [Partie 2]

Nombres Complexes - Cours sur les Nombres Complexes - 2 Bac SM S.ex- [Partie 2]

121.3K

Euler's formula states that e^(iθ) = cos θ + i sin θ. This allows conversion between exponential and trigonometric forms. For example, e^(iπ/2) = cos(π/2) + i sin(π/2) = 0 + i(1) = i. The real part of e^(iθ) is cos θ and the imaginary part is sin θ.

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Identidade de Euler

Identidade de Euler

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Euler's formula states that e^(ix) = 1 + (ix)/1! + (ix)²/2! + (ix)³/3! + (ix)⁴/4! + ... This infinite series expansion shows how complex exponentials can be expressed as a sum of terms involving powers of i (the imaginary unit) divided by factorials. The formula connects exponential functions with trigonometric functions.

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О жизни двух главных констант математики

О жизни двух главных констант математики

405.2K

Euler's formula states that e^(ix) = cos(x) + i sin(x) for any real number x. This connects exponential functions with trigonometric functions through complex numbers. When x = π, we get e^(iπ) = cos(π) + i sin(π) = -1 + 0i = -1, which is Euler's identity, one of the most beautiful equations in mathematics.

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Lec 6 | MIT 18.03 Differential Equations, Spring 2006

Lec 6 | MIT 18.03 Differential Equations, Spring 2006

165.9K

Euler's formula, e^(iθ) = cos(θ) + i sin(θ), is a fundamental definition that connects exponential functions with trigonometric functions in the complex plane. This formula satisfies the exponential law (e^(iθ₁) × e^(iθ₂) = e^(i(θ₁+θ₂))) and the differential equation d/dt(e^(it)) = i × e^(it), making it essential for solving differential equations involving trigonometric functions. The polar form of complex numbers, z = re^(iθ), simplifies multiplication by multiplying moduli and adding arguments, and enables elegant solutions to problems like finding nth roots of unity and integrating expressions like e^(-x)cos(x) through complexification.

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ECE2026 L6: Phasor Representations of Sinusoids (Introduction to Signal Processing, Georgia Tech)

ECE2026 L6: Phasor Representations of Sinusoids (Introduction to Signal Processing, Georgia Tech)

2.2K

Euler's formula states that e^(jθ) = cos(θ) + j·sin(θ), connecting exponential functions with trigonometric functions. This can be proven using Taylor series expansions. Special cases include e^(jπ) = -1. Complex numbers with magnitude R are written as R·e^(jθ) = R·cos(θ) + j·R·sin(θ), where real and imaginary parts are sinusoids 90° out of phase. The real part of Z is (Z + Z*)/2, and the imaginary part is (Z - Z*)/(2j). The inverse Euler formulas give cos(θ) = (e^(jθ) + e^(-jθ))/2 and sin(θ) = (e^(jθ) - e^(-jθ))/(2j).

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Deep Dive

Cauchy's Residue Theorem: Mastery of the theorem itself, including how to locate singularities (poles) and calculate residues of complex functions.

Complex Analysis: Residue Theorem Proof

Complex Analysis: Residue Theorem Proof

13.7K

Cauchy's Residue Theorem states that for a function f(z) analytic inside and on a simple closed contour C except for a finite number of singularities (poles) inside C, the contour integral ∮_C f(z) dz equals 2πi times the sum of the residues of f(z) at each pole inside C; this is proven by deforming the contour to avoid singularities, applying Laurent series expansions, and using the fact that the contour integral of (z-z_k)^j around a closed path is zero for all integers j ≠ -1, leaving only the residue term b₁ₖ which equals the residue at each pole.

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Cauchy Residue Theorem, Introduction

Cauchy Residue Theorem, Introduction

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The Cauchy Residue Theorem states that the integral of an analytic function around a closed contour equals 2πi times the sum of the residues enclosed by that contour; this video demonstrates the special case of integrating 1/z over a circular contour, showing that despite the function being undefined at the origin (a singularity), the integral evaluates to exactly 2πi, where the residue at the origin is 1.

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Cauchy Residue Theorem | Easiest Way | Complex Analysis

Cauchy Residue Theorem | Easiest Way | Complex Analysis

4.9K

This section presents the complete mathematical proof of the Cauchy Residue Theorem. The proof begins by considering a closed curve C containing n poles z₁, z₂, ..., zₙ. Small disjoint circles γ₁, γ₂, ..., γₙ are drawn around each pole with radius r, ensuring all circles lie entirely within C. By Cauchy's theorem, the integral over the original curve minus the sum of integrals over these small circles equals zero. Using the residue definition Res(f, z_k) = (1/(2πi))∫γ_k f(z)dz, each integral equals 2πi times the residue at that pole. Substituting gives the final result ∫_C f(z)dz = 2πiΣRes(f, z_k), completing the proof in approximately four lines.

