Evaluation of $ \int_{0}^{\infty}\frac{\sin (kx)\cdot (\cos x)^k}{x}dx\;,$ where $k\in \mathbb{Z^{+}}$ 
Evaluation of $\displaystyle \int_{0}^{\infty}\frac{\sin (kx)\cdot (\cos x)^k}{x}dx\;,$ where $k\in \mathbb{Z^{+}}$

$\bf{My\; Try::}$ Let $$I(k) = \int_{0}^{\infty}\frac{\sin (kx)\cdot (\cos x)^k}{x}dx$$
Then $$I'(k) = \int_{0}^{\infty}\frac{d}{dk}\left[\sin (kx)\cdot (\cos x)^k\right]\frac{1}{x}dx$$
So $$I'(k) = \int_{0}^{\infty}\left[\sin (kx)\cdot (\cos x)^{k}\cdot \ln(\cos x)+(\cos x)^k\cdot \cos (kx)\cdot x\right]\frac{1}{x}dx$$
Now How can I calculate after that, Help me
Thanks
 A: \begin{align*}\cos^k{x}\sin{(kx)}&=\left(\dfrac{e^{ix}+e^{-ix}}{2}\right)^k\cdot\dfrac{e^{ikx}-e^{-ikx}}{2i}=\dfrac{1}{2^{k+1}}e^{-ikx}(e^{2ix}+1)^k\cdot\dfrac{e^{ikx}-e^{-ikx}}{i}\\
&=\dfrac{1}{2^{k+1}}\sum_{l=0}^{k}\binom{k}{l}(e^{2ilx})\cdot\dfrac{1-e^{-2ikx}}{i}\\
&=\dfrac{1}{2^{k+1}}\sum_{l=0}^{k}\binom{k}{l}\dfrac{e^{2ilx}-e^{2i(l-k)}}{i}\\
&=\dfrac{\displaystyle\sum_{i=1}^{k}\binom{k}{i}\sin{(2ix)}}{2^k}\\
\end{align*}
and $$\int_{0}^{+\infty}\dfrac{\sin{mx}}{x}dx=\dfrac{\pi}{2}$$
so
$$\int_{0}^{\infty}\dfrac{\sin{(kx)}\cos^k{x}}{x}dx=\dfrac{\pi}{2}-\dfrac{\pi}{2^{k+1}}$$
A: The following is an approach using contour integration.
First notice that $$ \begin{align}\int_{0}^{\infty} \frac{\sin(kx) \cos^{k}(x)}{x} \, dx &= \frac{1}{2^{k+1}} \, \text{Im} \, \text{PV} \int_{-\infty}^{\infty} \frac{e^{ikx}(e^{ix}+e^{-ix})^{k}}{x} \, dx \\ &= \frac{1}{2^{k+1}} \, \text{Im} \,  \text{PV} \int_{-\infty}^{\infty} \frac{(1+e^{2ix})^{k}}{x} \, dx. \end{align}$$
So consider the complex function $$f(z) = \frac{(1+e^{2iz})^{k}}{z}. $$
Integrating $f(z)$ around an infinitely-large semicircle in the upper half-plane indented at the origin, we get
$$\text{PV} \int_{-\infty}^{\infty} \frac{(1+e^{2ix})^{k}}{x}  \, dx + \lim_{R \to \infty}  \int_{C_{R}} \frac{(1+e^{2iz})^{k}}{z}\, dz = i \pi  \, \text{Res} [f(z), 0] = i \pi  \, 2^{k},$$ where $C_{R}$ is the upper half of the circle $|z|=R$.
But using the binomial theorem and Jordan's lemma, $$\begin{align} \lim_{R \to \infty}  \int_{C_{R}} \frac{(1+e^{2iz})^{k}}{z}\, dz &= \lim_{R \to \infty} \int_{C_{R}} \frac{dz}{z} + \lim_{R \to \infty} \sum_{j=1}^{k} \binom{k}{j} \int_{C_{R}} \frac{e^{2ijz}}{z} \, dz \\  &= i \pi + 0 = i \pi. \end{align}$$
Therefore, $$\text{PV} \int_{-\infty}^{\infty} \frac{(1+e^{2ix})^{k}}{x} \, dx =  i \pi \left(2^{k}-1 \right),$$ which means $$\int_{0}^{\infty} \frac{\sin(kx) \cos^{k}(x)}{x} \, dx = \frac{1}{2^{k+1}} \pi  \left(2^{k}-1 \right) = \frac{\pi}{2} \left(1- \frac{1}{2^{k}} \right).  $$
