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.

  • $\begingroup$ @Tyler6 : You can use $\displaystyle\int\frac{x\cos x - \sin x}{x^3}\cos\frac{x}{2}dx = -\frac{3}{16}\left( \int\limits_0^{x/2}\frac{\sin x}{x} + \int\limits_0^{3x/2}\frac{\sin x}{x} \right) -\frac{\cos^2\frac{x}{2}}{x^2}\left(\frac{x}{2}\cos\frac{x}{2} - \sin\frac{x}{2}\right) + C$ $~~$ Why Fourier Transforms ? $\endgroup$
    – user90369
    Mar 13, 2019 at 11:46

9 Answers 9


$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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: $\ds{\int_{0}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\dd x = \int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x = {1 \over 2}\,\pi}$.

$$ \begin{align} &\int_0^\infty\frac{x\cos(x)-\sin(x)}{x^3}\cos\left(\frac{x}{2}\right)\,\mathrm{d}x\tag{1}\\ &=\int_0^\infty\frac{x\left(\cos\left(\frac32x\right)+\cos\left(\frac12x\right)\right)-\left(\sin\left(\frac32x\right)+\sin\left(\frac12x\right)\right)}{2x^3}\,\mathrm{d}x\tag{2}\\ &=-\int_0^\infty\frac{x\left(\cos\left(\tfrac32x\right)+\cos\left(\tfrac12x\right)\right)-\left(\sin\left(\tfrac32x\right)+\sin\left(\tfrac12x\right)\right)}{4}\,\mathrm{d}x^{-2}\tag{3}\\ &=\int_0^\infty\frac{\left(\frac12\cos\left(\frac12x\right)-\frac12\cos\left(\frac32x\right)\right)-x\left(\frac32\sin\left(\frac32x\right)+\frac12\sin\left(\frac12x\right)\right)}{4x^2}\,\mathrm{d}x\tag{4}\\ &=\int_0^\infty\left(\frac{1-\cos\left(\frac32x\right)}{8x^2}-\frac{1-\cos\left(\frac12x\right)}{8x^2}-\frac{3\sin\left(\frac32x\right)}{8x}-\frac{\sin\left(\frac12x\right)}{8x}\right)\mathrm{d}x \tag{5}\\ &=\int_0^\infty\left(\frac{3(1-\cos(x))}{16x^2}-\frac{1-\cos(x)}{16x^2}-\frac{3\sin(x)}{8x}-\frac{\sin(x)}{8x}\right)\mathrm{d}x \tag{6}\\ &=\int_0^\infty\left(\frac{3\sin(x)}{16x}-\frac{\sin(x)}{16x}-\frac{3\sin(x)}{8x}-\frac{\sin(x)}{8x}\right)\mathrm{d}x \tag{7}\\ &=-\frac38\int_0^\infty\frac{\sin(x)}{x}\,\mathrm{d}x\tag{8}\\ &=-\frac{3\pi}{16}\tag{9} \end{align} $$ 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
$(9)$: $\int_0^\infty\frac{\sin(x)}{x}\,\mathrm{d}x=\frac\pi2$

  • $\begingroup$ I see that this may be similar to Felix Marin's answer, but I think the path may be different enough, and possibly simpler, to warrant leaving it. $\endgroup$
    – robjohn
    Oct 29, 2016 at 13:53

We can also use contour integration.

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

$(1)$ 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 $a \le 0$.


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}


Integration by parts can be performed for the indefinite integral, using relations

$$\dfrac{x\cos x-\sin x}{x^2} = \left(\dfrac{\sin x}{x}\right)',\quad \sin^3 z =\frac{3\sin z-\sin3z}4,\quad \cos^3z=\frac{3\cos z+\cos3z}4.$$