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B.Tech S3 | KTU | Maths | 25 Sure Questions & Answers | Double Pass ഉറപ്പിക്കാം | #btechmaths

B.Tech S3 | KTU | Maths | 25 Sure Questions & Answers | Double Pass ഉറപ്പിക്കാം | #btechmaths

20.5K

Cauchy's Residue Theorem states that if f(z) is analytic inside and on a simple closed contour C except for isolated singular points inside C, then ∮f(z) dz = 2πi Σ Res(f, zₖ), where the sum is over all singular points zₖ inside C. This theorem is fundamental for evaluating complex integrals.

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6. Cauchy Residue Theorem -Simple Poles Problem - Part - 1

6. Cauchy Residue Theorem -Simple Poles Problem - Part - 1

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The Cauchy Residue Theorem states that the integral of a function around a closed curve equals 2πi times the sum of residues at poles inside the curve. For simple poles (poles of order 1), the residue at z=a is calculated as lim(z→a) (z-a)f(z). In the example with f(z) = (sin(πz²) + cos(πz²))/(z-1)(z-2) and curve |z|=3, both poles z=1 and z=2 lie inside, each with residue 1, yielding the integral value of 4πi. When the curve is changed to |z|=3/2, only z=1 lies inside, giving the integral value of 2πi.

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Deep Dive

Contour Integration on the Unit Circle: Familiarity with the parametrization of the unit circle (z = e^{iθ}, dz = i e^{iθ} dθ) and integrating complex functions along a closed path.

CSIR NET DEC 2025 Complete Solutions & Answer Key | CSIR NET 2025 Physical Sciences | 100% Accuracy

CSIR NET DEC 2025 Complete Solutions & Answer Key | CSIR NET 2025 Physical Sciences | 100% Accuracy

333

For contour integration over the unit circle |z| = 1, the conjugate z* = 1/z. The integrand |1 + 2z|² = (1 + 2z)(1 + 2z*) = 1 + 2z + 2z* + 4z z* = 1 + 2z + 2/z + 4. The integral of analytic functions (1, 2z) is zero by Cauchy's theorem. The residue of 2/z is 2. Since the integration is clockwise (opposite to standard positive orientation), the result is -2πi × 2 = -4πi.

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Linear Transformations & Singularities | Linear Algebra & Complex Analysis | MH SET Maths | IFAS

Linear Transformations & Singularities | Linear Algebra & Complex Analysis | MH SET Maths | IFAS

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For contour integrals involving modulus on the unit circle, parameterize z = e^(iθ) for θ from 0 to 2π. Then dz = ie^(iθ)dθ and |z - 1| = √((cosθ - 1)² + sin²θ) = √(2 - 2cosθ) = 2|sin(θ/2)|. The integral becomes ∫₀^(2π) 2|sin(θ/2)| dθ = 4∫₀^(2π) |sin(θ/2)| dθ = 8. This technique of parameterization simplifies complex contour integrals into real integrals.

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16. Complex analysis ||Calculus of Residues Contour Integration || Examples ||Honours 3rd year || NU

16. Complex analysis ||Calculus of Residues Contour Integration || Examples ||Honours 3rd year || NU

15.9K

This section introduces contour integration for solving definite integrals. The instructor explains that this chapter covers methods for evaluating integrals using complex analysis, typically worth 10-15 marks in exams. Key concepts include: (1) Using a unit circle (|z| = 1) as the standard contour, where z = x + iy and |z| = √(x² + y²) = 1 gives x² + y² = 1; (2) Parametrizing the unit circle using Euler's formula z = e^(iθ) = cos(θ) + i sin(θ), with θ ranging from 0 to 2π; (3) The derivative dz = i e^(iθ) dθ; (4) Euler's formula identities: e^(iθ) + e^(-iθ) = 2 cos(θ) and e^(iθ) - e^(-iθ) = 2i sin(θ). These foundational concepts enable conversion of real integrals into complex contour integrals.

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SET Physics final rapid revision

SET Physics final rapid revision

925

When the contour is not specified in a complex integration problem, assume the contour is the unit circle (|z| = 1). For the function 1/(z(z+2)), only z = 0 lies inside the unit circle, so only this pole contributes to the residue. The residue at z = 0 is 1/2, so the integral value is 2πi × (1/2) = πi.

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MTH632 Objective , MTH632 final term preparation past papers with detail concept for getting 4cgpa

MTH632 Objective , MTH632 final term preparation past papers with detail concept for getting 4cgpa

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For a function f(z) that is continuous on the unit circle, the integral ∮f(z)dz around the unit circle can be computed using Cauchy's integral theorem. If f(z) is analytic inside and on the unit circle, then ∮f(z)dz = 0. This is a fundamental result in complex analysis.

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Deep Dive

Classification of Singularities: Knowing the difference between removable singularities, poles (and their orders), and essential singularities.

[s4 | 2026] Математический анализ, Горшанова Анастасия, практика 8

[s4 | 2026] Математический анализ, Горшанова Анастасия, практика 8

257

Singularities are classified as removable, poles, or essential. A removable singularity has a finite limit; a pole of order m has (z-a)^m f(z) → non-zero finite limit; an essential singularity has no limit. For cos(π/z)/(z+1), z=0 is essential (cos oscillates), while z=∞ is removable (limit is 0).

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Singularity - Types of Singularity | Isolated & Non-Isolated Singularity | Complex Analysis

Singularity - Types of Singularity | Isolated & Non-Isolated Singularity | Complex Analysis

445.4K

In complex analysis, singularities are classified into two main types: isolated singularities and non-isolated singularities. An isolated singularity occurs at a point z=a where the function is not analytic, but analytic in some deleted neighborhood around a. Examples include poles (like 1/z at z=0) and essential singularities (like sin(1/z) at z=0). A non-isolated singularity cannot be isolated from other singularities, such as the accumulation point of zeros or poles. Regular points are where the function is analytic, while zeros of a function are points where f(a)=0, with their order determined by the smallest m such that the m-th derivative at that point is non-zero.

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Complex Analysis | Marathon Series| Most Repeated Questions for CSIR NET, CUET & IIT JAM | By GP Sir

Complex Analysis | Marathon Series| Most Repeated Questions for CSIR NET, CUET & IIT JAM | By GP Sir

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Singularities of complex functions are classified into three types: removable singularities, poles, and essential singularities. For a function f(z) = h(z)/g(z), if both h(z) and g(z) have zeros of the same order at z₀, then z₀ is a removable singularity. If g(z) has a zero of higher order than h(z), then z₀ is a pole. For example, with f(z) = sin(z)/z, both numerator and denominator have zeros of order 1 at z = 0, so it's a removable singularity. For f(z) = cos(z)/(z²), the denominator has a zero of order 2 while the numerator has a zero of order 0, so it's a pole of order 2.

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Analytic & Entire Function (Part 2) & Singularity || Engineering Mathematics || PrepFusion

Analytic & Entire Function (Part 2) & Singularity || Engineering Mathematics || PrepFusion

294

Singularities are points where a function fails to be analytic. Isolated singularities have no other singularities in their neighborhood. Non-isolated singularities have other singularities arbitrarily close. For f(z) = 1/(z - 1)(z - 3)(z - (1 + i))², singularities are at z = 1, z = 3, and z = 1 + i (multiplicity 2). For f(z) = sin(1/z), singularities are at z = 1/(nπ) for n = ±1, ±2, ... and at z = 0. A singularity is isolated if there exists a neighborhood around it containing no other singularities.

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Complex analysis: Singularities

Complex analysis: Singularities

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Singularities of holomorphic functions are classified into isolated and non-isolated types; isolated singularities include removable singularities (where the function can be redefined to become holomorphic, characterized by boundedness near the point), poles (where the function behaves like 1/z^n for some positive integer n, making the function meromorphic), and essential singularities (where the Laurent series contains infinitely many negative powers, causing the function to take values arbitrarily close to any complex number near the singularity according to the Casorati-Weierstrass theorem); non-isolated singularities include branch points (where the function cannot be defined as single-valued in any punctured neighborhood, such as log(z) or z^s for non-integer s) and natural boundaries (where the function cannot be analytically continued across a curve, such as the unit circle for certain power series).

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Problem Setup

Integral Transformation

Function Derivation

Singular Point Analysis

Residue Calculation

Computing Residue

Integral Value

Final Solution

Problem Setup

0:00

Playing Section

1
Introduces a real integral to be solved using residue theory.
2
Outlines the substitution method: e^(iθ) = z.
3
Converts the integral to a complex contour integral over the unit circle.

Deep Dive

Evaluating Improper Real Integrals: Applying the residue theorem to evaluate integrals over the entire real line (from -∞ to +∞) using semicircular contours.

1.1 Integrales Impropias

1.1 Integrales Impropias

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For first-kind improper integrals (infinite intervals), express as a limit: ∫ₐ^∞ f(x) dx = limₜ→∞ ∫ₐᵗ f(x) dx. For integrals over all real numbers, split at any convenient point c: ∫₋∞^∞ f(x) dx = ∫₋∞ᶜ f(x) dx + ∫ᶜ^∞ f(x) dx. The original converges if and only if both parts converge. For second-kind integrals (discontinuities), use limits approaching the discontinuity point from the appropriate direction.

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Intégrale impropre

Intégrale impropre

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To evaluate improper integrals, use substitution to find the primitive function. For 1/(x-4)^(2/3), let u = x-4, then du = dx, transforming the integral to ∫u^(-2/3) du = 3u^(1/3) + C. Apply the Fundamental Theorem of Calculus by evaluating the primitive at the limits. For the integral from 0 to 4: [3(x-4)^(1/3)] from 0 to 4 = 3(4)^(1/3) - 3(-4)^(1/3) = 3(4)^(1/3). For the integral from 4 to 6: [3(x-4)^(1/3)] from 4 to 6 = 3(2)^(1/3) - 0 = 3(2)^(1/3). Both integrals converge to finite values.

