Laplace transforms of powers of cosine (solved!)

Question

During the past several hours, while studying the Laplace transform, I've started wondering what

\begin{equation} \mathcal{L} \{ \cos^n(at)\}(s) \end{equation}

would look like – since it won't appear on your typical transform table. After some thinking and tedious pattern-finding and rearranging (and typing on Wolfram Alpha), I've come up with the following (interesting, yet messy) two formulae:

(1) For odd $n$'s ($n=2k+1, k \in \mathbb{N}$),

\begin{equation} \mathcal{L}\{\cos^{2k+1} (at)\}(s) = \frac{1}{2^{2k}}\sum_{j=0}^{k} \frac{\binom{2k+1}{k-j}s}{ (2j+1)^2 a^2 + s^2} \tag{1} \end{equation}

(2) For even $n$'s ($n=2k, k \in \mathbb{N}$),

\begin{equation} \mathcal{L}\{\cos^{2k} (at)\}(s) = \frac{\binom{2k}{k}}{2^{2k}s} + \frac{1}{2^{2k-1}}\sum_{j=1}^{k} \frac{\binom{2k}{k-j}s}{(2j)^2 a^2 + s^2} \tag{2} \end{equation}

So,

1. First of all, are these statements even correct?
2. If they are, how can I prove it using the definition of the transform? I already have a few ideas in my mind, which involve the exponential definition of cosine, "complexifying" the Laplace integral (but isn't $s$ already complex?), applying the inverse Laplace transform on each of the series terms, or heck, maybe even the Maclaurin expansion of cosine or some trig identities... it'd be lovely to read some suggestion from you all!
3. I'm also looking at merging the two formulae into a single one that works for every $n$, but the major obstacle seems to be the $ \frac{\binom{2k}{k}}{2^{2k}s} $ term in the even-$n$ formula: where does its weird binomial coefficient come from (1/2 of what the sum's general binomial coefficient would predict)?
4. A similar approach has been tried for $ \sin^n (at) $, with similar results — messy sums which only work for one or the other parity of $n$. How do the two relate to each other, and possibly to the (albeit trivial) transform of $\exp^n{(at)}=\exp{n(at)}$? And how about $ \sinh^n (at) $ and $ \cosh^n (at) $?

Thanks in advance, hope this has provided some interesting input

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EDIT. Following Alex R.'s suggestion, I've tried using

\begin{equation} \cos^n (at) = \frac{\left(e^{iat} + e^{-iat}\right)^n}{2^n} \end{equation}

The Laplace transform then becomes:

\begin{equation} \mathcal{L}\{\cos^{n} (at)\}(s) = \int_{0}^{\infty}2^{-n}\left(e^{iat} + e^{-iat}\right)^n e^{-st} dt \\ = 2^{-n} \int_{0}^{\infty} \sum_{j=0}^{n} \binom{n}{j} e^{(iat)j} e^{(-iat)(n-j)} e^{-st} dt \\ = 2^{-n} \sum_{j=0}^{n} \binom{n}{j} \int_{0}^{\infty} e^{-\big(s+ia(2j-n)\big)t} dt \\ = 2^{-n} \sum_{j=0}^{n} \binom{n}{j} \frac{1}{-\big(s+ia(2j-n)\big)} \left[\lim_{\beta \to \infty} e^{-\big(s+ia(2j-n)\big)\beta} - e^{-\big(s+ia(2j-n)\big)(0)}\right] \end{equation} \begin{equation} = 2^{-n} \sum_{j=0}^{n} \binom{n}{j} \frac{[-1]}{-\big(s+ia(2j-n)\big)} = 2^{-n} \sum_{j=0}^{n} \binom{n}{j} \frac{1}{s+ia(2j-n)} \tag{3} \end{equation}

I'm not very satisfied with what I've got — not even sure it's right. And besides that, I can't figure out how it connects with the two formulae given above. Please let me know what you think

