Krdytkyu

I'm looking for methods to solve the following integral:

$$I = int_{0}^{infty} frac{x - sin(x)}{x^3left(x^2 + 4right)} :dx$$

Write$$I = int_{0}^{infty} underbrace{frac{x - sin(x)}{x^3left(x^2 + 4right)}}_{:=g(x)} ~mathrm dx=frac 12 int_{-infty}^{infty} g(x) ~mathrm dx.$$
Let $f(z)=dfrac{z+ie^{iz}}{z^3(z^2+4)}$, so $Re f(x)=g(x).$

Take the contour

Then we have two poles ($z=0$ on the contour and $z=2i$ inside the contour),
$$int_{C_R}f(z)~mathrm dz+int_{-R}^R f(z)~mathrm dz=pi i operatorname{Res}_{z=0}f(z)+2pi i operatorname{Res}_{z=2i}f(z).tag{*}$$
By ML lemma,
$$lim_{Rtoinfty}int_{C_R}f(z)~mathrm dz=0.$$
We also calculate the residue of the function at $z=0$ by means of power series:
begin{align*}
f(z)=frac14frac{z+ie^{iz}}{z^3}frac{1}{1-(-frac{z^2}{4})}&=frac14frac1{z^3}left[z+ileft(1+iz-frac{z^2}{2}-frac{iz^3}{6}+cdotsright)right]left(1-frac{z^2}4+cdotsright)\
&=cdots+frac14left(-frac i2-frac i4right)frac 1z+cdots
end{align*}
implies
$$operatorname{Res}_{z=0}f(z)=frac14left(-frac i2-frac i4right)=-frac{3i}{16},$$
and the residue at $z=2i$ is
$$operatorname{Res}_{z=2i}f(z)=lim_{zto 2i}(z-2i)f(z)=frac{(2+e^{-2})i}{32}.$$
Thus, from $(*)$,
$$int_{-infty}^infty f(x) ~mathrm dx =frac{3pi}{16}-frac{(2+e^{-2})pi}{16}=frac{1}{16}(1-e^{-2}).$$

$$therefore I=frac 12 Reint_{-infty}^{infty} g(x) ~mathrm dx=frac{1}{32}(1-e^{-2}).$$

My approach:

Let

$$I(t) = int_{0}^{infty} frac{xt - sin(xt)}{x^3left(x^2 + 4right)} :dx$$

Where $I = I(1)$

Taking the first derivative:

$$ frac{dI}{dt} = int_{0}^{infty} frac{x - xcos(xt)}{x^3left(x^2 + 4right)} :dx = int_{0}^{infty} frac{1 - cos(xt)}{x^2left(x^2 + 4right)} :dx$$

Taking the second derivative:

$$ frac{d^2I}{dt^2} = int_{0}^{infty} frac{xsin(xt)}{x^2left(x^2 + 4right)} :dx = int_{0}^{infty} frac{sin(xt)}{xleft(x^2 + 4right)} :dx$$

Now, take the Laplace Transform w.r.t $t$:

begin{align}
mathscr{L}left[ frac{d^2I}{dt^2}right] &= int_{0}^{infty} frac{mathscr{L}left[sin(xt)right]}{xleft(x^2 + 4right)} :dx \
&= int_{0}^{infty} frac{x}{left(s^2 + x^2right)xleft(x^2 + 4right)}:dx \
&= int_{0}^{infty} frac{1}{left(s^2 + x^2right)left(x^2 + 4right)}:dx
end{align}

