real analysis - 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.

Answer

$$\color{blue}{\mathcal{I}=\frac{\Gamma(\frac{2}{3})\cos(1+\frac{\pi}{3})}{2^{2/3}}\approx-0.391190966503539\cdots}$$

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

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Explanation:
$(1)$: $\small{\displaystyle\frac{1}{x^n}=\frac{1}{\Gamma(n)}\int^\infty_0t^{n-1}e^{-xt}{\rm d}t}$
$(2)$: $\displaystyle\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$
$(2)$: $\small{\displaystyle\int^\infty_0e^{-ax}\sin(bx){\rm d}x=\frac{b}{a^2+b^2}}$
$(2)$: $\small{\displaystyle\int^\infty_0e^{-ax}\cos(bx){\rm d}x=\frac{a}{a^2+b^2}}$
$(3)$: $\displaystyle t\mapsto 2t$
$(4)$: $\displaystyle t\mapsto \sqrt{t}$
$(5)$: $\small{\displaystyle\int^\infty_0\frac{x^{p-1}}{1+x}{\rm d}x=\pi\csc(p\pi)}$
$(6)$: $\small{\displaystyle \Gamma(z)=\frac{\pi\csc(\pi z)}{\Gamma(1-z)}}$, $\small{\displaystyle a\cos{x}-b\sin{x}=\sqrt{a^2+b^2}\cos(x+\arctan{\frac{b}{a}})}$