Skip to main content
added 2 characters in body
Source Link
Glorfindel
  • 2.8k
  • 6
  • 28
  • 38

Your transformation (*) is related to the Mellin transform (with $\epsilon=s$). In particular, your result is obtainobtained as the limit $s\to0^+$ of Ramanujan's master theorem $$ \frac{\sin(\pi s)}{\pi}\int_0^\infty x^{s-1}\left(\,\lambda(0) - x\,\lambda(1) + x^2\,\lambda(2) -\,\cdots\,\right) dx = \,\lambda(-s)\,.$$

Your transformation (*) is related to the Mellin transform (with $\epsilon=s$). In particular, your result is obtain as the limit $s\to0^+$ of Ramanujan's master theorem $$ \frac{\sin(\pi s)}{\pi}\int_0^\infty x^{s-1}\left(\,\lambda(0) - x\,\lambda(1) + x^2\,\lambda(2) -\,\cdots\,\right) dx = \,\lambda(-s)\,.$$

Your transformation (*) is related to the Mellin transform (with $\epsilon=s$). In particular, your result is obtained as the limit $s\to0^+$ of Ramanujan's master theorem $$ \frac{\sin(\pi s)}{\pi}\int_0^\infty x^{s-1}\left(\,\lambda(0) - x\,\lambda(1) + x^2\,\lambda(2) -\,\cdots\,\right) dx = \,\lambda(-s)\,.$$

Source Link

Your transformation (*) is related to the Mellin transform (with $\epsilon=s$). In particular, your result is obtain as the limit $s\to0^+$ of Ramanujan's master theorem $$ \frac{\sin(\pi s)}{\pi}\int_0^\infty x^{s-1}\left(\,\lambda(0) - x\,\lambda(1) + x^2\,\lambda(2) -\,\cdots\,\right) dx = \,\lambda(-s)\,.$$