Ramanujan's master theorem states that if $f(x)$ has the expansion, $$f(x)=\sum_{k=0}^\infty\frac{\varphi(k)}{k!}(-x)^k,\quad x\in\mathbb{C}$$ then the Mellin transform of $f$ is given by, $$\int_0^{+\infty}x^{s-1}f(x)\ dx=\Gamma(s)\phi(-s).$$
This theorem is already very convenient, but I wonder if a slightly more general theorem can be established; namely when $f(x)$ has an expansion of the form, $$f(x)=\sum_{k=0}^\infty\frac{(-1)^k\varphi(k)}{k!}x^{ak+b},\quad a,b\in\mathbb{Z}.$$ For example, this is useful when evaluating, $$\int_0^{+\infty} \frac{\sin{(x)}-x\cos{(x)}}{x^\alpha}\ dx$$ as $f(x)=\sin(x)-x\cos(x)$ has the expansion, $$\sin(x)-x\cos(x)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}-x\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}x^{2n}=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\varphi(n)x^{2n+1}$$ where, $$\phi(n)=\frac{n!}{(2n+1)!}-\frac{n!}{(2n)!}=\frac{-2nn!}{(2n+1)!}$$ and the integral will follow if I can find the Mellin transform of $f(x)$.
Any ideas on how to derive such a theorem? Thank you in advance.
Assuming the conditions of Ramanujan's master theorem are met, the Mellin transform of $f$ is given by $$\int_0^\infty x^{s-1}f(x)dx=\frac1{|a|}\Gamma\left(\frac{s+b}a\right)\phi\left(-\frac{s+b}a\right)$$
To see this, note that $f(x)=x^bg(x^a)$ where $$g(x)=\sum_{k=0}^\infty\frac{\phi(k)}{k!}(-x)^k$$ so we may perform the substitution $u=x^a$, and then apply Ramanujan's master theorem to see that
$$\begin{split} \int_0^\infty x^{s-1}f(x)dx&=\int_0^\infty x^{b+s-1}g(x^a)dx\\ &=\int_0^\infty \frac1{|a|} u^{(s+b)/a-1}g(u)du\\ &=\frac1{|a|}\Gamma\left(\frac{s+b}a\right)\phi\left(-\frac{s+b}a\right) \end{split}$$ for values of $s$ where the integral converges.