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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?

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  • $\begingroup$ could you rewrite the integral as $\int_0^\infty \frac{\sin^4(u)}{u^{k}}$ where $k \in(1,3)$? $\endgroup$ – Brevan Ellefsen Nov 3 '15 at 23:57
  • $\begingroup$ Sure, I made the change $\endgroup$ – measure Nov 4 '15 at 0:05
  • $\begingroup$ I apologize, I wasn't clear. I was asking if you've tried that $\endgroup$ – Brevan Ellefsen Nov 4 '15 at 0:05
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    $\begingroup$ no du? anyway, IBP? $\endgroup$ – BCLC Nov 4 '15 at 0:05
  • $\begingroup$ First we have to show convergence for those bounds, unless that is assumed $\endgroup$ – Brevan Ellefsen Nov 4 '15 at 0:06
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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 $k=3$.

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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 $1\pm\cos2t)$.

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  • $\begingroup$ Nicely done Lucian. +1. You gave me a chance to look back on this gem of a question! $\endgroup$ – Brevan Ellefsen May 6 '18 at 7:16
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}{\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 $\ds{\to \infty}$.

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