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? 



Mellin Transform
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. 


