Asymptotics of $\int^\infty_0 e^{-(r+1)u}\cos\left(\frac{\pi}{2}e^{-u}\right)\,du$ as $r \to \infty$.

Question

Can someone explain the following asymptotics, through Watson’s Lemma or through another argument?

\begin{align} \int^\infty_0 e^{-(r+1)u}\cos\left(\frac{\pi}{2}e^{-u}\right)\,du \sim \frac{\pi}{2(r+1)^2},\qquad r\to \infty. \end{align}

Answer 1

By just integrating by parts twice one gets, as $r \to \infty$, $$ \begin{align} &\int^\infty_0 e^{-(r+1)u}\cos\left(\frac{\pi}{2}e^{-u}\right)\,du \\\\&=\left[ \frac{e^{-(r+1)u}}{-(r+1)}\cos\left(\frac{\pi}{2}e^{-u}\right)\right]_0^\infty +\frac{\pi}{2(r+1)}\int^\infty_0 e^{-(r+2)u}\sin\left(\frac{\pi}{2}e^{-u}\right)\,du \\\\&=\color{red}{0}+\frac{\pi}{2(r+1)}\left(\left[ \frac{e^{-(r+2)u}}{-(r+2)}\sin\left(\frac{\pi}{2}e^{-u}\right)\right]_0^\infty -\frac{\pi}{2(r+2)}\int^\infty_0 e^{-(r+3)u}\cos\left(\frac{\pi}{2}e^{-u}\right)\,du \right) \\\\&=\frac{\pi}{2(r+1)(r+2)}+o\left(\frac1{(r+1)^2}\right) \\\\&=\frac{\pi}{2(r+1)^2}+o\left(\frac1{(r+1)^2}\right). \end{align} $$

Answer 2

If we want to make it match the formula for Watson's lemma on wikipedia, we start by writing

$$ \int^\infty_0 e^{-(r+1)u}\cos\left(\frac{\pi}{2}e^{-u}\right)\,du = \int^\infty_0 e^{-ru} e^{-u}\cos\left(\frac{\pi}{2}e^{-u}\right)\,du = \int^\infty_0 e^{-ru} \varphi(u)\,du, $$

where

$$ \varphi(u) := e^{-u}\cos\left(\frac{\pi}{2}e^{-u}\right). $$

Now $\varphi(0) = 0$ and $\varphi'(0) = \pi/2$, so we can write

$$ \varphi(u) = ug(u), $$

where

$$ g(u) = u^{-1} e^{-u}\cos\left(\frac{\pi}{2}e^{-u}\right) $$

and $g(0) = \pi/2 \neq 0$. In the notation of wikipedia we have $\lambda = 1$.

Thus, taking only the first term ($n=0$) of the asymptotic expansion from the article, we get

$$ \int^\infty_0 e^{-(r+1)u}\cos\left(\frac{\pi}{2}e^{-u}\right)\,du \sim \frac{g^{(0)}(0)\, \Gamma(1+0+1)}{0!\, r^{1+0+1}} = \frac{\pi}{2r^2} $$

as $r \to \infty$ since

$$ g^{(0)}(0) = g(0) = \pi/2 \qquad \text{and} \qquad \Gamma(2) = 1! = 1. $$

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To conclude that

$$ \int^\infty_0 e^{-(r+1)u}\cos\left(\frac{\pi}{2}e^{-u}\right)\,du \sim \frac{\pi}{2(r+1)^2} $$

we just need to note that

$$ \frac{\pi}{2r^2} \sim \frac{\pi}{2(r+1)^2} $$

as $r \to \infty$.