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

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

• No matter how you look at it, you'll need the incomplete Gamma function for the antiderivative and the Gamma function for the exact result. – UserX Sep 10 '14 at 17:08

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

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

• Very details and easy to follow. I'm really familiar with the methods in your answer to this problem. Thanks. – Venus Sep 11 '14 at 9:45
• @Venus I am glad this answer is helpful. Thanks. – SuperAbound Sep 11 '14 at 10:01
• Nice answer and very well-organized. +1 – Tunk-Fey Sep 11 '14 at 12:59

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{\infty}{\cos\pars{2x + 1} \over \root[3]{x}}\,\dd x:\ {\large ?}}$.

\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}

• I love this blue color, I'm gonna steal it. – Lucas Zanella Sep 11 '14 at 8:15

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.

• But how to prove it the result of table? This is not Laplace/ Fourier transform course. Thanks anyway for your answer. +1 – Venus Sep 11 '14 at 9:43

Hint: $\cos(2x+1) = \Re e^{i(2x+1)}$ and $\int_0^\infty dx\, x^{a-1} e^{-x} = \Gamma(a)$. 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?