Evaluating the improper integral $\int_0^\infty \frac{x\cos x-\sin x}{x^3} \cos(\frac{x}{2}) \mathrm dx $

Question

I've been working through the following integral and am stumped:

$$\int_0^\infty \frac{x\cos x-\sin x}{x^3}\cos\left(\frac{x}{2}\right)\mathrm dx$$

Given the questions in my class that have proceeded and followed this integral, I believe that this is some form of Fourier transform/integral. However, it doesn't look like any of the content surrounding it. That is, there is no $e^{-ikx}$ or $g(k)$ or anything else that I'm familiar with.

I know that it's an even function, but that's about as far as I can get. If I try to split it over the subtraction, I get two non-converging integrals, so that wasn't much help either. I've been throwing lots of trig identities at it but nothing familiar has appeared yet.

Any help would be greatly appreciated. Thank you.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{\infty}{x\cos\pars{x} - \sin\pars{x} \over x^{3}}\, \cos\pars{x \over 2}\,\dd x \\[5mm] = & \int_{0}^{\infty}{x\bracks{1 - 2\sin^{2}\pars{x/2}} - \sin\pars{x} \over x^{3}}\,\cos\pars{x \over 2}\,\dd x \\[5mm] & = {1 \over 2}\int_{0}^{\infty}{2x\cos\pars{x/2} - 2x\sin\pars{x}\sin\pars{x/2} - 2\sin\pars{x}\cos\pars{x/2} \over x^{3}}\,\,\,\dd x \\[5mm] & = {1 \over 2}\int_{0}^{\infty}{x\cos\pars{x/2} + x\cos\pars{3x/2} - \sin\pars{3x/2} - \sin\pars{x/2} \over x^{3}}\,\,\,\dd x \\[1cm] & = -\,{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x - {1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x \\[5mm] & - {1 \over 4}\int_{x\ =\ 0}^{x\ \to\ \infty}\bracks{2x - \sin\pars{3x/2} - \sin\pars{x/2}}\,\dd\pars{1 \over x^{2}} \end{align}

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Integrating by parts the last integral: \begin{align} &\int_{0}^{\infty}{x\cos\pars{x} - \sin\pars{x} \over x^{3}}\, \cos\pars{x \over 2}\,\dd x = \\[5mm] & = -\,{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x - {1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x \\[5mm] & + {1 \over 4}\int_{x = 0}^{\infty}{2 - 3\cos\pars{3x/2}/2 - \cos\pars{x/2}/2 \over x^{2}}\,\dd x \\[1cm] & = -\,{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x - {1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x \\[5mm] & + {3 \over 8}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x + {1 \over 8}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x \\[1cm] & = -\,{3 \over 8}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x -\,{1 \over 8}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x \\[5mm] & = -\,{3 \over 16}\int_{0}^{\infty}{1 - \cos\pars{x} \over x^{2}}\,\dd x -\,{3 \over 16}\int_{0}^{\infty}{1 - \cos\pars{x} \over x^{2}}\,\dd x = -\,{3 \over 8}\ \int_{0}^{\infty}{1 - \cos\pars{x} \over x^{2}}\,\dd x \\[5mm] & = -\,{3 \over 4}\int_{0}^{\infty}{\sin^{2}\pars{x/2} \over x^{2}}\,\dd x = -\,{3 \over 8}\ \underbrace{\int_{0}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\dd x} _{\ds{=\ {\pi \over 2}}}\ = \ \bbox[#ffe,10px,border:1px dotted navy]{\ds{-\,{3 \over 16}\,\pi}} \end{align} >By integrating by parts: $\ds{\int_{0}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\dd x = \int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x = {1 \over 2}\,\pi}$.

Answer 2

Inspired by Felix Marin's calculation using integration by parts.

Observe \begin{align} \int^\infty_0 \frac{x\cos x-\sin x}{x^3}\cos\frac{x}{2}\ dx=&\ \frac{1}{2}\int^\infty_{-\infty} \frac{x\cos x-\sin x}{x^3}e^{ix/2}\ dx\\ =&\ \frac{-1}{4} \int^\infty_{-\infty} [x\cos x-\sin x] e^{ix/2}\ d\left(\frac{1}{x^2} \right). \end{align} Using integration by parts, we have \begin{align} \frac{-1}{4} \int^\infty_{-\infty} [x\cos x-\sin x] e^{ix/2}\ d\left(\frac{1}{x^2} \right)=&\ \frac{1}{4} \int^\infty_{-\infty}d([x\cos x-\sin x]e^{ix/2}) \frac{1}{x^2}\\ =&\ \frac{-1}{4} \int^\infty_{-\infty}\frac{\sin x}{x}e^{ix/2}\ dx + \frac{i}{8} \int^\infty_{-\infty} \frac{x\cos x-\sin x}{x^2}e^{ix/2}\ dx. \end{align} Now, observe \begin{align} \int^\infty_{-\infty}\frac{\sin x}{x}e^{ix/2}\ dx = \mathcal{F}^{-1}\left[\operatorname{sinc\left(\frac{x}{\pi}\right)}\right]\left(\frac{1}{2}\right) = \pi \mathcal{F}^{-1}[\operatorname{sinc}]\left( \frac{1}{2\pi}\right) = \pi. \end{align}

