$$f(x) = \mathop {\lim }\limits_{T \to \infty } {i}\int_{-1/2-i\,T}^{-1/2+i\,T} \frac{(x-1)^{s}}{2^{s+1}}\,\frac{1}{sin(\pi*s)\,}\,\frac{ds}{s}$$
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asked Oct 22, 2019 at 17:04
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1 Answer
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If I rewrite the integral in the usual form of an inverse Mellin transform, I get: $$-\frac{1}{2\pi}f(x+1)=\frac{1}{2\pi i}\int_{1/2-i\infty}^{1/2+i\infty}x^{-s}2^{1-s}\frac{1}{\sin\pi s}\frac{ds}{s}=\frac{2}{\pi} \ln \left(\frac{1}{2 x}+1\right)$$
answered Oct 22, 2019 at 19:20
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$\begingroup$ Is this formula right? Εxample for x = 12 something goes wrong. May be the first term, is -2/Pif(4x+1) = 1/(2*Pii).... and not -1/(2Pi)*f(x+1)? $\endgroup$ Commented Oct 27, 2019 at 17:20