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?

$\begingroup$ I took the liberty of fixing your formatting $\endgroup$ – Yemon Choi Jun 15 '13 at 21:13
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 $k=5.$

2$\begingroup$ I might be inclined to suspect that the interactions of the powers of $z$ and the derivatives of Dirac delta interact in a way that simplifies the numbers in the outcome, maybe getting a constant multiple of the $k$th derivative of Dirac delta at $1$. $\endgroup$ – paul garrett Jun 16 '13 at 17:02
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.

$\begingroup$ Up to en.wikipedia.org/wiki/Mellin_transform , this integral is not the inverse Mellin trasform of $(s1/2)^k$, assuming $k$ to be a natural number. $\endgroup$ – Mark Jun 17 '13 at 4:52

$\begingroup$ It depends how you define at the first place the Mellin trasform. In my case I have define it as: $M(I_k , s)=\int_{0}^{\infty} I_k (v) v^{s} dv , 0 < Re(s) < 1$ then its inverse has I have expressed should not be contradictory. $\endgroup$ – Arian Jun 17 '13 at 11:54
