Final answer:
The leading behavior of the given integrals as x approaches positive infinity using the method of stationary phase is determined by finding the dominant contribution to the integral through the stationary points of the phase function. For most of the given integrals, the leading behavior is 0 as the phase function does not have stationary points or the oscillating functions average out to 0 over a complete cycle. Only one integral exhibits a non-zero leading behavior.
Step-by-step explanation:
To find the leading behavior of the given integrals as x approaches positive infinity using the method of stationary phase, we need to find the dominant contribution to the integral by finding the stationary points of the phase function. Let's go through each part:
(a) ∫₀¹eᶦˣᵗ² cosh t²dt
The leading behavior of this integral as x approaches positive infinity is 0. This is because the phase function does not have any stationary points and the cosine hyperbolic function does not oscillate.
(b) ∫₀¹cos (xt⁴) tan t dt
The leading behavior of this integral as x approaches positive infinity is also 0. This is because the phase function does not have any stationary points and the tangent function oscillates but its average over a complete cycle is 0.
(c) ∫₀¹eᶦˣ ⁽ᵗ−ˢᶦⁿθ⁾ dt
The leading behavior of this integral as x approaches positive infinity is e^(-sin(θ)). This is because the phase function has a stationary point at t = sin(θ) and the exponential function does not oscillate.
(d) ∫₀¹sin [x(t+1/6t³−sinh t)] cos t dt
The leading behavior of this integral as x approaches positive infinity is 0. This is because the phase function does not have any stationary points and the sine function oscillates but its average over a complete cycle is 0.
(e) ∫−₁¹sin[x(t−sin t)]sinh t dt
The leading behavior of this integral as x approaches positive infinity is 0. This is because the phase function does not have any stationary points and the sine function oscillates but its average over a complete cycle is 0.