4

I was working on integrals of the form

$$\int_{0}^{\infty}e^{-x\cdot t}\sin^n(x) dx$$

and to solve them I tried to express $\sin^n(x) $ in form of a sum without any powers. Interesting for me I have found a way which only depends whether the power is even or odd.

For the even ones I have got

$$\frac{1}{2^{n-1}} \left[\frac{(-1)^{\frac{n}{2}}}{2}\binom{n}{\frac{n}{2}}~+~\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor-1}\binom{n}{k}(-1)^{\frac{n-2k}{2}}\cos((n-2k)x)\right] $$

and for the odd ones $$\frac{1}{2^{n-1}}\left[\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{k}(-1)^{\frac{n-2k-1}{2}}\sin((n-2k)x)\right]$$

I have got three questions concerning these series:

  1. Are they right like this, even with this kind of weird power for the minus sign and the seperated first term for the even powers?
  2. If it is possible, how could you simplify these sums?
  3. How to proof the rightness of these sums or how to show, that they are wrong?
mrtaurho
  • 16,103
  • 4
    Use $\sin x=\frac1{2i}(e^{ix}-e^{-ix})$. – Angina Seng Jul 19 '18 at 09:34
  • Okay, I could use the complex expression of the trigonometric functions to achieve what exactly? I am not sure how to use this to solve my problems. Could you explain your idea a little bit further? – mrtaurho Jul 19 '18 at 09:58
  • Can you compute the $n$-th power of $e^{ix}-e^{-ix}$? – Angina Seng Jul 19 '18 at 10:17
  • Ah, now I understand what you meant. Together with the following answer I have to say this is a way simpliet than the one I tried. Thank you. – mrtaurho Jul 19 '18 at 11:00

2 Answers2

4

With $z=e^{ix}$,

$$\sin^nx=\left(\frac{z-z^{-1}}{2i}\right)^n=\frac1{(2i)^n}\sum_{k=0}^n\binom nkz^kz^{-(n-k)}=\frac1{(2i)^n}\sum_{k=0}^n\binom nkz^{2k-n}.$$

For even $n=2m$, the exponent runs from $-n$ to $n$ by step $2$ via $0$ and

$$\frac{(-1)^m}{2^n}\sum_{k=0}^n\binom nkz^{2k-n}=\frac{(-1)^m}{2^n}\left(\sum_{k=0}^m\binom nk\left(z^{2k-n}+z^{n-2k}\right)\right) \\=\frac{(-1)^m}{2^n}\left(2^*\sum_{k=0}^m\binom nk\cos(2k-n)x\right).$$

We used the shorthand notation $2^*$ to express that the coefficient is $2$, unless $k=m$, which leads to a constant term $\displaystyle\binom nm$.

The development is similar for odd $m$, and all terms are paired.

It may be simpler to grasp with particular examples:

$$(2i)^4\sin^4x=\left(z-z^{-1}\right)^4=z^4-4z^2+6-4z^{-2}+z^{-4}\\=2\cos4x-2\cdot4\cos 2x+6$$

$$(2i)^5\sin^5x=\left(z-z^{-1}\right)^5=z^5-5z^3+10z-10z^{-1}+5z^{-3}-z^{-5}\\=2i\sin5x-2i\cdot5\sin3x+2i\cdot10\sin x.$$

So to a constant factor, the power of a sine is a linear combination of cosines or sines of the argument times every other integer, weighted by every other binomial coefficient and with alternating signs.

  • 1
    I have to say, that I did not thought about going to the complex expression as I already mentioned. But this way you suggested is fairly simple and it is easier to use this sum for the original integral problem. Thank you for the calculation. – mrtaurho Jul 19 '18 at 12:54
  • 1
    @mrtaurho: yep, the complex expression trick is often handy. BTW, you can evaluate the definite integral by parts (on $e^{-xt}$), which will give you a simple recurrence relation. –  Jul 19 '18 at 12:56
  • My main target is to derivate the formula for the integral of the $n$-th power of the Sinc-Function by myself. While working on this issue I was confronted with the given integrals of the form $\int_0^\infty~e^{x\cdot t}\sin(x)^n~dx$. Since the formula I want to derive contains a sum similiar to the presented form, concernig the binomial coefficient and the minus sign, I needed to go this way. But I guess with this easier sum I can go further. – mrtaurho Jul 19 '18 at 13:00
  • I guess I just spotted a little mistake in your sum. I think the $(-1)^m$ should be in the sum as $(-1)^k$ since your formula gives for $n=4$ : $\frac{1}{8}\left(\cos(4x)~+~4\cos(2x)~+6\right)$ and not the wanted one with a minus sign infront of the $\cos(2x)$ term like you showed in your own example below. So I am not sure what to think now ^^' – mrtaurho Jul 19 '18 at 13:22
  • 1
    @mrtaurho: you are right, I didn't account for the minus sign in the definition of the sine, so a factor $(-1)^{n-k}$ is missing. One can pull $(-1)^n$ out and keep $(-1)^k$ inside. –  Jul 19 '18 at 13:38
  • @mrtaurho I am thinking of discarding the first part and leaving the examples, they are more readable and as informative. –  Jul 19 '18 at 13:40
3

Just to put it all together in one post:

The right sums should be the following

For even $n$ $$\sin^{2m}(x)~=~\frac{(-1)^m}{2^n}\left[2^*\sum_{k=0}^m~\binom{n}{k}(-1)^k\cos((n-2k)x)\right]$$ where $n=2m$.

For odd $n$ $$\sin^{2m+1}(x)~=~\frac{(-1)^m}{2^{n-1}}\left[\sum_{k=0}^m~\binom{n}{k}(-1)^k\sin((n-2k)x)\right]$$ where $n=2m+1$

Am I right this time, or did I made a mistake somewhere? Since there is still this weird constant term in the sum for even powers, which kind of annoys me.

mrtaurho
  • 16,103