Next, observe \begin{align} \int^\infty_{-\infty} \frac{x\cos x-\sin x}{x^2} e^{ix/2}\ dx =&\ \int^\infty_{-\infty} \frac{d}{dx}\left( \frac{\sin x}{x}\right) e^{ix/2}\ dx\\ =&\ -\frac{i}{2}\int^\infty_{-\infty}\frac{\sin x}{x} e^{ix/2}\ dx = -\frac{i\pi}{2}. \end{align}

Hence combining everything yields \begin{align} \int^\infty_0 \frac{x\cos x-\sin x}{x^3} \cos \frac{x}{2}\ dx = -\frac{\pi}{4} + \frac{\pi}{16} = -\frac{3\pi}{16}. \end{align}

Answer 3

We can also use contour integration.

$$ \begin{align} &\int_{0}^{\infty} \frac{x \cos x - \sin x}{x^{3}} \, \cos \left(\frac{x}{2} \right) \, dx \\ &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{x \left(\frac{e^{ix}+e^{-ix}}{2} \right) -\frac{e^{ix}-e^{-ix}}{2i}}{x^{3}} \left(\frac{e^{ix/2}+e^{-ix/2}}{2} \right) \, dx \\ &= \frac{1}{2} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{x \left(\frac{e^{ix}+e^{-ix}}{2} \right) -\frac{e^{ix}-e^{-ix}}{2i}}{(x- i \epsilon)^{3}} \left(\frac{e^{ix/2}+e^{-ix/2}}{2} \right) \, dx \\ &= \frac{1}{8} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty}\frac{(x+i)(e^{3ix/2}+e^{ix/2})}{(x- i\epsilon)^{3}} \, dx + \frac{1}{8} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{(x-i)(e^{-ix/2}+e^{-3ix/2})}{(x- i\epsilon)^{3}} \, dx \\ &=\frac{1}{8} \lim_{\epsilon \to 0^{+}} 2 \pi i \, \text{Res} \left[\frac{(z+i)(e^{3iz/2}+e^{iz/2})}{(z- i \epsilon)^{3}} , i\epsilon \right] + 0 \tag{1} \\ &= \frac{1}{8} \lim_{\epsilon \to 0^{+}} \, 2 \pi i \, \frac{1}{2!} \lim_{z \to i \epsilon}\frac{d^{2}}{dz^{2}} \, (z+i)(e^{3iz/2}+e^{iz/2}) \\ &= \frac{1}{8} \lim_{\epsilon \to 0^{+}} \frac{\pi}{4} \, e^{-3 \epsilon/2} \left((\epsilon-3) e^{\epsilon} + 9 \epsilon -3 \right) \\ &= - \frac{3 \pi}{16} \end{align}$$

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$(1)$ The second integral vanishes since the function $ \displaystyle \frac{(z-i)(e^{-iz/2}+e^{-3iz/2})}{(z- i\epsilon)^{3}} $ is analytic in the lower half-plane where $\left| e^{iaz} \right| \le 1$ if $a \le 0$.

