Dtjytyk

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

$$int_0^infty frac{xcos x-sin x}{x^3}cosleft(frac{x}{2}right)mathrm dx$$

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.

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begin{align}
&int_{0}^{infty}{xcospars{x} - sinpars{x} over x^{3}},
cospars{x over 2},dd x =
int_{0}^{infty}{xbracks{1 - 2sin^{2}pars{x/2}} - sinpars{x} over x^{3}},cospars{x over 2},dd x
\[5mm] & =
{1 over 2}int_{0}^{infty}{2xcospars{x/2} - 2xsinpars{x}sinpars{x/2} -
2sinpars{x}cospars{x/2} over x^{3}},,,dd x
\[5mm] & =
{1 over 2}int_{0}^{infty}{xcospars{x/2} + xcospars{3x/2} -
sinpars{3x/2} - sinpars{x/2} over x^{3}},,,dd x
\[1cm] & =
-,{1 over 2}int_{0}^{infty}{1 - cospars{x/2} over x^{2}},dd x -
{1 over 2}int_{0}^{infty}{1 - cospars{3x/2} over x^{2}},dd x
\[5mm] & -
{1 over 4}int_{x = 0}^{x to infty}bracks{2x - sinpars{3x/2} - sinpars{x/2}},ddpars{1 over x^{2}}
end{align}

By integrating by parts:
$ds{int_{0}^{infty}{sin^{2}pars{x} over x^{2}},dd x =
int_{0}^{infty}{sinpars{x} over x},dd x = {1 over 2},pi}$.

$$
begin{align}
&int_0^inftyfrac{xcos(x)-sin(x)}{x^3}cosleft(frac{x}{2}right),mathrm{d}xtag{1}\
&=int_0^inftyfrac{xleft(cosleft(frac32xright)+cosleft(frac12xright)right)-left(sinleft(frac32xright)+sinleft(frac12xright)right)}{2x^3},mathrm{d}xtag{2}\
&=-int_0^inftyfrac{xleft(cosleft(tfrac32xright)+cosleft(tfrac12xright)right)-left(sinleft(tfrac32xright)+sinleft(tfrac12xright)right)}{4},mathrm{d}x^{-2}tag{3}\
&=int_0^inftyfrac{left(frac12cosleft(frac12xright)-frac12cosleft(frac32xright)right)-xleft(frac32sinleft(frac32xright)+frac12sinleft(frac12xright)right)}{4x^2},mathrm{d}xtag{4}\
&=int_0^inftyleft(frac{1-cosleft(frac32xright)}{8x^2}-frac{1-cosleft(frac12xright)}{8x^2}-frac{3sinleft(frac32xright)}{8x}-frac{sinleft(frac12xright)}{8x}right)mathrm{d}x
tag{5}\
&=int_0^inftyleft(frac{3(1-cos(x))}{16x^2}-frac{1-cos(x)}{16x^2}-frac{3sin(x)}{8x}-frac{sin(x)}{8x}right)mathrm{d}x
tag{6}\
&=int_0^inftyleft(frac{3sin(x)}{16x}-frac{sin(x)}{16x}-frac{3sin(x)}{8x}-frac{sin(x)}{8x}right)mathrm{d}x
tag{7}\
&=-frac38int_0^inftyfrac{sin(x)}{x},mathrm{d}xtag{8}\
&=-frac{3pi}{16}tag{9}
end{align}
$$
Explanation:

$(2)$: trigonometric product formulas

$(3)$: prepare to integrate by parts

$(4)$: integrate by parts

$(5)$: separate integrals

$(6)$: substitute $xmapsto2x$ and $xmapstofrac23x$

$(7)$: integrate by parts

$(8)$: combine

$(9)$: $int_0^inftyfrac{sin(x)}{x},mathrm{d}x=fracpi2$

We can also use contour integration.