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Tích phân Bất định, Xác định & Suy rộng (Lý thuyết nền tảng)

Tích phân Bất định, Xác định & Suy rộng (Lý thuyết nền tảng)

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To evaluate an improper integral ∫_a^∞ f(x)dx: (1) Express it as a limit: lim_(t→∞) ∫_a^t f(x)dx. (2) Evaluate the definite integral ∫_a^t f(x)dx. (3) Take the limit as t approaches infinity. For example, ∫_0^∞ dx/(x² + 2x + 2) = lim_(t→∞) [arctan(x+1)]_0^t = lim_(t→∞) [arctan(t+1) - arctan(1)] = π/2 - π/4 = π/4.

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[Old Syllabus] 2nd Semester B.Sc. Complementary Mathematics| Module- I |Part- 06 |Calicut University

[Old Syllabus] 2nd Semester B.Sc. Complementary Mathematics| Module- I |Part- 06 |Calicut University

5.4K

An integral ∫ₐᵇ f(x)dx is improper if: (1) one or both limits are infinite, or (2) f(x) has an infinite discontinuity within [a, b]. The improper integral ∫₋∞^∞ f(x)dx is defined as ∫₋∞^x₀ f(x)dx + ∫ₓ₀^∞ f(x)dx. Basic evaluation uses power rule ∫xⁿdx = xⁿ⁺¹/(n+1) + C and integration by parts ∫u dv = uv - ∫v du. Convergence criteria: ∫₁^∞ dx/xᵖ converges if p > 1 (value = 1/(p-1)) and diverges if p ≤ 1. For rational functions, use partial fraction decomposition. The improper integral ∫₂^∞ (x+3)/[(x-1)(x²+1)]dx converges to -π/2 + log(5) + tan⁻¹(2).

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Using the Residue Theorem for improper integrals involving multiple-valued functions (2/2)

Using the Residue Theorem for improper integrals involving multiple-valued functions (2/2)

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To evaluate difficult improper integrals of real variables using complex analysis, follow these steps: (1) Identify the integrand and choose a suitable complex extension by replacing real variables with complex variables z; (2) Select an appropriate closed contour in the complex plane that avoids branch cuts and singularities; (3) Apply the residue theorem to compute the complex contour integral; (4) Relate the computed complex integral back to the original real integral by analyzing contributions from different parts of the contour. This method transforms challenging real integrals into problems amenable to complex analysis techniques.

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Deep Dive

Jordan's Lemma and Fourier-type Integrals: Learning how to evaluate integrals of the form ∫ f(x)cos(ax)dx or ∫ f(x)sin(ax)dx using complex exponentials and bounding arcs.

Complex Analysis Chapter 7.80 & 7.81: Improper Integrals from Fourier Analysis & Jordan's Lemma

Complex Analysis Chapter 7.80 & 7.81: Improper Integrals from Fourier Analysis & Jordan's Lemma

169

Jordan's Lemma states that if a function f(z) is analytic in the upper half-plane exterior to a circle |z| = R₀, and if the maximum value M_R of |f(z)| on the semicircle C_R approaches zero as R → ∞, then the integral of f(z)e^(iaz) over C_R approaches zero for any constant a. This lemma enables the evaluation of improper integrals of the form ∫(-∞ to ∞) f(x)sin(ax)dx or ∫(-∞ to ∞) f(x)cos(ax)dx by converting them to complex exponential form f(z)e^(iaz), applying the residue theorem on a closed semicircular contour, and showing the semicircle integral vanishes via Jordan's Lemma. The procedure requires f(z) to be a rational function P(z)/Q(z) with real coefficients, no common factors, degree of Q greater than degree of P, and Q(z) having no real zeros but at least one zero above the real axis.

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How to Integrate Fourier Integrals | Complex Variables

How to Integrate Fourier Integrals | Complex Variables

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Jordan's Lemma states that ∫_{C_R} e^(iaz)f(z) dz → 0 as R→∞ if three conditions hold: (1) f(z) is analytic everywhere beyond a certain distance from origin; (2) integrating over semicircular arc C_R; (3) |f(z)| has an upper bound M(R) that approaches zero as R→∞. For f(z) = 1/(z² + α²), write z = Re^(iθ) on the arc. The magnitude |e^(iaz)| = e^(-aRsinθ). The denominator magnitude is √(R⁴ - 2R²α²cos(2θ) + α⁴). The maximum value of |f(z)| occurs when cos(2θ) = -1, giving upper bound 1/(R² - α²) → 0 as R→∞. Thus, Jordan's Lemma applies and the semicircular arc integral vanishes.

