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Asymptotic behavior of functions, asymptotic series and related topics
0
votes
Small parameter expansion of an integral
Some partial progress: following the suggestion in the comments, make the substitution $t\mapsto \varepsilon t$ to arrive at the integral
$$I(\varepsilon) = \int_0^\infty \frac{e^{-\varepsilon t}}{1+t …
2
votes
Accepted
Nonstationary phase method for oscillatory integral
For stationary phase, you usually consider the integral
$$I(\lambda)=\int_a^b f(t) e^{i\lambda g(t)}\,dt$$
with $\lambda>0$ a large parameter. If there are no stationary points inside $[a,b]$, then yo …