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\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\df}{\ds\frac}
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\begin{document}


{\bf Question}

Show that as $x\to\infty$

$$\ds\int_x^\infty \ds\frac{\cos t}{\sqrt{t}}\, dt \sim
\ds\frac{1}{\sqrt{x}}(f\cos x-g\sin x)$$

$$\ds\int_x^\infty \ds\frac{\sin t}{\sqrt{t}}\, dt \sim
\ds\frac{1}{\sqrt{x}}(f\cos x-g\sin x)$$

where

$$f\sim\ds\frac{1}{2x}-\ds\frac{1.3.5}{(2x)^3}+\ds\frac{1.3.5.7.9}{(2x)^5}$$

$$g\sim1-\ds\frac{1.3}{(2x)^2}+\ds\frac{1.3.5.7}{(2x)^4}$$

Comment on the asymptotic validity of the results.

\vspace{.5in}

{\bf Answer}

Consider
$J=\ds\int_x^\infty\ds\frac{e^{it}}{\sqrt{t}}\,dt=\ds\int_x^\infty\ds\frac{\cos
t}{\sqrt{t}}\,dt+i\ds\int_x^\infty\ds\frac{\sin t}{\sqrt{t}}\,dt$

so can do both integrals by just looking at $J$.

Use integration by parts

$\begin{array} {rclrcl} u &amp; = &amp; \df{1}{i\sqrt{t}} &amp; dv &amp; = &amp;
ie^{it}\\ du &amp; = &amp; -\ds\frac{1}{2i}t^{-\frac{3}{2}} &amp; v &amp; = &amp;
e^{it}\end{array}$

$J=\left[\df{e^{it}}{i\sqrt{t}}\right]_x^\infty+\df{1}{2i}\ds\int_x^\infty\df{e^{it}}
{t^{\frac{3}{2}}}\,dt=\df{e^{ix}}{i\sqrt{x}}+\df{1}{2i}\ds\int_x^\infty\df{e^{it}
{t^{\frac{3}{2}}}}\,dt$

Iterate integration by parts to get

\begin{eqnarray*} J &amp; = &amp;
+\df{ie^{ix}}{\sqrt{x}}-\df{1}{2}\left\{[e^{it}t^{-\frac{3}{2}}]_x^\infty
+\df{3}{2}\ds\int_x^\infty e^{it}t^{-\frac{5}{2}}\,dt\right\}\\ &amp;
= &amp;
ie^{ix}x^{-\frac{1}{2}}+\df{1}{2}e^{ix}x^{-\frac{3}{2}}-\df{3}{4}
\left\{[-ie^{it}t^{-\frac{5}{2}}]_x^\infty-\df{5}{2}\ds\int_x^\infty
ie^{it} t^{-\frac{7}{2}}\,dt\right\}\\ &amp; = &amp;
ie^{ix}{x^{-\frac{1}{2}}}+\df{1}{2}e^{ix}x^{-\frac{3}{2}}
-\df{3}{4}ie^{ix}x^{-\frac{5}{2}}\\ &amp; &amp;
+\df{15}{8}\left\{\left[e^{it}
t^{-\frac{7}{2}}\right]_x^\infty+\df{7}{2}\ds\int_x^\infty
e^{it}t^{-\frac{9}{2}}\,dt \right\}\\ &amp; = &amp;
ie^{ix}x^{-\frac{1}{2}}+\df{1}{2}e^{ix}x^{-\frac{3}{2}}
-\df{3}{4}ie^{ix}x^{-\frac{5}{2}}-\df{15}{8}e^{ix}x^{-\frac{7}{2}}\\
&amp; &amp;
+\df{105}{16}\left\{[-ie^{-it}t^{-\frac{9}{2}}]_x^\infty-\df{9}{2}\ds\int_x^\infty
ie^{it} t^{-\frac{11}{2}}\,dt\right\}\\ &amp; = &amp;
ie^{ix}x^{-\frac{1}{2}}+\df{1}{2}e^{ix}x^{-\frac{3}{2}}-\df{3}{4}ie^{ix}x^{-\frac{5}{2}}-\df{15}{8}e^{ix}e^{-\frac{7}{2}}+\df{105}{16}ie^{ix}x^{-\frac{9}{2}}\\
&amp; &amp;
-\df{945}{32}\left\{[e^{it}t^{-\frac{11}{2}}]_x^\infty+\df{11}{2}\ds\int_x^\infty
e^{it}t^{-\frac{13}{2}}\,dt\right\}\\ &amp; &amp; \rm{etc...}
\end{eqnarray*}

Now take $Re$ part for the cos integral, $Im$ for the sine and get

\begin{eqnarray*} \ds\int_x^\infty \df{\cos t}{\sqrt{t}}\,dt &amp; \sim &amp;
\df{1}{\sqrt{x}}\left\{\df{1}{2x}-\df{1\cdot 3\cdot
5}{(2x)^3}+\df{1 \cdot 3\cdot 5 \cdot 7\cdot
9}{(2x^5)}+\cdots\right\}\cos x\\ &amp; &amp;
-\df{1}{\sqrt{x}}\left\{1-\df{1\cdot 3}{(2x)^2}+\left(\df{1 \cdot
3\cdot 5\cdot 7}{2x^4}\right)+\cdots \right\}\sin x
\end{eqnarray*}

and $\ds\int_x^\infty \df{\sin t}{\sqrt{t}}\,dt \sim
\df{1}{\sqrt{x}}\{\cdots\}\cos x+\df{1}{\sqrt{x}}\{\cdots\}\sin x$

The asymptotic validity is suspect! sin $x$ and cos $x$ have an
infinite number of zeros as $x\to\infty$. Thus any truncation has
an infinite number of zeros so any remainder term will also have
zeros. This makes finding an implied constant difficult since you
may have to show a remainder is $O(0)$! It's best to consider $f$
and $g$ themselves as the asymptotic expansion, extracting the
oscillatory terms before bounding.

\end{document}
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