What is the inverse Mellin transform of (s1/2)^k on the vertical line Re(s)=a where 0 < a <1 and k is a natural number?

Maple command $with(inttrans):invmellin((s1/2)^k,s,z,0..1)$ (see http://www.maplesoft.com/support/help/Maple/view.aspx?path=inttrans/invmellin for info) calculates it for concrete values of $k.$ For example, Maple produces $$ 1/32\,{\it Dirac} \left( z1 \right) {\frac {121}{16}}\,z{\it Dirac} \left( 1,z1 \right) {\frac {165}{4}}\,{\it Dirac} \left( 2,z1 \right) {z}^{2}$$ $${\frac {85}{2}}\,{\it Dirac} \left( 3,z1 \right) {z} ^{3}{\frac {25}{2}}\,{\it Dirac} \left( 4,z1 \right) {z}^{4}{\it Dirac} \left( 5,z1 \right) {z}^{5} $$ in the case $k=5.$ 


In fact I was aiming at evaluating the expression: $I_k=\frac{1}{2\pi i}\int_{ai\infty}^{a+i\infty} (s\frac{1}{2})^k x^{s1} ds, 0 < a < 1, k\in N $ changing the variable $s\frac{1}{2}=z$ one can rewrite the integral $I_k$ as: $I_k=\frac{1}{2\pi i}\int_{a\frac{1}{2}i\infty}^{a\frac{1}{2}+i\infty} z^k x^{z\frac{1}{2}} dz$ Recall that the Dirac delta function is defined in the integral form as: $\delta(u)=\frac{1}{2\pi}\int_{\infty}^{+\infty} e^{ipu} dp$ One could rewrite it as: $\delta(u)=\frac{e^{cu}}{2\pi i}\int_{ci\infty}^{c+i\infty} e^{zu}dz , Re(z)=c$ which would imply that: ${e^{cu}}\delta(u)=\frac{1}{2\pi i}\int_{ci\infty}^{c+i\infty} e^{zu}dz , Re(z)=c$ this in turn yields: $\frac{d^k}{d u^k}{e^{cu}}\delta(u)=\frac{1}{2\pi i}\int_{ci\infty}^{c+i\infty} z^k e^{zu}dz , Re(z)=c, k\in N$ Now we could express the integral $I_k$ as a Fourier transform by making the change of variable $x=e^u$, hence: $I_k=\frac{1}{2\pi i}\int_{a\frac{1}{2}i\infty}^{a\frac{1}{2}+i\infty} z^k e^{u(z\frac{1}{2})} dz$ which is the same as: $e^{u/2}I_k=\frac{1}{2\pi i}\int_{a\frac{1}{2}i\infty}^{a\frac{1}{2}+i\infty} z^k e^{zu} dz$ set $c=a\frac{1}{2}$ then: $e^{u/2}I_k=\frac{d^k}{d u^k}{e^{cu}}\delta(u)$ Going to our old variable $x$ the above expression becomes: $\sqrt x\ I_k=\frac{d^k}{d (ln(x))^k}{x^{c}}\delta(ln(x))$ Since $ln(1)=0$ then $f(x)=ln(x)$ has a root at $x=1$. This in turn would imply: $\delta(ln(x))=\delta(f(x))=\frac{\delta(x1)}{f'(1)}=\delta(x1)$ Also $\frac{d}{d ln(x)}=x\frac{d}{dx}\rightarrow \frac{d^k}{d (ln(x))^k}=[x\frac{d}{dx}]^k$ Thus finally we get that: $\sqrt x\ I_k=[x\frac{d}{dx}]^k(x^{c}\delta(x1))\leftrightarrow I_k=\frac{1}{\sqrt x}[x\frac{d}{dx}]^k(x^{c}\delta(x1))$ I hope this is a correct solution as it is an important step in a work that I have to complete. Any suggestions if this happens to be wrong would be more than appreciated. 

