TITLE: Calculus Question: Improper integral $\int_{0}^{\infty}\frac{\cos(2x+1)}{\sqrt[3]{x}}\text dx$
QUESTION [11 upvotes]: How to evaluate integral $$\int_{0}^{\infty}\frac{\cos(2x+1)}{\sqrt[3]{x}}\text dx?$$ I tried substitution $x=u^3$ and I got $3\displaystyle\int_{0}^{\infty}u \cos(2u^3+1)\text du$. After that I tried to use integration by parts but I don't know the integral $\displaystyle\int \cos(2u^3+1)\text du$. Any idea? Thanks in advance.

REPLY [6 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{0}^{\infty}{\cos\pars{2x + 1} \over \root[3]{x}}\,\dd x:\ {\large ?}}$.

\begin{align}
&\color{#c00000}{\int_{0}^{\infty}{\cos\pars{2x + 1} \over \root[3]{x}}\,\dd x}
=\Re\bracks{\expo{\ic}\ \overbrace{\int_{0}^{\infty}x^{-1/3}\expo{2x\ic}\,\dd x}
^{\ds{x \equiv \ic t/2=\expo{\pi\ic/2}t/2\ \imp\ t = -2\ic x}}}\ =\
\\[3mm]&=\Re\bracks{\expo{\ic}\int_{0}^{-\infty\ic}%
\pars{\expo{\pi\ic/2}t \over 2}^{-1/3}\expo{2\pars{\ic t/2}\ic}\,{\ic \over 2}
\,\dd t}
\\[3mm]&=-2^{-2/3}\,\Im\bracks{\expo{\pars{1 - \pi/6}\ic}
\int_{0}^{-\infty\ic}t^{-1/3}\expo{-t}\,\dd t}
\\[3mm]&=-2^{-2/3}\,\Im\braces{\expo{\pars{1 - \pi/6}\ic}
\lim_{R\ \to\ \infty}\bracks{-\int_{R}^{0}t^{-1/3}\expo{-t}\,\dd t
-\left.\int_{-\pi/2}^{0}\,z^{-1/3}\expo{-z}\,\dd z
\right\vert_{z\ =\ R\expo{\ic\theta}}}}
\end{align}

However,
\begin{align}
&\color{#c00000}{\large%
\verts{\int_{-\pi/2}^{0}\,z^{-1/3}\expo{-z}\,\dd z}_{\,z\ =\ R\expo{\ic\theta}}}
\leq \verts{\int_{-\pi/2}^{0}R^{-1/3}\expo{-R\cos\pars{\theta}}R\,\dd\theta}
=R^{2/3}\int_{0}^{\pi/2}\expo{-R\sin\pars{\theta}}\,\dd\theta
\\[3mm]& < R^{2/3}\int_{0}^{\pi/2}\expo{-2R\theta/\pi}\,\dd\theta
={\pi \over 2}\,R^{-1/3}\pars{1 - \expo{-R}}
\color{#c00000}{\large\stackrel{R \to \infty}{\to} 0}
\end{align}

Then,
  \begin{align}
&\color{#66f}{\large\int_{0}^{\infty}{\cos\pars{2x + 1} \over \root[3]{x}}\,\dd x}
=-2^{-2/3}\sin\pars{1 - {\pi \over 6}}\int_{0}^{\infty}t^{-1/3}\expo{-t}\,\dd t
\\[3mm]&=\color{#66f}{\large%
-2^{-2/3}\sin\pars{1 - {\pi \over 6}}\Gamma\pars{2 \over 3}}
\approx {\tt -0.3911}
\end{align}