1

$\int_{0}^{+\infty}\left(\frac{sin(x)}{x}\right)^ndx=n\int_{0}^{+\infty}\frac{x^{n-2}}{(x^2+2^2)(x^2+4^2)...(x^2+n^2)}$ if n is even

and

$\int_{0}^{+\infty}\left(\frac{sin(x)}{x}\right)^ndx=n\int_{0}^{+\infty}\frac{x^{n-1}}{(x^2+1^2)(x^2+3^2)...(x^2+n^2)}$ if n is odd

I found these equalities in the Integral and Series website.

I already know there exists a not-so-pretty closed form as a sum for the integral above, I am just interested in these equalities.

The right integrands being always positive, they are most likely not equal to the integrands on the left

Replacing $sin(x)$ by its expansion $sin(x)=\sum_{k=0}^{k=+\infty}\frac{(-1)^kx^{2k+1}}{(2k+1)!}$, dividing by $x$ and expanding is somewhat confusing, and using the residue theorem to compute the right integral in the hope of finding it is equal to the final form expected (Closed form for integral of integer powers of Sinc function) might work (I found something similar, but didn't simplify it enough yet), but it is far from ideal as a proof as it implies I know the end result before actually solving the problem... which is a bit cheap.

Is there an easier way to derive this equality without actually solving both integrals and comparing?

@user1952009 Thanks, I tried to apply it for n=2, but I don't get much out of it

$I=1_{[-\frac{1}{2},+\frac{1}{2}]}*1_{[-\frac{1}{2},+\frac{1}{2}]}=\int_{-\infty}^{+\infty}1_{[-\frac{1}{2},+\frac{1}{2}]}(t)1_{[-\frac{1}{2},+\frac{1}{2}]}(x-t)dt$

$x-t \in [-\frac{1}{2},+\frac{1}{2}] \leftrightarrow t-x \in [-\frac{1}{2},+\frac{1}{2}] \leftrightarrow t \in [x-\frac{1}{2},+\frac{1}{2}+x]$

So

$I=Length([-\frac{1}{2},+\frac{1}{2}]\cap[x-\frac{1}{2},+\frac{1}{2}+x])=0$ if $|x|>1, =1-x$ if $1\geq x \geq 0,=1+x$ if $-1\leq x \leq 0$

So evalutating at $0$ (inverse transform of fourier transform of $(sinx/x)^2$ at $0$...), I get $\int_{-\infty}^{+\infty}\left(\frac{sin(x)}{x}\right)^2dx=1$

What am I doing wrong please ? ... A factor pi is missing, and I'm not seeing anything close to the formula I'm looking for

Community
  • 1
Evariste
  • 2,511
  • It's easy to make a case that the integrals on the left are manifestly positive: since $\sin x\geq 0$ for $0\leq x\leq \pi$ and $\sin x\leq 0$ for $\pi\leq x\leq 2\pi$ and $\sin(x+\pi)=-\sin x$, then $\left|\frac{\sin(x+\pi)}{x+\pi}\right|\lt \left|\frac{\sin x}{x}\right|$ and so the 'negative' piece of each $2\pi$ period is smaller than the positive piece just before it - i.e., break the integral into integrals from $0$ to $\pi$, $\pi$ to $2\pi$, etc. and make the case that each integral is smaller in absolute value than the one before and they're alternating in sign with the first positive. – Steven Stadnicki Apr 24 '16 at 21:23
  • @StevenStadnicki Thanks for your input, I know the integrals are positive. I was talking about the integrands (meaning it is probably not worth bothering to compare that what's inside the integral on the left with what's inside the integral on the right). It could have been otherwise (i.e. both integrands are equal over R+), in which case the equality would have been solved. – Evariste Apr 24 '16 at 21:39
  • I would solve it by using the fact that $\frac{\sin x}{x}$ is the Fourier transform of $1_{[-1/2,1/2]}(\xi)$ hence $ \left(\frac{\sin x}{x}\right)^n$ is the Fourier transform of $\underbrace{1_{[-1/2,1/2]} \ast \ldots \ast 1_{[-1/2,1/2]}}{n}(\xi)$ and $\int{-\infty}^\infty \left(\frac{\sin(x)}{x}\right)^n dx = \underbrace{1_{[-1/2,1/2]} \ast \ldots \ast 1_{[-1/2,1/2]}}_{n}(0)$, by induction on $n$ you should get your formula – reuns Apr 24 '16 at 23:12
  • @user1952009 Thanks, I tried it, but I got nothing out of it. I edited my post because it is too long for a comment. – Evariste Apr 25 '16 at 16:14

0 Answers0