Calculus Question: Improper integral $\int_{0}^{\infty}\frac{\cos(2x+1)}{\sqrt[3]{x}}\text dx$ 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.
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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}

A: Here is how. The approach is based on Mellin transform.

$$I =  \int_{0}^{\infty}\frac{\cos(2x+1)}{\sqrt[3]{x}}dx = \cos(1)\int_{0}^{\infty}\frac{\cos(2x)}{\sqrt[3]{x}}dx -\sin(1)\int_{0}^{\infty}\frac{\sin(2x)}{\sqrt[3]{x}}dx . $$

To evaluate the integrals on the right hand side make the change of variables $t=2x$ and use the Mellin transform (see tables) of $\cos(x)$ and $\sin(x)$ and then take the limit as $s \to 2/3 $. I believe you can finish the problem. See related techniques. 
A: $$\color{blue}{\mathcal{I}=\frac{\Gamma(\frac{2}{3})\cos(1+\frac{\pi}{3})}{2^{2/3}}\approx-0.391190966503539\cdots}$$

\begin{align}
\int^\infty_0\frac{\cos(2x+1)}{x^{1/3}}{\rm d}x
&=\int^\infty_0\frac{\cos(2x+1)}{\Gamma(\frac{1}{3})}\int^\infty_0t^{-2/3}e^{-xt} \ {\rm d}t \ {\rm d}x\tag1\\
&=\frac{1}{\Gamma(\frac{1}{3})}\int^\infty_0t^{-2/3}\int^\infty_0e^{-xt}\cos(2x+1) \ {\rm d}x \ {\rm d}t\\
&=\frac{\cos(1)}{\Gamma(\frac{1}{3})}\int^\infty_0\frac{t^{1/3}}{t^2+4}{\rm d}t-\frac{2\sin(1)}{\Gamma(\frac{1}{3})}\int^\infty_0\frac{t^{-2/3}}{t^2+4}{\rm d}t\tag2\\
&=\frac{\cos(1)}{2^{2/3}\Gamma(\frac{1}{3})}\int^\infty_0\frac{t^{1/3}}{1+t^2}{\rm d}t-\frac{\sin(1)}{2^{2/3}\Gamma(\frac{1}{3})}\int^\infty_0\frac{t^{-2/3}}{1+t^2}{\rm d}t\tag3\\
&=\frac{\cos(1)}{2^{5/3}\Gamma(\frac{1}{3})}\int^\infty_0\frac{t^{-1/3}}{1+t}{\rm d}t-\frac{\sin(1)}{2^{5/3}\Gamma(\frac{1}{3})}\int^\infty_0\frac{t^{-5/6}}{1+t}{\rm d}t\tag4\\
&=\frac{\pi\left(\cos(1)-\sqrt{3}\sin(1)\right)}{2^{2/3}\Gamma(\frac{1}{3})\sqrt{3}}\tag5\\
&=\frac{2\pi\cos(1+\frac{\pi}{3})}{2^{2/3}\frac{2\pi}{\Gamma(\frac{2}{3})\sqrt{3}}\sqrt{3}}\tag6\\
&=\frac{\Gamma(\frac{2}{3})\cos(1+\frac{\pi}{3})}{2^{2/3}}
\end{align}

Explanation:
$(1)$: $\small{\displaystyle\frac{1}{x^n}=\frac{1}{\Gamma(n)}\int^\infty_0t^{n-1}e^{-xt}{\rm d}t}$
$(2)$: $\displaystyle\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$
$(2)$: $\small{\displaystyle\int^\infty_0e^{-ax}\sin(bx){\rm d}x=\frac{b}{a^2+b^2}}$
$(2)$: $\small{\displaystyle\int^\infty_0e^{-ax}\cos(bx){\rm d}x=\frac{a}{a^2+b^2}}$
$(3)$: $\displaystyle t\mapsto 2t$
$(4)$: $\displaystyle t\mapsto \sqrt{t}$
$(5)$: $\small{\displaystyle\int^\infty_0\frac{x^{p-1}}{1+x}{\rm d}x=\pi\csc(p\pi)}$
$(6)$: $\small{\displaystyle \Gamma(z)=\frac{\pi\csc(\pi z)}{\Gamma(1-z)}}$, $\small{\displaystyle a\cos{x}-b\sin{x}=\sqrt{a^2+b^2}\cos(x+\arctan{\frac{b}{a}})}$
A: Hint: $\cos(2x+1) = \Re e^{i(2x+1)}$ and $\int_0^\infty \text dx\, x^{a-1} e^{-x} = \Gamma(a)$. The latter equation can be analytically continued to complex exponentials, as long as the integral converges.
Also, why does the new Latex font look terrible?
