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 $$ \int_0^\infty x^{s-1}\operatorname{Si}(x)\;dx = -\frac{\Gamma(s)\sin(\frac{\pi s}{2})}{s} $$