Integral of $\int_0^\infty \frac{\sin^4(u)}{u^{k}}du$ where $k\in(1,3)$ My task is to

Evaluate $$\int_0^\infty \frac{\sin^4(u)}{u^{k}}\,du$$ where $k\in(1,3).$

I've tried a few things, but nothing seems to be working. Any help?
 A: A chance is given by switching to Laplace transforms. We have:
$$\mathcal{L}(\sin^4 x) = \frac{24}{s \left(4+s^2\right) \left(16+s^2\right)},\qquad \mathcal{L}^{-1}\left(\frac{1}{x^k}\right)=\frac{s^{k-1}}{\Gamma(k)}$$
and the equivalent integral
$$ \frac{1}{\Gamma(k)}\int_{0}^{+\infty}\frac{24\, s^{k-2}}{(4+s^2)(16+s^2)}\,ds $$
can be computed through partial fraction decomposition and the residue theorem.

Assuming $1<k<3$, we get:
  $$ \int_{0}^{+\infty}\frac{\sin^4(x)}{x^k}\,dx = \frac{\pi\, 2^k(2^k-8)}{64\,\Gamma(k)}\cdot \sec\left(\frac{\pi k}{2}\right)$$
  and $\log(2)$ when $k=3$.

A: First, write $\sin^4x=\dfrac{3-4\cos2x+\cos4x}8~.~$ Now, $3=3\cos(0~x)$, so our integral is basically  a 
linear combination of terms of the form $~\displaystyle\int_0^\infty\frac{\cos(ax)}{x^k}~dx.~$ Of course, the honest reader will 
immediately object that the aforementioned expression diverges for $k>1$. True indeed, but we 
will pretend to ignore such issues of convergence, and evaluate the three cosine integrals as if 
$k\in(0,1)$. How do we do that ? By using Euler's formula in conjunction with the well-known 
integral expression for the $\Gamma$ function. $($Doing that will once again stretch the norm of rigor, 
since we will pretend that the upper limit, following a linear substitution involving imaginary 
numbers, is real infinity, instead of complex infinity$)$. The final result will be $$a^{k-1}\cdot(-k)!~\cdot\sin\bigg(k~\dfrac\pi2\bigg),$$ where $a\in\{0,~2,~4\}.~$ Adding them all together, we have $I=\dfrac{2^k~(2^k-8)}{32}~(-k)!~\sin\bigg(k~\dfrac\pi2\bigg),~$ 
which can be shown to be the same as Jack's result, using Euler's reflection formula for the $\Gamma$ 
function. Also, for $k=2,~$ taking the limit, we have $I=\dfrac\pi4.~$ $($The expression for $\sin^4x$ was 
obtained by making use of the two famous trigonometric identities for $1\pm\cos2t)$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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)}
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\begin{align}
\int_{0}^{\infty}{\sin^4\pars{u} \over u^{k}}\,\dd u & \stackrel{\mrm{IBP}}{=}
{1 \over k - 1}\int_{0}^{\infty}{4\sin^{3}\pars{u}\cos\pars{u} \over
u^{k - 1}}\,\dd u =
{2 \over k - 1}\int_{0}^{\infty}{\sin^{2}\pars{u}\sin\pars{2u} \over
u^{k - 1}}\,\dd u
\\[5mm] & 
{2 \over k - 1}\int_{0}^{\infty}{\braces{\rule{0pt}{5mm}\bracks{1 - \cos\pars{2u}}/2}\sin\pars{2u} \over
u^{k - 1}}\,\dd u
\\[5mm] & =
{1 \over 2\pars{k - 1}}\int_{0}^{\infty}
{2\sin\pars{2u} - \sin\pars{4u} \over u^{k - 1}}\,\dd u
\\[5mm] & =
{1 \over k - 1}\int_{0}^{\infty}
{\sin\pars{2u} - 2u \over u^{k - 1}}\,\dd u -
{1 \over 2\pars{k - 1}}\int_{0}^{\infty}
{\sin\pars{4u} - 4u \over u^{k - 1}}\,\dd u
\\[5mm] & =
{2^{k - 2} \over k - 1}\int_{0}^{\infty}
{\sin\pars{u} - u \over u^{k - 1}}\,\dd u -
{2^{2k - 5} \over k - 1}\int_{0}^{\infty}
{\sin\pars{u} - u \over u^{k - 1}}\,\dd u
\\[5mm] & =
\bbx{{2^{k - 2} - 2^{2k - 5} \over k - 1}\int_{0}^{\infty}
{\sin\pars{u} - u \over u^{k - 1}}\,\dd u}
\\[5mm] & =
{2^{k - 2} - 2^{2k - 5} \over k - 1}\,\Im\int_{0}^{\infty}
{\expo{\ic u} - 1 - \ic u + u^{2}/2 \over u^{k - 1}}\,\dd u
\label{1}\tag{1}
\\[5mm] & =
-\,{2^{k - 2} - 2^{2k - 5} \over k - 1}\,\Im\int_{\infty}^{0}
{\expo{-y} - 1 + y - y^{2}/2 \over
y^{k - 1}\expo{\ic\pars{k - 1}\pi/2}}\,\ic\,\dd y\label{2}\tag{2}
\\[5mm] & =
-\,{2^{k - 2} - 2^{2k - 5} \over k - 1}\,
\Im\bracks{\expo{-\ic k\pi/2}\int_{0}^{\infty}y^{1 - k}\expo{-y}\,\dd y}
\\[5mm] & =
{2^{k - 2} - 2^{2k - 5} \over k - 1}\,\sin\pars{k\pi \over 2}
\Gamma\pars{2 - k} =
\pars{2^{2k - 5} - 2^{k - 2}}\,\sin\pars{k\pi \over 2}\Gamma\pars{1 - k}
\\[5mm] & =
\pars{2^{2k - 5} - 2^{k - 2}}\,\sin\pars{k\pi \over 2}
{\pi \over \Gamma\pars{k}\sin\pars{\pi k}} =
{2^{2k - 6} - 2^{k - 3} \over \Gamma\pars{k}}\,\sec\pars{k\pi \over 2}
\\[5mm] & =
\bbx{{2^{k}\pars{2^{k} - 8} \over 64\,\Gamma\pars{k}}\,\sec\pars{k\pi \over 2}}
\end{align}

In \eqref{1} and \eqref{2}, I'll 'closed' a contour which is a quarter circle in the first quadrant. The integrals along the arc vanishes out as the arc radius $\ds{\to \infty}$.