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Complex analysis: Jordan's lemma

Complex analysis: Jordan's lemma

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Jordan's lemma provides a method to show that integrals of the form ∫e^(iαz)f(z)dz over large semicircular contours in the upper half-plane tend to zero as the radius R→∞, which is essential for evaluating real integrals like ∫sin(x)/x dx from -∞ to ∞ using complex analysis. Unlike the ML estimate, which fails for such integrals because the exponential factor e^(iαz) has modulus e^(-αR sinθ) that is exponentially small over most of the contour except near the endpoints, Jordan's lemma exploits this exponential decay by replacing sinθ with the linear approximation 2θ/π on [0, π/2], yielding an upper bound of π/R that tends to zero. This lemma applies generally when f(z) is continuous and its maximum magnitude on the contour tends to zero as R→∞, regardless of how slowly f(z) decays, making it indispensable for Fourier transform-type integrals where the ML estimate would be insufficient.

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Most Important Questions | Calculus of Residue | Sem-2 | Unit-2 | B.Tech | IPU | Part-8

Most Important Questions | Calculus of Residue | Sem-2 | Unit-2 | B.Tech | IPU | Part-8

177

This segment introduces Jordan's Lemma for evaluating trigonometric integrals. For integrals involving cos(mx) or sin(mx), we use f(z) = e^(imz)/(z² + a²). Jordan's Lemma states that if f(z) is analytic except at finite singularities and f(z) → 0 uniformly as z → ∞, then lim(r→∞) ∫(semicircle) e^(imz)f(z)dz = 0 for m > 0. The poles of f(z) are found by solving z² + a² = 0, giving z = ±ia. Only z = ia lies in the upper half-plane. The residue at z = ia is calculated as lim(z→ia) [(z-ia)f(z)] = e^(-ma)/(2ia). By the Residue Theorem, ∮f(z)dz = 2πi × residue = πe^(-ma)/a.

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Лекция 10. Теоремы о вычетах

Лекция 10. Теоремы о вычетах

315

On the semicircle CR, z = Re^(it) for t ∈ [0, π], so dz = iRe^(it) dt. The integral is ∫₀^π f(Re^(it)) iRe^(it) dt. Since |z f(z)| → 0, for any ε > 0, there exists R₀ such that for R > R₀, |z f(z)| < ε. Then |∮_CR f(z) dz| ≤ ∫₀^π |f(Re^(it))| R dt ≤ ∫₀^π ε dt = επ. Since ε is arbitrary, the integral tends to 0 as R → ∞.

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Deep Dive

Branch Cut Integrals: Handling multi-valued functions (like logarithms or fractional powers) using specialized contours such as keyhole contours.

Снова об интеграле от функции Бесселя. Доказательство для нецелых значений индекса. Интеграл века.

Снова об интеграле от функции Бесселя. Доказательство для нецелых значений индекса. Интеграл века.

117

This section derives the final expression for the original integral. The sum of the integrals over the upper and lower banks of the branch cut equals 2i sin(πν) ∫ from ε to 1 of R^ν / (R² - 2aR + 1) dR. As ε → 0, this becomes 2i sin(πν) times the transformed integral B. Combining all parts, the original integral equals (1/π) times the residue expression minus the integral over the unit circle. As ε → 0, the small circle integral vanishes. The integral over the unit circle tends to 0 as a → 0. The final expression shows that the integral equals 1 for any ν > -1, matching the result for integer indices.

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$Complex Analysis: keyhole, fractional residues, Jordan Lemma, examples, 10-31-23 part 1$

Complex Analysis: keyhole, fractional residues, Jordan Lemma, examples, 10-31-23 part 1

191

When integrating functions with branch cuts (such as z^a/(1+z²)² where -1 < a < 1), a keyhole contour is used. This contour consists of an outer circle Γ_R, an inner circle γ_ε, and two line segments L+ and L- just above and below the positive real axis. The branch cut for log(z) is placed along the positive real axis, and the function is holomorphic inside the contour except at the singularity z = -1.

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Лекция 1.22 Поле квантуется два! | Введение в КТП | Наумов Д.В.

Лекция 1.22 Поле квантуется два! | Введение в КТП | Наумов Д.В.

162

The contour consists of the real axis, a large semicircle in the upper half-plane, and paths around the branch cut along the imaginary axis. The integral around the closed contour is zero since there are no poles inside. The integral along the large semicircle vanishes due to the exponential decay. The integrals along the two sides of the branch cut differ by a phase factor e^(iπ) = -1 due to the square root branch cut.

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Complex analysis Branch cut method (residue theorem)

Complex analysis Branch cut method (residue theorem)

2.3K

This section introduces complex analysis and the branch cut technique for evaluating integrals. The instructor explains how to draw diagrams with real and imaginary axes, set up the integral J from 0 to infinity, and represent the integrand using exponential notation. The branch cut technique involves creating a contour gamma consisting of four parts: a large circle of radius R, a small circle of radius epsilon around the origin, and two straight lines connecting them above and below the branch cut. This contour deformation enables evaluation of otherwise difficult integrals by avoiding branch points and properly accounting for multi-valued functions like square roots.

