I have a question about an Inverse Mellin Transform. The Mellin transform of $\sin(x)$ is $$ \Gamma(z)\sin((\pi z)/2) $$ with a strip of definition of $-1 < \Re z < 1 $
My question is about taking the inverse Mellin transform of this function along the line $\Re z= a$ if $a > 1/2$: $$ \int_{a-i\infty}^{a+i\infty} x^{-z} \Gamma(z)\sin((\pi z)/2)dz $$
How can I show that this integral converges if $\Re z > 1/2$? Along a line parallel to the imaginary axis, the absolute value of the Gamma function has the asymptotic expression, $$ |\Gamma(a+ib)|\sim \sqrt{2\pi} |b|^{a-(1/2)} e^{-|b|\pi/2} $$
So the modulus of the integrand as a whole is asymptotic to,
$$|\Gamma(a+ib)\sin((\pi (a+ib))/2)|\sim \sqrt{2\pi} |b|^{a-(1/2)}$$
along this line. Therefore, if $a>1/2$, the modulus of the integrand actually grows along this line, and in general, I believe this integral does not converge absolutely if $$ a\geq-1/2 $$
So I'd like to know how to show that the integral converges conditionally in this case and also how to justify using a Bromwich type contour so as to use residue theory to recover the power series for $\sin(x)$. Any help would be much appreciated.