Proof of sine integral How to prove:

$$\operatorname{Si}(x) =\sum_{k=0}^\infty { \frac { -\sin\left( \frac{\pi k}{2} \right) }{ k\times k! } {(-x)}^k} $$

Here $\operatorname{Si}(x)$ is Sine Integral. 
Actually I wanted to prove the mellin transform of $\operatorname{Si}(x)$. But I couldn't. I used power series for $\sin(a)$ but couldn't proceed from there. Please help. 

We can write
$$
\operatorname{Si}(x) =\sum_{k=0}^\infty { \frac { -\sin\left( \frac{\pi k}{2} \right) }{ k\times k! } {(-x)}^k} = \sum_{k=0}^\infty { \frac {\phi(k) }{ k! } {(-x)}^k}
$$
then according to the Ramanujan Master theorem the Mellin transform of $\operatorname{Si}(x)$ is given by
$$
\mathcal{M}[\operatorname{Si}(x)] = \Gamma(s)\phi(-s)
$$
so taking care of the signs from the definition of $\phi(k)$ above directly gives
$$
\int_0^\infty x^{s-1}\operatorname{Si}(x)\;dx = -\frac{\Gamma(s)\sin(\frac{\pi s}{2})}{s}
$$
