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

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?

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

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 $k=3$.

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 $1\pm\cos2t)$.

• Nicely done Lucian. +1. You gave me a chance to look back on this gem of a question! – Brevan Ellefsen May 6 '18 at 7:16


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 $\ds{\to \infty}$.