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1. Consider the integral *° −kt I (k) = √ e²kt 4k dt = [° with phase function f(t) = t — ln t. -kf(t) dt, k > 0, (a) Determine the saddle point of ƒ (t) and use it to find the first term in the asymptotic expansion of I(k) as k Note that log(1+x) = x − 1½ x² + 1/3 x ³ +... Then T(x+1)~xx+1 Le-x-12x(1-8)² ds e-x = x²+¹²² √ e¯½² (1-8)² ds. We cannot evaluate this integral, set y = s-1 and the lower limit goes to y = -1. However, the integrand is dominated for x large near s = 1. We can then write T(x+1)x+1-x ~ e-2 (1-8)² ds لسيسيليا ... = xx+1-x x- e¯ ½ xy² dy = x²+¹e¯x 2πT X as the integrand is negligible at s = 0, so that anything away from s = 1 makes no contribution to the integral.

1. Consider the integral *° −kt I (k) = √ e²kt 4k dt = [° with phase function f(t) = t — ln t. -kf(t) dt, k > 0, (a) Determine the saddle point of ƒ (t) and use it to find the first term in the asymptotic expansion of I(k) as k Note that log(1+x) = x − 1½ x² + 1/3 x ³ +... Then T(x+1)~xx+1 Le-x-12x(1-8)² ds e-x = x²+¹²² √ e¯½² (1-8)² ds. We cannot evaluate this integral, set y = s-1 and the lower limit goes to y = -1. However, the integrand is dominated for x large near s = 1. We can then write T(x+1)x+1-x ~ e-2 (1-8)² ds لسيسيليا ... = xx+1-x x- e¯ ½ xy² dy = x²+¹e¯x 2πT X as the integrand is negligible at s = 0, so that anything away from s = 1 makes no contribution to the integral.

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter2: Graphical And Tabular Analysis
Section2.1: Tables And Trends
Problem 1TU: If a coffee filter is dropped, its velocity after t seconds is given by v(t)=4(10.0003t) feet per...
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Hi, when solve the first integral, the range from [0,\inf] becomes [-inf, inf] and doesn't times 1/2. I thought that the most of the second integral's contribution comes from the saddle point t=1 so it doesn't need to times 1/2, too. However the solution of the second one does need to times 1/2. I wonder the reason and how to determine whether should we times a 1/2 when change integral range from [0,\inf] to [-inf, inf]

1. Consider the integral *° −kt I (k) = √ e²kt 4k dt = [° with phase function f(t) = t — ln t. -kf(t) dt, k > 0, (a) Determine the saddle point of ƒ (t) and use it to find the first term in the asymptotic expansion of I(k) as k
Transcribed Image Text:1. Consider the integral *° −kt I (k) = √ e²kt 4k dt = [° with phase function f(t) = t — ln t. -kf(t) dt, k > 0, (a) Determine the saddle point of ƒ (t) and use it to find the first term in the asymptotic expansion of I(k) as k
Note that log(1+x) = x − 1½ x² + 1/3 x ³ +... Then T(x+1)~xx+1 Le-x-12x(1-8)² ds e-x = x²+¹²² √ e¯½² (1-8)² ds. We cannot evaluate this integral, set y = s-1 and the lower limit goes to y = -1. However, the integrand is dominated for x large near s = 1. We can then write T(x+1)x+1-x ~ e-2 (1-8)² ds لسيسيليا ... = xx+1-x x- e¯ ½ xy² dy = x²+¹e¯x 2πT X as the integrand is negligible at s = 0, so that anything away from s = 1 makes no contribution to the integral.
Transcribed Image Text:Note that log(1+x) = x − 1½ x² + 1/3 x ³ +... Then T(x+1)~xx+1 Le-x-12x(1-8)² ds e-x = x²+¹²² √ e¯½² (1-8)² ds. We cannot evaluate this integral, set y = s-1 and the lower limit goes to y = -1. However, the integrand is dominated for x large near s = 1. We can then write T(x+1)x+1-x ~ e-2 (1-8)² ds لسيسيليا ... = xx+1-x x- e¯ ½ xy² dy = x²+¹e¯x 2πT X as the integrand is negligible at s = 0, so that anything away from s = 1 makes no contribution to the integral.
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