nth power of sine as sum of sine and cosine terms

Question

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?

Answer 1

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.

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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.

Answer 2

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.