One can get \begin{align} &\int \dfrac{x\cos x-\sin x}{x^3}\,\cos\dfrac x2\, \mathrm dx = \int\dfrac1{4\sin \frac x2}\,\mathrm d\left(\dfrac{\sin x}{x}\right)^2 \\[4pt] &=\dfrac1{4\sin \frac x2}\left(\dfrac{\sin x}{x}\right)^2 +\int\left(\dfrac{\sin x}{x}\right)^2\dfrac{\cos\frac x2}{8\sin^2 \frac x2}\,\mathrm dx =\dfrac1{x^2}\sin\frac x2\,\cos^2 \frac x2 +\dfrac12\int\dfrac{\cos^3\frac x2}{x^2}\,\mathrm dx\\[4pt] &=\dfrac1{x^2}\sin\frac x2 -\dfrac1{4x^2}\left(3\sin\frac x2-\sin \frac {3x}2\right) +\dfrac18\int\dfrac{3\cos\frac x2+\cos\frac{3x}2}{x^2}\,\mathrm dx\\[4pt] &=\dfrac1{4x^2}\left(\sin\frac x2+\sin \frac {3x}2\right) -\dfrac18\int\left(3\cos\frac x2+\cos\frac{3x}2\right)\,\mathrm d\frac1x\\[4pt] &=\dfrac1{8x^2}\left(2\sin\frac x2+2\sin\frac{3x}2-3x\cos\frac {x}2-x\cos \frac {3x}2\right) -\dfrac3{16}\int\frac1x\left(\sin\frac x2+\sin\frac{3x}2\right)\,\mathrm dx. \end{align} Since $$\lim\limits_{x\to0}\dfrac{2\sin\frac x2+2\sin\frac{3x}2-3x\cos\frac {x}2-x\cos\frac {3x}2}{8x^2} = \lim\limits_{x\to0}\dfrac{2\frac x2+2\frac{3x}2-3x-x+O(x^3)}{8x^2}=0,$$

$$\int\limits_{0}^{\infty} \dfrac{\sin x}x\,\mathrm dx =\frac\pi2,$$

then $$\int\limits_{0}^{\infty} \dfrac{x\cos x-\sin x}{x^3}\,\cos\dfrac x2\, \mathrm dx =-\frac3{16}\left(\int\limits_{0}^{\infty} \dfrac{\sin\frac x2}{\frac x2}\,\mathrm d\frac x2+\int\limits_{0}^{\infty} \dfrac{\sin \frac{3x}2}{\frac{3x}2}\,\mathrm d\frac{3x}2\right) = \color{green}{\mathbf{-\frac{3\pi}{16}}}.$$


I use the Laplace transform method instead of Fourier because the first one is the main tool for an electrician (who I am) and the Laplace transform is very similar to the Fourier transform.

By definition of the Laplace transform:



Now we use the following equation from the Laplace transform theory:


and apply it to the required integral.

We take


$g(x)=x\cos x\cos \frac{x}{2}-\sin x\cos \frac{x}{2}$

and get after taking the Laplace and inverse Laplace transforms



Placing these results into the relationship above we arrive at the next expression for the required integral:

$$I=\frac{1}{2}\int_{0}^{\infty}x^2\left [ \frac{4x^2-3}{(4x^2+1)^2}-\frac{4x^2+45}{(4x^2+9)^2} \right ]dx$$

$I=\left [- \frac{3}{16}\arctan 2x- \frac{3}{16}\arctan\frac{2x}{3}+\frac{9x}{4(4x^2+9)^2}+\frac{x}{4(4x^2+1)^2}\right ]_{0}^{\infty}=$

$=- \frac{3}{16}\frac{\pi}{2}- \frac{3}{16}\frac{\pi}{2}=- \frac{3\pi}{16}$


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

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


$$\displaystyle{\displaylines{I=\int_{0}^{\infty }\frac{xcos(x)-sin(x)}{x^{2}}.\frac{cos(\frac{x}{2})}{x}dx=\int_{0}^{\infty }\frac{cos(\frac{x}{2})}{x}\left( \frac{cos(x)}{x}-\frac{sin(x)}{x^{2}} \right)dx}}$$ 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 $$\displaystyle{\displaylines{I=\frac{-1}{2}\left[\int_{0}^{\frac{1}{2}} \left( y(\frac{\pi}{2}-\frac{\pi}{2}) \right)dy+\int_{\frac{1}{2}}^{1}\left(y(\frac{\pi}{2}+\frac{\pi}{2}) \right) dy\right]}}$$

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


This is very old exercise has been evaluated by applying MAZ Identity

This is very old exercise has been evaluated by applying MAZ Identity

  • 2
    $\begingroup$ What's MAZ identity ?. Do you have any reference for it ?. $\endgroup$ Mar 2, 2021 at 8:51
  • $\begingroup$ @felixmarin I had never heard of it until today (saw it in a video, tried to find a reference, found this problem instead), but it appears to be the identity used in the answer of Martin Gales. If $K(x,y)=\overline{K(y,x)}$, and we define an integral transform by integrating against K, then if $f(x)K(x,y)\overline{g(y)}$ is integrable, we can apply Fubini. Since the Laplace kernel is real, bounded, and symmetric, it defines a self adjoint operator between appropriate spaces, but can be applied in a wider context where everything still makes sense. $\endgroup$
    – Aaron
    Mar 8, 2021 at 1:12
  • 1
    $\begingroup$ @uniquesailor Can you please enlighten us all by telling us where does the name "Maz" come from and what does it stand for? It is a person's name or some sort of acronym? $\endgroup$
    – omegadot
    Jul 16, 2021 at 3:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .