Asymptotic behavior of $\int_{0}^{\infty} \frac{\cos(x)}{\sqrt{x+r}} dx$ as $r \to \infty$

Question

Sometimes I come across integrals that seem like they should have a nice asymptotic expansion, but I get nervous when the terms of the expansion seem to fail to converge. The following is an example integral that captures the features I want to better understand (variants of this integral appear in the radiation problem for 2d electrodynamics).

Consider the integral $$I(r) = \int_{0}^{\infty} \frac{\cos(x)}{\sqrt{x+r}} dx$$ as $r \to \infty$. I want to consider an asymptotic expansion in powers of $\sqrt{\frac{1}{r}}$.

What is the asymptotic expansion? What is the "leading" non-analytic piece?

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Here are my thoughts. I'll play fast and loose: I'm Taylor expanding past radii of convergence, exchanging integrals and sums, and renormalizing non-convergent integrals.

First, if I attempt to naively expand the denominator, there is trouble already at the lowest nontrivial order:

$$I(r) \sim \sum_{n=0}^{\infty} \frac{1}{r^{n+\frac{1}{2}}} (-1)^n \frac{\binom{2n}{n}}{2^{2n}} \int_{0}^{\infty} \cos(x) x^{n} dx$$

This makes me want to regularize the integrals, such as considering $\lim_{\epsilon \to 0^+} \int_{0}^{\infty} \cos(x) x^{n} e^{-\epsilon x} dx $. This limit results in $(-1)^{m+1} (2m+1)!$ for $n=2m+1$ is and $0$ for even $n$.

This results in

$$I(r) \sim \sum_{m=0}^{\infty} \frac{1}{r^{2m+\frac{3}{2}}} (-1)^m \frac{\binom{4m+2}{2m+1}}{2^{4m+2}} (2m+1)!$$

If I naively go by "the smallest term in the sum sets the scale of the hyperasymptotics," then the leading non-analytic piece would appear to be $e^{-r}=e^{-1/(1/r)}$, dropping many factors. However, this almost seems too good to be true, since this is quite a small piece.

It's possible there are typos above; please let me know if there are any algebra mistakes.

Answer 1

Performing the change of variables from $x$ to $t$ via $t=rx$ in the integral representation provided by MathArt, we obtain: $$\tag{1}\label{1} I(r) = \frac{1}{{\sqrt \pi }}\int_0^{ + \infty } {\frac{{{\rm e}^{ - rx} x^{1/2} }}{{1 + x^2 }}{\rm d}x} = \frac{1}{{\sqrt \pi }}\frac{1}{{r^{3/2} }}\int_0^{ + \infty } {\frac{{{\rm e}^{ - t} t^{1/2} }}{{1 + (t/r)^2 }}{\rm d}t} $$ provided $|\arg r|<\frac{\pi}{2}$. For any $N\ge 0$, $t>0$ and $r$ with $|\arg r|<\frac{\pi}{2}$, we substitute the identity $$ \frac{1}{{1 + (t/r)^2 }} = \sum\limits_{n = 0}^{N - 1} {t^{2n} \frac{{( - 1)^n }}{{r^{2n} }}} + \frac{{( - 1)^N }}{{r^{2N} }}\frac{{t^{2N} }}{{1 + (t/r)^2 }} $$ into \eqref{1}. Integrating term-by-term, we get $$ I(r) = \frac{1}{{r^{3/2} }}\left( {\sum\limits_{n = 0}^{N - 1} {( - 1)^n \frac{{\left( {\frac{1}{2}} \right)_{2n + 1} }}{{r^{2n} }}} +R_N (r) } \right) $$ for any $N\ge0$ and $r$ with $|\arg r|<\frac{\pi}{2}$. Here $(w)_p=\Gamma(w+p)/\Gamma(w)$ is the Pochhammer symbol. The remainder term $R_N(r)$ is given by $$\tag{2}\label{2} R_N (r) = \frac{{( - 1)^N }}{{r^{2N} }}\frac{1}{{\sqrt \pi }}\int_0^{ + \infty } {\frac{{{\rm e}^{ - t} t^{2N + 1/2} }}{{1 + (t/r)^2 }}{\rm d}t} = ( - 1)^N \frac{{\left( {\frac{1}{2}} \right)_{2N + 1} }}{{r^{2N} }}\Pi _{2N + 3/2} (r), $$ where $\Pi _p (w)$ is one of Dingle's basic terminants: $$ \Pi _p (w) = \frac{1}{{\Gamma (p)}}\int_0^{ + \infty } {\frac{{{\rm e}^{ - t} t^{p - 1} }}{{1 + (t/w)^2 }}{\rm d}t} ,\quad |\arg w|<\frac{\pi}{2}, \quad p>0. $$ Through analytic continuation, the function $\Pi _p (w)$ and the final expression in \eqref{2} can be extended across the entire Riemann surface associated with the logarithm. The Stokes phenomenon associated with $I(r)$ is fully determined by the basic terminant. Specifically, since $$ \Pi _p (w) = \frac{1}{{\Gamma (p)}}\int_0^{ + \infty } {\frac{{{\rm e}^{ - t} t^{p - 1} }}{{1 + (t/w)^2 }}{\rm d}t} \pm \pi {\rm i}\frac{{{\rm e}^{ \mp \frac{\pi }{2}{\rm i}p} }}{{\Gamma (p)}}w^p {\rm e}^{ \pm {\rm i}w} $$ for $\frac{\pi }{2} < \pm \arg w < \frac{{3\pi }}{2}$, we deduce $$ r^{3/2} I(r) \sim \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{\left( {\frac{1}{2}} \right)_{2n + 1} }}{{r^{2n} }}} +\begin{cases}0 & \text{if }\; |\arg r| < \frac{\pi}{2},\\[1ex] \frac{{\sqrt \pi }}{2}{\rm e}^{ \mp \frac{\pi }{4}{\rm i}} {\rm e}^{ - \left| r \right|} & \text{if }\; \arg r = \pm \frac{\pi }{2},\\[1ex] \sqrt \pi {\rm e}^{ \mp \frac{\pi }{4}{\rm i}} {\rm e}^{ \pm {\rm i}r} & \text{if }\; \frac{{\pi }}{2} < \pm\arg r < \frac{3\pi }{2}, \end{cases} $$ as $|r|\to+\infty$. From the bounds for $\Pi _p (w)$ given in this paper, we immediately obtain $$ \left| {R_N (r)} \right| \le \frac{{\left( {\frac{1}{2}} \right)_{2N + 1} }}{{\left| r \right|^{2N} }} \times \begin{cases} 1 & \text{ if } \; |\arg r| \le \frac{\pi}{4}, \\[1ex] \min\! \left(|\csc ( 2\arg r)|,\frac{1}{2}\chi\!\left(2N+\frac{3}{2}\right)+1\right) & \text{ if } \; \frac{\pi}{4} < |\arg r| \le \frac{{\pi }}{2}, \\[1ex] \cfrac{\sqrt {2\pi \left( 2N + \frac{3}{2} \right)} }{2\left| \sin (\arg r)\right|^{2N + 3/2} } + \frac{1}{2}\chi\!\left(2N+\frac{3}{2}\right)+1 & \text{ if } \; \frac{\pi}{2} < |\arg r| < \pi . \end{cases} $$ Here $\chi (w) = \sqrt \pi \Gamma \left( {\frac{w}{2} + 1} \right)/\Gamma \left( {\frac{w}{2} + \frac{1}{2}} \right)$ for $w>0$.

