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## Integral now

Integral now exponentiall small, but still wrong (can be checked by expanding the cos and comparing the order of the omitted term with the retained term. ) iii) Use method of steepest descent. Rewrite with I(x) = 1 2 � +1 −1 e −4xt2 +5ixt−ixt3 dt = 1 2 ρ(t) = −it 3 − 4t 2 + 5it Phases at the end points differ from each other ℑ (ρ(t = ±1) = ±4 Identify contours C1,2 of constant ℑ (ρ) using t = u + iv Note: � +1 −1 e xρ(t) dt ρ = −i � u 3 + 3iu 2 v − 3uv 2 − iv 3� − 4 � u 2 − v 2 + 2iuv � + 5iu − 5v = −v 3 + 3u 2 v − 4u 2 + 4v 2 − 5v � �� � φ • φ is even in u and ψ is odd in u. ψ = 4σ at t = σ with σ = ±1: 3uv 2 − 8uv + 5u − u 3 − 4σ = 0 +i � −u 3 + 3uv 2 − 8uv + 5u � � �� � ψ v = 1 � 8u ± 6u � 64u2 − 12u (5u − u3 � − 4σ) = 1 � 4u ± 3u √ u2 + 3u4 � + 12σu We need the contours passing through t = σ, i.e. u = σ and v = 0 σ = +1 : v2 = 1 � 4u − 3u √ u2 + 3u4 � + 12u σ = −1 : v1 = 1 � 4u + 3u √ u2 + 3u4 � − 12u Saddles Up Down v Down Up Down Up Down Up Down Down Up Figure 12: Contours with ψ = const and saddle points for e−4xt2 +5ixt−ixt3 . 70 u

Limiting behavior of the contours v ± 1,2 σ = +1 : v2 → − u √ 3 σ = −1 : v1 → + u √ 3 for u → +∞ v2 → − 2 1 √ √ 3 u for u → 0 + for u → −∞ v1 → − 2 1 √ √ 3 −u for u → 0 − Limiting behavior of real part φ along the contours v ± 1,2 for u → ±∞ and v → −∞ Use u 2 → 3v Thus σ = ±1 : φ → −v 3 + 3u 2 v → −v 3 + 9v 3 = 8v 3 → −∞ for v → −∞ • If we deformed the integration contour to include the singularity at u = 0 the main contribution to the integral would arise at the singularity. Poor choice • Integrate from t = ±1 outward to u → ±∞ – the main contribution to this contour integral arises near t = ±1 – we need to connect the end points with an additional contour C3 Identify all saddle points of ρ(t) dρ dt = −3it2 − 8t + 5i = 0 � t (s) 5 1 = 3i t (s) 2 = i t (s) 1,2 = − 8 ± √ 64 − 60 6i Along the contours with ψ = const. the real part φ can only have a local maximum at one of the two saddle points ⇒ φ does not have a local maximum along the contours v1,2 Identify contours with ψ = const. through the saddle point t = i ψ(t = i) = 0 ⇒ u = 0 or 3v 2 − 8v + 5 − u 2 = 0 ⇒ v3,4 = 4 ± √ 1 + 3u 2 The contour v4 is asymptotic to the contours v1,2. The integral along C3 does not vanish: φ reaches a maximum at the saddle at t = i. Therefore the dominant contribution to that integral arises near that saddle. � 2I(x) ∼ e xρ � dt + e xρ � dt + e xρ dt ∼ C1 C3 � t=−1−O(ǫ) e xρ � t=i+O(ǫ) dt + e xρ � t=1 dt + t=−1 C2 t=i−O(ǫ) t=1+O(ǫ) e xρ dt We can deform the contours slightly away from the contour ψ = const. to simplify the integrals C1 : t = −1 − s 0 ≤ s ≤ ǫ 71 3

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