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.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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