Answer 2

This will not be an answer but might be useful.

Your original integral can be transformed into one form with decaying factor of $e^{-rx}$ by Laplace transform

\begin{align} I(r)&=\int_{0}^{\infty} \frac{\cos(x)}{\sqrt{x+r}}\mathrm dx\\ &=\int_{0}^{\infty}\mathcal{L}[\cos(t)](x)\mathcal{L}^{-1}\left[\frac{1}{\sqrt{s+r}}\right](x)\mathrm dx\\ &=\int_{0}^{\infty}\frac{x}{x^2+1}\frac{e^{-rx}}{\sqrt{\pi x}}\mathrm dx\\ &=\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}e^{-rx}\frac{\sqrt{x}}{x^2+1}\mathrm dx. \end{align} The last integral is evaluated here by Wolfram.

Answer 3

Here's a partial answer to my question.

As Pavel R. pointed out, integration by parts gives a nice way to generate the asymptotic series.

$$\int_0^\infty \frac{\cos(x)}{\sqrt{x+r}} dx \sim \frac{\sin(x)}{(x+r)^{1/2}}\bigg|_{0}^\infty - \frac{1}{2}\frac{\cos(x)}{(x+r)^{3/2}}\bigg|_{0}^\infty - \frac{1}{2} \frac{3}{2} \frac{\sin(x)}{(x+r)^{3/2}}\bigg|_{0}^\infty + \frac{1}{2} \frac{3}{2} \frac{5}{2} \frac{\cos(x)}{(x+r)^{5/2}}\bigg|_{0}^\infty + \, ...$$

We can quickly see the result by noting that every other contribution to the series vanishes by virtue of $\sin(0)=0$, and alternates by virtue of the double integral of $\cos(x)$ being $-\cos(x)$.

That is, we have

$$\int_0^\infty \frac{\cos(x)}{\sqrt{x+r}} \sim \sum_{m=0}^{\infty} \frac{1}{r^{2m+\frac{3}{2}}}(-1)^m \frac{(4m+1)!!}{2^{2m+1}}$$

The first couple terms are $\frac{1}{2}\frac{1}{r^{3/2}}-\frac{15}{8} \frac{1}{r^{7/2}} + \, ...$.

It's straightforward to verify that this matches the series found in the original post via unorthodox means! The integration by parts is more straightforward and much more sensible, so thanks Pavel!

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Integration by parts also allows us to find bounds on the error term. The error term after truncating the sum at $\frac{1}{r^{2m+\frac{3}{2}}}$

$$E_{2m+\frac{3}{2}} = (-1)^m \frac{(4m+3)!!}{2^{2m+2}} \int_0^\infty \frac{\cos(x)}{(x+r)^{2m+5/2}}$$

I'll stop here, but I'll note that the error term where $m$ is order $r$ appear to be exponentially bounded above by $r$, pointing to the $e^{-r}$-style leading "non-analytic-in-$\frac{1}{r}$" piece.

Answer 4

Since $$\int \frac{e^{i x}}{\sqrt{x+r}}\,dx=\sqrt[4]{-1} \,e^{-i r} \,\,\Gamma\left(\frac{1}{2},-i (r+x)\right)$$ $$\sqrt{\frac{2}{\pi }}\int_0^\infty \frac{e^{i x}}{\sqrt{x+r}}\,dx=\left(1-2 C\left(\sqrt{\frac{2r}{\pi }} \right)\right) \cos (r)+$$ $$\left(1-2 S\left(\sqrt{\frac{2}{\pi }} \sqrt{r}\right)\right) \sin(r)$$ where appear Fresnel integrals.

Asymptotically $$\color{blue}{I=\int_0^\infty \frac{\cos(x)}{\sqrt{x+r}}\,dx= \frac{1}{\sqrt{\pi }}\sum_{n=0}^\infty (-1)^n\,\frac{\Gamma \left(2 n+\frac{3}{2}\right) } {r^{2 n+\frac{3}{2}} }}$$