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Contour integration: A really cool example for the keyhole contour

Contour integration: A really cool example for the keyhole contour

11.5K

This section evaluates the integrals over the straight lines L₁ and L₂. As Δ→0, the lines approach the real axis. On L₁ (upper), z = x with argument 0, so z^(1/3) = x^(1/3). On L₂ (lower), z = x with argument 2π, so z^(1/3) = x^(1/3)e^(i2π/3). Since L₂ is traversed leftward from ∞ to 0, the integral becomes ∫₀^∞ [x^(1/3)e^(i2π/3)]/(x²+1) dx with a negative sign. Combining both lines gives I(1 - e^(i2π/3)), where I = ∫₀^∞ x^(1/3)/(1+x²) dx.

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Deep Dive

Physical Applications in Physics and Engineering: Using these integration techniques to solve problems in fluid dynamics, electrostatics, and performing inverse Laplace or Fourier transforms.

9.SINIF FİZİK 1. DÖNEM 1. YAZILI | 1.SENARYO | Maarif Modeli | 2025-2026 #yazılıyahazırlık

9.SINIF FİZİK 1. DÖNEM 1. YAZILI | 1.SENARYO | Maarif Modeli | 2025-2026 #yazılıyahazırlık

26.6K

Physics sub-disciplines are essential in engineering: Computer Engineering uses Solid State Physics, Artificial Intelligence connects to Atomic Physics, Climate Cooling involves Thermodynamics, Machine Engineering applies Mechanics, and Ophthalmology uses Optics. The energy unit kg·m²/s² is derived from mass (kg), length (m), and time (s). Measurement tools include balance scales for mass, tape measures for length, and stopwatches for time. All fundamental quantities are scalar.

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Path integrals - How to integrate over curves. Chris Tisdell UNSW

Path integrals - How to integrate over curves. Chris Tisdell UNSW

29.4K

Scalar line integrals solve real-world problems in physics and engineering. For thin wires or springs in 3D space, total mass M = ∫C ρ ds, where ρ(x,y,z) is density (mass per unit length). Center of mass coordinates are x̄ = (∫C xρ ds)/M, ȳ = (∫C yρ ds)/M, z̄ = (∫C zρ ds)/M. Symmetry simplifies calculations—for wires in the YZ-plane, X-coordinate of center of mass is zero. Example: wire parameterized by C(t) = (0, t²-1, t) with ρ = 15√(t²+1) yields mass 14 and center of mass at (0, -3/5, 7/5). These applications connect abstract mathematics to tangible engineering problems like calculating spring masses.

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Scope of physics chapter 1 physics and measurement class 11 new physics book | branches of physics

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Problem Setup

Integral Transformation

Function Derivation

Singular Point Analysis

Residue Calculation

Computing Residue

Integral Value

Final Solution

Problem Setup

0:00

Playing Section

1
Introduces a real integral to be solved using residue theory.
2
Outlines the substitution method: e^(iθ) = z.
3
Converts the integral to a complex contour integral over the unit circle.

11.8K views23018:59@ranjankhatuOriginal Release: 2024-07-13

To evaluate real integrals of the form ∫₀²π f(cosθ, sinθ) dθ using the residue theorem, substitute e^(iθ) = z, which transforms the integral into a complex contour integral over the unit circle |z|=1; identify the singular points (poles) of the resulting complex function, compute residues only at poles lying inside the unit circle, and apply the residue theorem formula I = 2πi × sum(residues) to find the integral value.

