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Jul 27, 2016 at 4:13 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Jun 29, 2016 at 3:28 comment added reuns you are allowed to upvote and accept my answer :) (the second part makes everything rigorous) and you easily get $\int_0^\infty \frac{\Xi(t)}{t^2 + \frac{1}{4}} \cos(xt) dt = \frac{1}{2} \pi (e^{\frac{1}{2}x} - 2e^{-\frac{1}{2}x} \psi(e^{-2x}))$ from it, since $\pi\int_0^\infty \frac{\Xi(t)}{t^2 + \frac{1}{4}} \cos(xt) dt =\frac{1}{2 i \pi} \int_{1/2-i\infty}^{1/2+i \infty} E(s) e^{x (s-1/2)} ds$
Jun 24, 2016 at 4:41 answer added reuns timeline score: 3
Jun 23, 2016 at 16:13 history edited fje CC BY-SA 3.0
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Jun 23, 2016 at 15:49 history edited fje CC BY-SA 3.0
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Jun 23, 2016 at 15:43 history edited fje CC BY-SA 3.0
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Jun 23, 2016 at 15:37 history asked fje CC BY-SA 3.0