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EDIT 2. We could actually check the result in $(3)$ by the use of the Fourier-Mellin integral below:

\begin{equation} f(t)=\mathcal{L}^{-1}\{\hat{f}(s)\} = \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty} e^{st} \hat{f}(s) ds \end{equation} where $\gamma \in \mathbb R$ is greater than the real part of all singularities of $\hat{f}(s)$. Applying this formula to the sum in $(3)$, we get

\begin{equation} \begin{split} f(t) &= \frac{2^{-n}}{2\pi i} \int_{\gamma-i\infty}^{\gamma+i\infty} e^{st} \sum_{j=0}^{n} \binom{n}{j} \frac{1}{s+ia(n-2j)} ds \\ &= \frac{2^{-n}}{2\pi i} \sum_{j=0}^{n} \binom{n}{j} \int_{\gamma-i\infty}^{\gamma+i\infty} \frac{e^{st}}{s+ia(n-2j)}ds \end{split} \end{equation}

All the poles of our sum-defined $\hat{f}(s)$ lie on the imaginary axis, so we can choose any $\gamma > 0 $.

Let us now consider the integral alone. We may now close the contour with a semicircle lying on the left side of the plane and call it $\Gamma$.

\begin{equation} \newcommand{\Res}{\mathop{\text{Res}}} \oint_\Gamma \frac{e^{st}}{s+ia(n-2j)}ds = 2\pi i \lim_{s\to -ia(n-2j)} e^{st} = 2\pi i e^{-ia(n-2j)t} \tag{4} \end{equation} which can be demonstrated to be equal to the integral over the line $\gamma+iq, \quad -\infty < q <\infty$.

We can plug $(4)$ into our formula for $f(t)$ \begin{equation} f(t) = \frac{2^{-n}}{2\pi i} \sum_{j=0}^{n} \binom{n}{j} (2\pi i e^{-ia(n-2j)t}) = 2^n \sum_{j=0}^{n} \binom{n}{j} e^{-(n-j)(iat)}e^{j(iat)} \end{equation} which of course we recognize as being the binomial expansion of \begin{equation} f(t) = \left(\frac{e^{iat} + e^{-iat}}{2}\right)^n = \cos^n(at) \end{equation}

I will soon try to prove $(1)$ and $(2)$ by the same means.

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FINAL EDIT. I just had to expand the result in $(3)$ to obtain

\begin{equation} \bbox[5px,border:2px solid red]{ \mathcal{L} \{ \cos^n(at)\}(s) = \frac{1}{2^n}\sum_{j=0}^{n} \binom{n}{j} \frac{s+(n-2j)a}{s^2+(n-2j)^2a^2}} \tag{5} \end{equation}

(a big duhhh on my behalf!) We can try out $(5)$ with even or odd $n$'s and it will work, connecting beautifully to the formulae in $(1)$ and $(2)$. The painful asymmetry in $(2)$ can now be explained: it was actually caused by the fact that, if $n=2k$ and $k$ is integer, then the $n$th row in Pascal's triangle contains an odd number of terms – of which $\binom{2k}{k}$ is the central one. Since $\binom{n}{q} = \binom{n}{n-q}$, every other term besides that middle one were appearing twice in the sum, and so they could be reduced to $2\binom{n}{q}$. The asymmetry in $(2)$ was just concealing the underlying symmetry expressed in $(5)$.

Finding an expression for the Laplace transform of the powers of sine was trickier, but possible:

\begin{equation} \mathcal{L} \{ \sin^n(at)\}(s) = \frac{1}{2^n}\sum_{j=0}^{n} \binom{n}{j} (-1)^{j+\lfloor\frac{n}{2}\rfloor} \frac{s+(n-2j)a}{s^2+(n-2j)^2a^2} \tag{6} \end{equation}

It works in a similar fashion for the hyperbolic functions: \begin{equation} \mathcal{L} \{ \cosh^n(at)\}(s) = \frac{1}{2^n}\sum_{j=0}^{n} \binom{n}{j} \frac{s+(n-2j)a}{s^2-(n-2j)^2a^2} \tag{7} \end{equation} \begin{equation} \mathcal{L} \{ \sinh^n(at)\}(s) = \frac{1}{2^n}\sum_{j=0}^{n} \binom{n}{j} (-1)^j \frac{s+(n-2j)a}{s^2-(n-2j)^2a^2} \tag{8} \end{equation}

Answer 1

Perhaps the dirtiest but most straightforward way would be to write $\cos^n(kx)=\left(\frac{e^{ikx}+e^{-ikx}}{2}\right)^n$, use the binomial theorem on this and then you'll have an easy sum of integrals of an exponential.