Answer 4

$$ \begin{align} &\int_0^\infty\frac{x\cos(x)-\sin(x)}{x^3}\cos\left(\frac{x}{2}\right)\,\mathrm{d}x\tag{1}\\ &=\int_0^\infty\frac{x\left(\cos\left(\frac32x\right)+\cos\left(\frac12x\right)\right)-\left(\sin\left(\frac32x\right)+\sin\left(\frac12x\right)\right)}{2x^3}\,\mathrm{d}x\tag{2}\\ &=-\int_0^\infty\frac{x\left(\cos\left(\tfrac32x\right)+\cos\left(\tfrac12x\right)\right)-\left(\sin\left(\tfrac32x\right)+\sin\left(\tfrac12x\right)\right)}{4}\,\mathrm{d}x^{-2}\tag{3}\\ &=\int_0^\infty\frac{\left(\frac12\cos\left(\frac12x\right)-\frac12\cos\left(\frac32x\right)\right)-x\left(\frac32\sin\left(\frac32x\right)+\frac12\sin\left(\frac12x\right)\right)}{4x^2}\,\mathrm{d}x\tag{4}\\ &=\int_0^\infty\left(\frac{1-\cos\left(\frac32x\right)}{8x^2}-\frac{1-\cos\left(\frac12x\right)}{8x^2}-\frac{3\sin\left(\frac32x\right)}{8x}-\frac{\sin\left(\frac12x\right)}{8x}\right)\mathrm{d}x \tag{5}\\ &=\int_0^\infty\left(\frac{3(1-\cos(x))}{16x^2}-\frac{1-\cos(x)}{16x^2}-\frac{3\sin(x)}{8x}-\frac{\sin(x)}{8x}\right)\mathrm{d}x \tag{6}\\ &=\int_0^\infty\left(\frac{3\sin(x)}{16x}-\frac{\sin(x)}{16x}-\frac{3\sin(x)}{8x}-\frac{\sin(x)}{8x}\right)\mathrm{d}x \tag{7}\\ &=-\frac38\int_0^\infty\frac{\sin(x)}{x}\,\mathrm{d}x\tag{8}\\ &=-\frac{3\pi}{16}\tag{9} \end{align} $$ Explanation:
$(2)$: trigonometric product formulas
$(3)$: prepare to integrate by parts
$(4)$: integrate by parts
$(5)$: separate integrals
$(6)$: substitute $x\mapsto2x$ and $x\mapsto\frac23x$
$(7)$: integrate by parts
$(8)$: combine
$(9)$: $\int_0^\infty\frac{\sin(x)}{x}\,\mathrm{d}x=\frac\pi2$

Answer 5

$$\displaystyle{\displaylines{I=\int_{0}^{\infty }\frac{xcos(x)-sin(x)}{x^{2}}.\frac{cos(\frac{x}{2})}{x}dx=\int_{0}^{\infty }\frac{cos(\frac{x}{2})}{x}\left( \frac{cos(x)}{x}-\frac{sin(x)}{x^{2}} \right)dx}}$$ so we have : $$I=\int_{0}^{\infty }\frac{cos(\frac{x}{2})}{x}\left( \frac{cos(x)}{x} -\int_{0}^{1}\frac{cos(xy)}{x}dy\right)dx=\int_{0}^{\infty }\frac{cos(\frac{x}{2})}{x}\left(\int_{0}^{1} (-ysin(yx)) dy\right)dx$$ $$=\int_{0}^{\infty }\frac{-cos(\frac{x}{2})}{2x}\left(\int_{0}^{1} (-ysin(yx)) dy\right)dx$$$$=\displaystyle{\displaylines{\frac{-1}{2}\int_{0}^{1}y\int_{0}^{\infty }\frac{sin(y+\frac{1}{2})x-sin(\frac{1}{2}-y)x}{x}dxdy}}$$ Hence $$\displaystyle{\displaylines{I=\frac{-1}{2}\left[\int_{0}^{\frac{1}{2}} \left( y(\frac{\pi}{2}-\frac{\pi}{2}) \right)dy+\int_{\frac{1}{2}}^{1}\left(y(\frac{\pi}{2}+\frac{\pi}{2}) \right) dy\right]}}$$

$$\displaystyle{\displaylines{=\frac{-1}{2}\int_{\frac{1}{2}}^{1}\pi y dy=\frac{-3\pi}{16}}}$$

Answer 6

Simple attempt:

$$let\ \ f(x)=\left\{\begin{matrix} & 1-x^{2}\ \ \ ,\ |x|< 1& \\ & 0\ \ \ \ \ \ \ \ \ \ , \ \ |x|> 1 & \end{matrix}\right \\$$

by fouier transform of f(x) we have :

$$F(a)=\int_{0}^{\infty }f(t).cos(at)dt=\int_{0}^{1}\ (1-t^{2}).cos(at)\ dt=\int_{0}^{1}cos(at)dt-\int_{0}^{1}t^{2}cos(at)dt$$

$$\therefore F(a)=\frac{sin(a)}{a}-\frac{2acos(a)+(a^{2}-2).sin(a)}{a^{3}}=-\frac{2acos(a)-2sin(a)}{a^{3}}$$

$$\therefore f(a)=1-a^{2}=\frac{2}{\pi }\int_{0}^{\infty }F(x).cos(ax)\ dx=-\frac{4}{\pi }\int_{0}^{\infty }\frac{xcos(x)-sin(x)}{x^{3}}.\ cos(ax)dx$$

Now let put a=1/2 then we have

$$putting\ a=\frac{1}{2}\ \ \ \Rightarrow \int_{0}^{\infty }\frac{xcos(x)-sin(x)}{x^{3}}.cos\left ( \frac{x}{2} \right )dx\ =\frac{-3\pi }{16}$$