Applying the Partial Fraction Decomposition we may find the integral

begin{align}
mathscr{L}left[ frac{d^2I}{dt^2}right] &= int_{0}^{infty} frac{1}{left(s^2 + x^2right)left(x^2 + 4right)}:dx \
&= frac{1}{s^2 - 4} int_{0}^{infty} left[frac{1}{x^2 + 4} - frac{1}{x^2 + s^2} right]:dx \
&= frac{1}{s^2 - 4} left[frac{1}{2}arctanleft(frac{x}{2}right) - frac{1}{s}arctanleft(frac{x}{s}right)right]_{0}^{infty} \
&= frac{1}{s^2 - 4} left[frac{1}{2}frac{pi}{2} - frac{1}{s}frac{pi}{2} right] \
&= frac{pi}{4sleft(s + 2right)}
end{align}

We now take the inverse Laplace Transform:

$$ frac{d^2I}{dt^2} = mathscr{L}^{-1}left[frac{pi}{4sleft(s + 2right)} right] = frac{pi}{8}left(1 - e^{-2t} right) $$

We now integrate with respect to $t$:

$$ frac{dI}{dt} = int frac{pi}{8}left(1 - e^{-2t} right):dt = frac{pi}{8}left(t + frac{e^{-2t}}{2} right) + C_1$$

Now

$$ frac{dI}{dt}(0) = int_{0}^{infty} frac{1 - cos(xcdot 0)}{x^2left(x^2 + 4right)} :dx = 0 = frac{pi}{8}left(frac{1}{2} right) + C_1 rightarrow C_1 = -frac{pi}{16}$$

Thus,

$$ frac{dI}{dt} = int frac{pi}{8}left(1 - e^{-2t} right):dt = frac{pi}{8}left(t + frac{e^{-2t}}{2} right) - frac{pi}{16}$$

We now integrate again w.r.t $t$

$$ I(t) = int left[frac{pi}{8}left(t + frac{e^{-2t}}{2} right) - frac{pi}{16} right] :dt = frac{pi}{8}left(frac{t^2}{2} - frac{e^{-2t}}{4} right) - frac{pi}{16}t + C_2 $$

Now

$$I(0) = int_{0}^{infty} frac{xcdot0 - sin(xcdot0)}{x^3left(x^2 + 4right)} :dx = 0 = frac{pi}{8}left( -frac{1}{4} right) + C_2 rightarrow C_2 = frac{pi}{32}$$

And so we arrive at our expression for $I(t)$

$$I(t)= frac{pi}{8}left(frac{t^2}{2} - frac{e^{-2t}}{4} right) - frac{pi}{16}t + frac{pi}{32}$$

Thus,

$$I = I(1) = frac{pi}{8}left(frac{1}{2} - frac{e^{-2}}{4} right) - frac{pi}{16} + frac{pi}{32} = frac{pi}{32}left(1 - e^{-2}right)$$

Using the expansion near $x=0$,
$$
frac1{{x^3left(x^2+4right)}}=frac1{4x^3}-frac1{16x}+O(1)
$$
and the contours
$$
gamma^+=[-R-i,R-i]cup Re^{i[0,pi]}-i
$$
which contains $0$ and $2i$, and
$$
gamma^-=[-R-i,R-i]cup Re^{-i[0,pi]}-i
$$
which contains $-2i$, we get
$$
begin{align}
int_0^inftyfrac{x-sin(x)}{x^3left(x^2+4right)},mathrm{d}x
&=frac12int_{-infty-i}^{infty-i}frac{x-sin(x)}{x^3left(x^2+4right)},mathrm{d}x\
&=color{#C00}{frac12int_{gamma^+}frac{x-frac{e^{ix}}{2i}}{x^3left(x^2+4right)},mathrm{d}x}\
&color{#090}{+frac12int_{gamma^-}frac{frac{e^{-ix}}{2i}}{x^3left(x^2+4right)},mathrm{d}x}\
&=color{#C00}{underbrace{quadfrac{3pi}{32}quad}_{x=0}underbrace{-frac{4+e^{-2}}{64}pi}_{x=2i}}color{#090}{underbrace{ -frac{e^{-2}}{64}pi }_{x=-2i}}\
&=frac{1-e^{-2}}{32}pi
end{align}
$$