 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-336/1/a/solution

Factor out the elementary exponential and write

I (λ) = λ^{- 1/2} e^{2 λ} J (λ), J (λ) = \int_{0}^{\infty} e^{- s - λ^{2} / s} d s .

0 < λ ≪ δ ≪ 1

. In the outer part,

J_{out} = \int_{δ^{2}}^{\infty} e^{- s} (1 - \frac{λ ^{2}}{s} + \dots) d s .

The given identity for the Euler--Mascheroni constant, integrated by parts, implies

\int_{δ^{2}}^{\infty} \frac{e ^{- s}}{s} d s = - γ_{e} - 2 lo g δ + O (δ^{2}) .

Hence

J_{out} = 1 - δ^{2} + λ^{2} (γ_{e} + 2 lo g δ) + o (λ^{2}),

apart from pure higher matching powers in

δ

In the inner part, use

t = λ^{2} / s

J_{in} = λ^{2} \int_{λ^{2} / δ^{2}}^{\infty} t^{- 2} e^{- t} e^{- λ^{2} / t} d t .

Since

s = λ^{2} / t \leq δ^{2}

, expand the last exponential. At the required order,

\int_{a}^{\infty} t^{- 2} e^{- t} d t = \frac{e ^{- a}}{a} - in t_{a}^{\infty} \frac{e ^{- t}}{t} d t = \frac{1}{a} - 1 + γ_{e} + lo g a + O (a) .

With

a = λ^{2} / δ^{2}

, this gives

J_{in} = δ^{2} + λ^{2} [- 1 + γ_{e} + 2 lo g λ - 2 lo g δ] + o (λ^{2}) .

The overlap terms

δ^{2}

and

lo g δ

cancel, leaving the logarithmic switchback term

J (λ) = 1 + λ^{2} (2 lo g λ + 2 γ_{e} - 1) + o (λ^{2}) .

Finally,

e^{2 λ} = 1 + 2 λ + 2 λ^{2} + O (λ^{3}),

I (λ) \sim λ^{- 1/2} + 2 λ^{1/2} + λ^{3/2} (1 + 2 lo g λ + 2 γ_{e}) .

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