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

$(1)$ The second integral vanishes since the function $ displaystyle frac{(z-i)(e^{-iz/2}+e^{-3iz/2})}{(z- iepsilon)^{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{xcos x-sin x}{x^3}cosfrac{x}{2} dx=& frac{1}{2}int^infty_{-infty} frac{xcos x-sin x}{x^3}e^{ix/2} dx\
=& frac{-1}{4} int^infty_{-infty} [xcos x-sin x] e^{ix/2} dleft(frac{1}{x^2} right).
end{align}
Using integration by parts, we have
begin{align}
frac{-1}{4} int^infty_{-infty} [xcos x-sin x] e^{ix/2} dleft(frac{1}{x^2} right)=& frac{1}{4} int^infty_{-infty}d([xcos 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{xcos 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{sincleft(frac{x}{pi}right)}right]left(frac{1}{2}right) = pi mathcal{F}^{-1}[operatorname{sinc}]left( frac{1}{2pi}right) = pi.
end{align}

Next, observe
begin{align}
int^infty_{-infty} frac{xcos 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{ipi}{2}.
end{align}

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

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

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

One can get
begin{align}
&int dfrac{xcos x-sin x}{x^3},cosdfrac x2, mathrm dx
= intdfrac1{4sin frac x2},mathrm dleft(dfrac{sin x}{x}right)^2 \[4pt]
&=dfrac1{4sin frac x2}left(dfrac{sin x}{x}right)^2
+intleft(dfrac{sin x}{x}right)^2dfrac{cosfrac x2}{8sin^2 frac x2},mathrm dx
=dfrac1{x^2}sinfrac x2,cos^2 frac x2
+dfrac12intdfrac{cos^3frac x2}{x^2},mathrm dx\[4pt]
&=dfrac1{x^2}sinfrac x2
-dfrac1{4x^2}left(3sinfrac x2-sin frac {3x}2right)
+dfrac18intdfrac{3cosfrac x2+cosfrac{3x}2}{x^2},mathrm dx\[4pt]
&=dfrac1{4x^2}left(sinfrac x2+sin frac {3x}2right)
-dfrac18intleft(3cosfrac x2+cosfrac{3x}2right),mathrm dfrac1x\[4pt]
&=dfrac1{8x^2}left(2sinfrac x2+2sinfrac{3x}2-3xcosfrac {x}2-xcos frac {3x}2right)
-dfrac3{16}intfrac1xleft(sinfrac x2+sinfrac{3x}2right),mathrm dx.
end{align}
Since
$$limlimits_{xto0}dfrac{2sinfrac x2+2sinfrac{3x}2-3xcosfrac {x}2-xcosfrac {3x}2}{8x^2}
= limlimits_{xto0}dfrac{2frac x2+2frac{3x}2-3x-x+O(x^3)}{8x^2}=0,$$

$$intlimits_{0}^{infty} dfrac{sin x}x,mathrm dx =fracpi2,$$

then
$$intlimits_{0}^{infty} dfrac{xcos x-sin x}{x^3},cosdfrac x2, mathrm dx
=-frac3{16}left(intlimits_{0}^{infty} dfrac{sinfrac x2}{frac x2},mathrm dfrac x2+intlimits_{0}^{infty} dfrac{sin frac{3x}2}{frac{3x}2},mathrm dfrac{3x}2right) = color{green}{mathbf{-frac{3pi}{16}}}.$$

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:

$mathcal{L}f(x)=int_{0}^{infty}e^{-sx}f(x)dx=F(s)$

$mathcal{L}g(x)=int_{0}^{infty}e^{-sx}g(x)dx=G(s)$

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

$$int_{0}^{infty}F(x)g(x)dx=int_{0}^{infty}G(x)f(x)dx$$

and apply it to the required integral.

We take

$F(x)=frac{1}{x^3}$

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

and get after taking the Laplace and inverse Laplace transforms

$f(x)=frac{x^2}{2}$

$G(x)=frac{4x^2-3}{(4x^2+1)^2}-frac{4x^2+45}{(4x^2+9)^2}$