[00:00:00]hi everyone in this video we are going to discuss this example okay so what have they given to us one real integration integration of one real valued function we have to find its value with the help of residue actually the residue concept is related with complex valued function but we have to use this concept to find the value of this integral okay this real integral so let us see how to find the value I will explain the procedure first and the same procedure I'm going to use to solve this problem the first step is we use this substitution e to I Theta isal Z whatever the problem we have doesn't matter our substitution is fixed e to I Theta is equal to Z so in this case generally we have the integral like this 0 to 2 pi F of theta D Theta and what we put e to I Theta is equal to Z this substitution will convert the given integral into complex integration integration over c f of Z DZ f is a function of Z it's a complex valued function and here C is a curve mod Z isal 1 because of this substitution always we get the same curve mod Z is equal to 1 it is a unit circle okay circle with Center 0 0 radius 1 after that what we do we find the singular points of f of Z singular points of f of Z singular points that means a point where the function is not analytic or generally we say the denominator is zero so those points can locate anywhere in a complex Place those can be anywhere getting this circle we have at the center this circle right some points some singular points can lie inside the circle some can lie outside a circle right so we are not interested in the singular points which are lying outside we are more interested in the singular points which are lying inside Circle getting the point so at those singular points we find residues and after getting residue okay residue only at these points which are lying inside the circle then the value of this integration is 2 pi I sum of residues so this is a simple formula we have to find the value of this integral okay but for that we need to follow the entire process so the same procedure I'm going to follow to solve the given problem also so let us call it as I whatever the integral we have let us call it as I so let me write we have I is equal to see same limits we have 0 to 2 pi 0 to 2 pi but the problem is that here we have cos 2 Theta which we should have just cos Theta sin Theta e to I Theta like that but see what we have cos 2 Theta so twice angle okay 2 Theta we have so for that I will not take directly cos 2 Theta instead of that I will write e to 2 I Theta 2 I Theta getting if you have cos 3 Theta or sin 3 Theta then I should write e to 3 I Theta okay please keep in your mind in denominator I will keep as it it is 5 + 4 cos Theta D Theta that means the integral which we assume it is different from this one what is the difference instead of cos 2 Theta I wrote e to 2 I Theta then it will be very simple to solve and at the end with the help of value of this integral definitely and easily we can find the value of this integral also let us move further I will follow the procedure first of all I should put e IA is equal to Z let us do putting e to I Theta is equal to Z okay so when we put anything first task is to find derivative and the second task to find a curve okay over this integral we are finding or sometimes we find the limits also so let us take derivative with respect to Theta derivative of e to I Theta derivative of Z with respect to Theta derivative of exponential function is same its derivative will be e to I Theta but by Chain rule again we need to find derivative of I Theta which is i d z upon D Theta e to I Theta means Z so this is i z is equal to DZ upon D Theta so let us take all terms of theta on one side of this equality okay and this all terms of Z will be shift on other side so we will have D Theta is = to DZ upon i z okay so finally we got the value of D Theta this value I'm going to put in this integral I okay so let us continue here see we want some more space to write so let me remove this part and then we'll go further so as you can see easily we have cos Theta so let us find the value of cos Theta also in terms of Z okay so let me write now cos Theta is equal to see cos can be expressed in the form of exponential function in this way e i theta plus eus I Theta upon 2 it means e to I theta plus see minus sign is there in the power so that's why we can write 1 upon e to I Theta so power will be positive if you shift at denominator e to Y Theta means Z here also I can put we can cross multiply right Z into z z s + 1 upon Z divided by 2 actually Z and 2 both are in denominator so we can write simply cos Theta is equal to Z 2 + 1 upon 2 Z so finally we got the value of cos Theta okay so let us talk about the curve now okay so what I do I consider mod Z I'm going to find its value mod Z so Z is e to I Theta e to I Theta means cos Theta + I sin Theta so we know well how to find the value of mod of any complex number okay so the formula is root square of real part plus square of imaginary part cos square + sin square is 1 so root 1 is one that means we get we started with mod mod Z and we got one so this is a curve we have right so this is the Circle Center 0 0 and radius 1 radius one okay so let me draw a circle here getting so this type of region we have real imaginary AIS real AIS right let us have some scaling okay right so we get a circle here with Center 0 0 radius 1 so this circle we have got here right okay so now the everything is ready we have we found out the value of D Theta we found out the value of cos Theta as well as we have got the curve also which is a circle with Center 0 0 radius 1 okay so now I'm going to put all those things here in I okay so just make a screenshot of it then I will go further so let us write therefore therefore okay therefore this I is equal to integration over C this curve we have right mod Z is equal to 1 E to uh 2 I Theta actually e to I Theta means what z e 2 I Theta means z s upon 5 + 4 this is cos Theta Z 2 + 1 upon 2 Z D Theta is DZ upon i z okay so I have substituted all values so let us simplify it further so what will happen this z z will get cancel integration over c z will get cancel so Z upon 5 + here 2 two will get cancel that means 2 Z 2 + 1 upon Z DZ upon I so this is equal to integration over see let me remove all this part so we'll get a more space to write okay so what will I do Z as it is I will cross multiply in denominator so 5 Z + 2 Z 2 + 2 upon Z DZ upon I see this Zed is in denominator of denominator we shift to the numerator okay so let