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Asymptotic Analysis and Singular Perturbation Theory

Asymptotic Analysis and Singular Perturbation Theory

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3.2 Perturbed Gaussian integrals<br />

Consider the following integral<br />

I(a, ε) =<br />

∞<br />

−∞<br />

<br />

exp − 1<br />

2 ax2 − εx 4<br />

<br />

dx, (3.4)<br />

where a > 0 <strong>and</strong> ε ≥ 0. For ε = 0, this is a st<strong>and</strong>ard Gaussian integral, <strong>and</strong><br />

I(a, 0) = 1<br />

√ 2πa .<br />

For ε > 0, we cannot compute I(a, ε) explicitly, but we can obtain an asymptotic<br />

expansion as ε → 0 + .<br />

First, we proceed formally. Taylor exp<strong>and</strong>ing the exponential with respect to ε,<br />

<br />

exp − 1<br />

2 ax2 − εx 4<br />

<br />

1 −<br />

= e 2 ax2<br />

<br />

1 − εx 4 + 1<br />

2! ε2x 8 + . . . + (−1)n<br />

ε<br />

n!<br />

n x 4n <br />

+ . . . ,<br />

<strong>and</strong> integrating the result term-by-term, we get<br />

I(a, ε) ∼ 1<br />

<br />

√<br />

2πa<br />

where<br />

1 − ε〈x 4 〉 + . . . + (−1)n<br />

ε<br />

n!<br />

n 〈x 4n 〉 + . . .<br />

〈x 4n 〉 =<br />

∞<br />

−∞ x4n 1 − e 2 ax2<br />

∞ 1<br />

e− 2<br />

−∞ ax2<br />

dx<br />

dx .<br />

We use a special case of Wick’s theorem to calculate these integrals.<br />

Proposition 3.3 For m ∈ N, we have<br />

where<br />

Proof. Let<br />

〈x 2m 〉 =<br />

(2m − 1)!!<br />

am ,<br />

(2m − 1)!! = 1 · 3 · 5 . . . (2m − 3) · (2m − 1).<br />

J(a, b) =<br />

∞ 1<br />

e− 2<br />

−∞ ax2 +bx dx<br />

∞ 1<br />

e− 2<br />

−∞ ax2 dx .<br />

<br />

, , (3.5)<br />

Differentiating J(a, b) n-times with respect to b <strong>and</strong> setting b = 0, we find that<br />

〈x n 〉 = dn<br />

<br />

<br />

J(a, b) .<br />

dbn Writing<br />

b=0<br />

1 −<br />

e 2 ax2 +bx −<br />

= e 1<br />

2 a(x−b)2 + b2<br />

2a<br />

32

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