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 Incoming links: Moving-cutoff correction to a symmetric exponential integral

The expansion is singular near the moving lower limit, so expanding

e^{- ε / x}

throughout the original integral would lose its constant term. Introduce the symmetric exponential integral

J (ε) = \int_{0}^{\infty} \frac{e ^{- x - ε / x}}{x} d x .

Under

x = ε / t

, the discarded piece becomes

J - I = \int_{1}^{\infty} \frac{e ^{- t - ε / t}}{t} d t = β - ε \int_{1}^{\infty} \frac{e ^{- t}}{t ^{2}} d t + O (ε^{2}) = β - ε (e^{- 1} - β) + O (ε^{2}) .

This is a uniform Taylor expansion because

t \geq 1

. To derive the logarithmic expansion of a symmetric exponential integral without needing another special function, first exploit the same substitution to write

J = 2 \int_{ε}^{\infty} \frac{e ^{- x - ε / x}}{x} d x = 2 E (ε) + O (ε) = - lo g ε - 2 γ + o (1) .

The estimate follows from

0 \leq 1 - e^{- ε / x} \leq ε / x

. Here

γ

Differentiate under the integral and integrate the derivative of

e^{- x - ε / x} / x

over

(0, \infty)

; the exponentially small endpoint terms vanish. This gives the ordinary differential equation

ε J^{''} + J^{'} = J .

J = A lo g ε + B + ε (C_{1} lo g ε + D) + \dots

. Substitution gives

C_{1} = A

and

D = B - 2 A

. The preceding leading calculation sets

A = - 1

B = - 2 γ

, hence

J = - lo g ε - 2 γ + ε [- lo g ε + 2 - 2 γ] + O (ε^{2} ∣ lo g ε ∣) .

Subtracting the moving-cutoff correction to a symmetric exponential integral yields the requested answer, including the logarithmically enhanced first-order term:

I (ε) = - lo g ε - 2 γ - β + ε [- lo g ε + 2 - 2 γ + e^{- 1} - β] + O (ε^{2} ∣ lo g ε ∣) .

Thus the

ε lo g ε

switchback term must be retained when asking for all contributions through order

ε