Mellin Transform What is the inverse Mellin transform of (s-1/2)^k on the vertical line Re(s)=a where 
0 < a <1 and k is a natural number? 
 A: Maple command $with(inttrans):invmellin((s-1/2)^k,s,z,0..1)$ (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
$$ -1/32\,{\it Dirac} \left( z-1 \right) -{\frac {121}{16}}\,z{\it Dirac}
 \left( 1,z-1 \right) -{\frac {165}{4}}\,{\it Dirac} \left( 2,z-1
 \right) {z}^{2}-$$   $${\frac {85}{2}}\,{\it Dirac} \left( 3,z-1 \right) {z}
^{3}-{\frac {25}{2}}\,{\it Dirac} \left( 4,z-1 \right) {z}^{4}-{\it
Dirac} \left( 5,z-1 \right) {z}^{5} $$ in the case $k=5.$
A: In fact I was aiming at evaluating the expression:
$I_k=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty} (s-\frac{1}{2})^k x^{s-1} 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:
$\delta(u)=\frac{e^{-cu}}{2\pi i}\int_{c-i\infty}^{c+i\infty} e^{zu}dz , Re(z)=c$ 
which would imply that:
${e^{cu}}\delta(u)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} e^{zu}dz , Re(z)=c$ 
this in turn yields:
$\frac{d^k}{d u^k}{e^{cu}}\delta(u)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} z^k e^{zu}dz , Re(z)=c, k\in N$
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:
$\delta(ln(x))=\delta(f(x))=\frac{\delta(x-1)}{|f'(1)|}=\delta(x-1)$
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:
$\sqrt x\ I_k=[x\frac{d}{dx}]^k(x^{c}\delta(x-1))\leftrightarrow I_k=\frac{1}{\sqrt x}[x\frac{d}{dx}]^k(x^{c}\delta(x-1))$
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. 
