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Unformatted text preview: Chapter 7: Evaluation of Improper Integrals • Advice 1: • Page No. 257: Q. Nos.: 1  5 exists. RHS on limit the provided ) ( lim ) ( then 0, x all for continuous is f(x) Let ) 1 ( ∫ ∫ ∞ ∞ → = ≥ R R dx x f dx x f ( 29 ∫ ∞ ∞ ) ( then x. all for continuous is f(x) Let 2 dx x f exist. RHS on limits both the provided , ) ( lim ) ( lim 1 2 2 1 ∫ ∫ ∞ → ∞ → + = R R R R dx x f dx x f exist. RHS. on limit the provided , ) ( lim dx f(x) P.V. number the is (2) integral the of (P.V.) value principal Cauchy R R R ∫ ∫ ∞ ∞ ∞ → = dx x f ∫ ∫ ∞ ∞ ∞ ∞ dx x f dx x f ) ( P.V. of existence the implies ) ( integral improper of Existence (1) : Remark not true. is converse But 2 x lim xdx lim dx ) x ( f P.V. x.Then f(x) Let . Ex R R 2 R R R R = = = = ∞ → ∞ ∞ ∞ → ∫ ∫ 2 lim 2 lim lim lim ) ( 2 2 2 2 1 1 2 2 1 1 R R xdx xdx dx x f But R R R R R R ∞ → ∞ → ∞ → ∞ → ∞ ∞ + = + = ∫ ∫ ∫ ∫ ∞ ∞ ⇒ ∴ . exist to fails dx ) x ( f integral improper The exist to fails RHS on Limit If the function f(x) ( ) is an even function i.e. f(x) =f(x) for all x, then the symmetry of the graph of y = f(x) with respect to y axis leads to ∞ < < ∞ x ∫ ∫ = R R R dx x f dx x f ) ( 2 ) ( When f(x) is an even function and the Cauchy pricncipal value exists, then ∫ ∫ ∫ ∞ ∞ ∞ ∞ ∞ = = ) ( 2 ) ( ) ( . . dx x f dx x f dx x f V P To evaluate improper integral of Even Rational Functions f(x)=p(x)/ q(x) • p(x) and q(x) are polynomials with real coefficients and no factors in common • q(z) has no real zeros but has at least one zero above the real axis. Method • Identify all distinct zeros of the polynomial q(z) that lie above the real axis • They will be finite in number • May be labeled as z 1 , z 2 ,…..z n where n is less than or equal to the degree of q(z) • Now, integrate the quotient f(z)=p(z)/q(z) around the positively oriented boundary of the semicircular region. The simple closed contour consists of • The segment of the real axis from z = R to z =R and • The top half of the circle R z = described counterclockwise and denoted by C R . Remark: • The positive number R is large enough that the points z 1 , z 2 ,… z n all lie inside closed path. ....
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This note was uploaded on 05/14/2010 for the course MATHEMATIC mathe taught by Professor Xyz during the Spring '10 term at Birla Institute of Technology & Science.
 Spring '10
 XYZ
 Math, Improper Integrals, Integrals

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