On the other hand, what I saw here:
http://en.wikipedia.org/wiki/Mellin_inversion_theorem
suggests that the Mellin inverse transform requires only the right boundedness in vertical strips, not on a whole half-plane. So it should work...
... but then I tried a numerical test to compare the value of \( \vartheta(w) \) against the inverse Mellin transform of \( \frac{\Gamma(s)}{^{-s} e} \) using the Kneser tetrational function as obtained from sheldonison's excellent Kneser-method code (which satisfies the required boundedness criteria). The approximation to the inverse Mellin transform is done by numerically integrating
\( IM(x) \approx \frac{1}{2\pi i} \int_{c - iA}^{c + iA} x^{-s} \frac{\Gamma(s)}{^{-s} e} ds \)
with \( A \) some large real value (here, I chose 15) and \( c \) is any number in \( (0, 1) \) (here, I chose 0.5). The following graph shows \( IM(x) \) and \( \vartheta(x) \) on the positive reals from near-0 to 20:
As you can see, \( IM(x) \) decays toward 0, but \( \vartheta(x) \) does not. It seems that \( IM(x) \approx \vartheta(x) \) for small \( x \), but the approximation gets worse at larger values of \( x \).
This suggests that there is an error in your derivation of \( \vartheta(x) \) from the inverse Mellin transform, since you should be able to Mellin-transform it back, which means your \( \vartheta(x) \) should decay to 0 along the real axis, but it doesn't. I suspect a formal disproof could be had if you can show that \( \mathrm{lim\ sup}_{x \rightarrow \infty} \ \vartheta(x) x^{s-1} = +\infty \). Whatever your \( \vartheta(x) \) is, it seems most assuredly not to be the inverse Mellin transform of \( \frac{\Gamma(s)}{^{-s} e} \), at least not for the Kneser tetrational (and considering it doesn't seem to provide a convergent Mellin transform, perhaps not the inverse Mellin of anything).
http://en.wikipedia.org/wiki/Mellin_inversion_theorem
suggests that the Mellin inverse transform requires only the right boundedness in vertical strips, not on a whole half-plane. So it should work...
... but then I tried a numerical test to compare the value of \( \vartheta(w) \) against the inverse Mellin transform of \( \frac{\Gamma(s)}{^{-s} e} \) using the Kneser tetrational function as obtained from sheldonison's excellent Kneser-method code (which satisfies the required boundedness criteria). The approximation to the inverse Mellin transform is done by numerically integrating
\( IM(x) \approx \frac{1}{2\pi i} \int_{c - iA}^{c + iA} x^{-s} \frac{\Gamma(s)}{^{-s} e} ds \)
with \( A \) some large real value (here, I chose 15) and \( c \) is any number in \( (0, 1) \) (here, I chose 0.5). The following graph shows \( IM(x) \) and \( \vartheta(x) \) on the positive reals from near-0 to 20:
As you can see, \( IM(x) \) decays toward 0, but \( \vartheta(x) \) does not. It seems that \( IM(x) \approx \vartheta(x) \) for small \( x \), but the approximation gets worse at larger values of \( x \).
This suggests that there is an error in your derivation of \( \vartheta(x) \) from the inverse Mellin transform, since you should be able to Mellin-transform it back, which means your \( \vartheta(x) \) should decay to 0 along the real axis, but it doesn't. I suspect a formal disproof could be had if you can show that \( \mathrm{lim\ sup}_{x \rightarrow \infty} \ \vartheta(x) x^{s-1} = +\infty \). Whatever your \( \vartheta(x) \) is, it seems most assuredly not to be the inverse Mellin transform of \( \frac{\Gamma(s)}{^{-s} e} \), at least not for the Kneser tetrational (and considering it doesn't seem to provide a convergent Mellin transform, perhaps not the inverse Mellin of anything).
