Question

Integrate:
\[\int_0^\infty \frac{(\sin x)^{2k+1}}{x}\,dx\]
for non-negative integer \(k\).

Answer 1

Attempts so far:

(1) Differentiating under the integral sign, with parameterization
\[I_a=\int_0^\infty \frac{(\sin(ax))^{2k+1}}{x}\,dx\]
which gives
\[\frac{\mathrm d}{\mathrm{d} a}I_a=(2k+1)\int_0^\infty (\sin(ax))^{2k}\cos (ax)\,dx\]
but then subbing \(u=\sin(ax)\) makes that upper limit a bit tricky...

(2) Laplace transform (a work in progress): Same parameterization as before,
\[I_a=\int_0^\infty \frac{(\sin(ax))^{2k+1}}{x}\,dx\]
then taking the transform,
\[\begin{align*}{\large\mathcal{L}\left\{I_a\right\}_s}&=\int_0^\infty\int_0^\infty \frac{(\sin(ax))^{2k+1}}{x}e^{-as}\,da\,dx\\\\
&=\int_0^\infty \frac{1}{x}\mathcal{L}\left\{(\sin(ax))^{2k+1}\right\}\,dx\end{align*}\]
Computing the transform is ... exhausting, but presumably can be done with extensive application of power reduction, a task I'm not looking forward to, needless to say.

That being said, I'm open to simpler approaches!

Answer 2

For \(k\in\{1,2,\ldots,5\}\), we get
\[\begin{array}{c|c}
k&\text{value of integral}\\
\hline
1&\frac{\pi}{4}\\
2&\frac{3\pi}{16}\\
3&\frac{5\pi}{32}\\
4&\frac{35\pi}{256}\\
5&\frac{63\pi}{512}
\end{array}\]
(Evaluation courtesy of Mathematica.) I found the somewhat-periodic pattern in the denominator pretty interesting, so I'm curious to determine a closed form.

Answer 3

We might be able to make use of the fact that
\[\sin(ax)=\Im\left\{e^{iax}\right\}\]

Answer 4

Oh and that table should have started with \(k=0\), so the next column would have been \(\dfrac{\pi}{2}\).

Answer 5

The transforms for \(k=0,\ldots,5\) are a bit unwieldy I'm afraid, so I think I'll stop attempting that approach...

Answer 6

I'm trying something like this\[\int_0^\infty \frac{(\sin x)^{2k+1}}{x}\,dx = \int_0^{\infty} \left( \int_0^{\infty} e^{-ax} (\sin x)^{2k+1}da\right)dx\]

Answer 7

hopefully we can change the order of integration as the bounds are just numbers

\[= \int_0^{\infty} \left( \int_0^{\infty} e^{-ax} (\sin x)^{2k+1}dx\right)da\]

Answer 8

I see thats the same thing as you had ^

Answer 9

Alright hold on tight, cause where I'm about to show you is not a pretty place: \[\large \int\limits_0^\infty x^{-1}\sin^{2k+1}x dx \\ \large x^{-1}=u, \ \ -x^{-2}dx = du \\ \large \sin^{2k+1}xdx = dv, \ \ v = ...\] Ok v is going to get scary, so stay with me. \[\Large \sin^{2k+1}x =\sin^{2k}x * \sin x =(1-\cos^2x)^k \sin x \\ \Large dv= \sin x \sum_{n=0}^k \left(\begin{matrix}k \\ n\end{matrix}\right)(-1)^k \cos^{2k}x dx \\ r= \cos x, -dr = \sin x dx \\ \Large \sum_{n=0}^k \left(\begin{matrix}k \\ n\end{matrix}\right)(-1)^{k+1} r^{2k}dr \\\Large \sum_{n=0}^k \left(\begin{matrix}k \\ n\end{matrix}\right)(-1)^{k+1} r^{2k+1} \\ \Large v= \sum_{n=0}^k \left(\begin{matrix}k \\ n\end{matrix}\right)(-1)^{k+1} \cos^{2k+1}x \] Now we've finished a step of IBP (in the wrong direction? Not quite. =P)

\[\left[ \sum_{n=0}^k \left(\begin{matrix}k \\ n\end{matrix}\right)(-1)^{k+1}\frac{ \cos^{2k+1}x}{x} \right]_0^\infty + \sum_{n=0}^k \left(\begin{matrix}k \\ n\end{matrix}\right)(-1)^{k+1} \int\limits_0^\infty \frac{\cos^{2k+1}x}{x^2}dx\] Now before we evaluate this problematic limit on the left, let's talk about what it can be. At infinity all of these terms are zero, so no problem. But evaluating it at 0 will cause us problems. However, I think if we consider this sum with 0 plugged in to the cosine part and not to the denominator, just to make the argument reasonable, notice that cosine to all those weird powers is just 1. So really we might end up with a bunch of terms that are nearly identical, and that -1 term there is alternating, so maybe with some luck the terms will cancel each other out! This seems fairly obvious for when k is even, since the binomial theorem is symmetrical, but what about when it's odd? Still, it cancels out it appears. 1-2+1=0, 1-4+6-4+1=0, etc...? So we can probably drop this term off and consider the integral with the sum equal to our old integral we're trying to solve. Why is this useful? Well, there are two options here I want to talk about, but I have to go to dinner, so I'll say them when I return. =)

Answer 10

A pattern emerges for the numerators: http://oeis.org/search?q=1%2C3%2C5%2C35%2C63&language=english&go=Search

while the denominators are not so clear... A lot of strange skipping going.

Answer 11

There seems to be some pattern to it all. Clearly the denominator has some form of \(2^t\), but they show up in this strange, seemingly non-repeating pattern.
\[\begin{array}{c|cccc}
t&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&\ldots\\
\hline
2^t&2&4&&16&32&&&2^8&2^9&&2^{11}&2^{12}&&&&\ldots
\end{array}\]
So we see that whenever we do take powers of 2, we take pairs that are partitioned by varying intervals.

I've been cross-examining the integral values and powers of 2 up to \(k=67\) and \(t=132\), and this general skipping pattern I'll try to model with a diagram.

Since the \(T\)'s are all the same, I'll just isolate the \(S\)'s arguments, denoting the sequence as \(S_n\):
\[\begin{align*}S_n&=\{1,2,1,3,1,2,1,\color{red}4,1,2,1,3,1,2,1,\color{red}5,\\
&\quad\quad1,2,1,4,1,2,1,\color{red}4,1,2,1,3,1,2,1,\color{red}6,\ldots\}\end{align*}\]
I don't think my laptop can take much more of these computations, unfortunately, so there's no telling if this pattern exists.