me do that thing here this is equal to integration / c z s this I in denominator I'm just writing it here I 2 z s + 5 Z + 2 DZ okay so this is a given integral this is our I so we are interested in that function f of Z so let us write the F of Z here here F of Z is z s upon I 2 Z 2 + 5 Z + 2 now we have to find a singular point of this function that means a point where the function is not analytic generally we say the point where the denominator is zero if we have the factors of this denominator we can easily find find a singular point so I will try to find the factors of denominator okay let us see how to find the factors Z 2 upon I 2 into 2 4 okay what we have 2 into 2 4 so we have to express four as a product of two numbers whose sum is 5 okay whose sum is 5 so 4 into 1 4 and 4 + 1 5 okay so that's why we can express 5 Z as 4 Z + 1 Z 2 Z 2 + 4 Z + 1 Z that means Z + 2 so this is equal to z s upon I tell me what can I take common from first two terms what can I take common from these two terms we can take 2 Z common 2 Z common so if you take 2 Z common Z + 2 here we cannot take anything common so I'm simply writing 1 Z + 2 did you notice one one thing that is z + 2 Z + 2 okay at both places we have got the same bracket we can take that bracket common so that means Z 2 upon i z + 2 first bracket and this is 2 Z + 1 2 Z + 1 second bracket so this is our F of Z so I express the denominator as a product of two brackets we found out the factors so the singular Point means a point where the denominator is zero if I put Z is equal to -2 denominator will be zero or if I put Z equal to -1 by 2 getting the point or you can simply solve in rough 2 Z + 1 is = 0 that means 2 Z is = -1 that means Z is = -1 by in that way also you can find okay so Z is equal to -1 by2 and Z is equal to -2 these are two singular points so let me clearly mention here okay so let me remove this part it is not required so here here Z is = -2 and Z = -1 by2 are singular points actually both are simple poles huh both are simple poles I'm clearly mentioning here so let us find those points are lying inside or outside - 1 by2 now - 1 by2 will be here and -2 -2 will be here -1 this is -2 so that min-2 lying outside a circle and -1 by2 lying inside a circle we are more interested in this point which is lying inside so at Min - 1x2 we have to find race let me mention but only Z is equal - 1x2 lies inside C okay the second point lies outside C so we have to find residue only at Z is equal to - 1 by2 okay there is no more space to write make a screenshot of it first so now our Target is to find the residue of F at Z is equal to - 1 by2 but now the question is how to find find residu the definition of residu involves Lauren series expansion it will take more time to find a residu with help of definition but fortunately both are simple poles that means Z is equal to minus1 by2 is also simple pole if you have a simple pool we have very simple formula to calculate residue and same formula I'm going to use so let us write residue of f of Z at Z is = -1 by2 that formula is limit z t to -1 by 2 Z minus actually Z minus z z is - 1 by2 minus- plus and I can write + 1 by2 getting the point F of Z actually we have a minus sign here but it has already minus sign so double minus you will have minus minus plus this is equal to limit z t to -1 2 we can easily cross multiply so 2 Z + 1 upon 2 F of Z this F of Z we can easily put since we found out the factors no so it is very easy to calculate Z + 2 and 2 Z + 1 so will you tell me what can we cancel here yes anyone can easily say that 2 Z + 1 2 Z + 1 will get cancel now in the remaining part of this function we can apply the limit apply the limit that means at the place of Z I'm going to put this - 1 by2 let us see what will happen so here at numerator - 1x2 s in denominator 2 i - 1 by 2 + 2 - 1X 2 s means 1X 4 2 I right here we can cross multiply 2 into 2 4 -1 + 4 upon 2 so this is 1x 4 getting 2 2 cancel i 4 - 1 3 3 I so 4 basically it is in denominator so that 4 into this 3 that means 12 I okay so this is a value of residue only see only two singular points we had one point lied outside the circle second Point Li inside a circle we finded the resid at that point then we can easily find the value of integration let us see how to find okay just make a screenshot of it then I will go further then the value of integral will be okay so let me write there therefore that I is equal to 2 pi I sum of residues so this is 2 pi I only one residue we got which is 1 upon 12 I so I I will get cancel right 2 two cancel so we'll get < by 6 so this is value of integral I but actually this integral is different from the given integral since instead of cos 2 pi we consider e to 2 I Theta so that thing also I should mention so I means what 0 to 2 pi we had consider e to 2 I Theta upon 5 + 4 cos Theta D Theta is equal to < / 6 we know that how to express e to I Theta in terms of cos and sin using that I can write 0 to 2 pi cos 2 Theta + I sin 2 2 Theta cos 2 I Theta can be expressed in this way denominator will be same 5 + 4 cos Theta D Theta is equal to < by 6 my mistake huh < by 6 so let us divide separately cos Theta cos 2 Theta upon denominator sin 2 Theta upon denominator and we can make separate separate integrals that means those will look like this 0 to 2 pi cos 2 Theta upon 5 + 4 cos Theta D Theta + i 0 to 2 pi sin 2 Theta upon 5 + 4 cos Theta D Theta so I divided separately I consider when you know when we have addition we can easily take separate separate integration so with the help of that I got this one < by 6 which is purely real number so < by 6 + 0 I since in left hand side we have got a complex number I is there so I tried to express the right hand side also in the same form see did you notice one thing whatever the integral we have that is nothing but the real part of this left hand side so that means we are more interested in a real part instead of imaginary part so if you compare the real part getting and if you compare the imaginary part so the real part is equal to < by 6 and the imaginary part is equal to Zer but we are not interested at all in imaginary part so that's why simply compare real part only just make a screenshot of it so if you equate real parts we get 0 to 2 pi right 0 to 2 pi cos 2 Theta upon 5 + 4 cos Theta D Theta is equal to the real part is < by 6 that means the value of given integral is < by 6 so in this way we solve this problem make a screenshot of it then we will stop thank you see you in next video you

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