Application of Ramanujan's master theorem

Mellin transform of the Bessel function

We show the following result $$\int_{0}^{\infty} \left(\frac{x}{2}\right)^{s} J_{s}(x)dx = \Gamma(s)$$ Proof
We will use the Ramanujan's master theorem to prove this result: Assume that the function $f$ has an expansion of the form $$f(x) = \sum_{k=0}^{\infty} \left(\frac{\phi(k)}{k!}\right)(-x)^{k}$$ for some analytic function $\phi(k)$, then the Mellin transform of $f(x)$ is given by $$\{\mathscr{M}f(x)\} = \int_{0}^{\infty } x^{s-1} f(x) dx = \Gamma(s)\phi(-s)$$ Back to our problem, recall the expansion series of the Bessel function: $$J_{s}(x) = \left(\frac{x}{2}\right)^{s} \sum_{j=0}^{\infty} \frac{(-x^2/4)^j}{j!\Gamma(1+j+s)}$$ Then $$\int_{0}^{\infty} \left(\frac{x}{2}\right)^{s-1} J_{s}(x)dx=\int_{0}^{\infty} \left(\frac{x}{2}\right)^{2s-1} \sum_{j=0}^{\infty} \frac{(-x^2/4)^j}{j!\Gamma(1+j+s)}dx$$ let $\displaystyle w=\frac{x^2}{4}$ then $\displaystyle dw= \frac{x}{2}dx$, therefore $$\int_{0}^{\infty} \left(\frac{x}{2}\right)^{2s-1} \sum_{j=0}^{\infty} \frac{(-x^2/4)^j}{j!\Gamma(1+j+s)}dx = \int_{0}^{\infty} w^{s-1} \sum_{j=0}^{\infty} \frac{(-w)^j}{j!\Gamma(1+j+s)}dw$$ Clearly $\displaystyle f(w)=\sum_{j=0}^{\infty} \frac{(-w)^j}{j!\Gamma(1+j+s)}$ and $\displaystyle \phi(j) =\frac{1}{\Gamma(1+j+s)}$ By Ramanujan's master theorem : $$\int_{0}^{\infty} w^{s-1} \sum_{j=0}^{\infty} \frac{(-w)^j}{j!\Gamma(1+j+s)}dw = \Gamma(s)\phi(-s) =\frac{\Gamma(s)}{\Gamma(1)} =\Gamma(s)$$ Then $$\boxed{ \int_{0}^{\infty} \left(\frac{x}{2}\right)^{s} J_{s}(x)dx = \Gamma(s)}$$