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 Feb19 comment Integrate: $\int_0^{\infty} \frac{\sin (ax)}{e^{\pi x} \sinh(\pi x)}dx$ Thanks Ron. So in the first integral in the denominator, I have $1 - 1 + 2\pi \epsilon e^{i\phi}$ + higher order terms. Then the $e^{i \phi}$ in the numerator would cancel the one in the first order term. For the rest of the numerator, I would have $1 + ia\epsilon e^{i\Phi}$ + higher order terms. So I am stuck how to deal with the higher order terms, and what happens to the $a$. Thanks, sorry to be so dense. Feb19 comment Integrate: $\int_0^{\infty} \frac{\sin (ax)}{e^{\pi x} \sinh(\pi x)}dx$ Dear Ron, I saw this on your blog as well. Didn't know where you would prefer to be asked a question (maybe neither - but hope not since you are always so generous). Sorry to be a serial questioner, yet your work is so compelling. Could I please ask you how you applied the residue theorem to get from the fifth and sixth integrals to the "combine to form" line. On second thought, this is the better place to ask so I can +50 for this and your untiring generosity in helping others. All the best, Andrew Feb19 comment Prove $\int^\infty_0\frac x{e^x-1}dx=\frac{\pi^2}{6}$ Thanks for you help. All the best, Andrew Feb18 comment Prove $\int^\infty_0\frac x{e^x-1}dx=\frac{\pi^2}{6}$ so am unable to get his results as well. Thanks very much for your consideration. All the best, Feb18 comment Prove $\int^\infty_0\frac x{e^x-1}dx=\frac{\pi^2}{6}$ Dear Raymond, Maybe I could please ask you to elaborate as to how you get (2) & (3). I know you showed how for the half circle you get "half" a residue here math.stackexchange.com/questions/478534/…. But that example is in a nice convenient form with the $\frac{C_{-1}}{z-z_0}$ of a Laurent series. So how do you get the terms in the ( ) in each of (2) and (3)? I tried using $z = \epsilon e^{i \theta}$, but that got pretty messy and is far from the simplicity of your terms. Also looked at the link in Ron's blog where he uses $z$ as I mentioned Feb18 comment Prove $\int^\infty_0\frac x{e^x-1}dx=\frac{\pi^2}{6}$ Thanks, Ron. As you know, you are a key All-Pro on my fantasy math team. All the best, Andrew Feb18 comment Prove $\int^\infty_0\frac x{e^x-1}dx=\frac{\pi^2}{6}$ Dear Ron, Could I please trouble you for a hint as to what looks how to isolate $e^{-kx}$ outside the integral so you have $\Gamma(u)$ times the summation. Actually I fear that posing this question is naive. But that's what it looks like. Actually, probably much better is a hint how to get from the second expression to the third. Thanks very much. With regards, Andrew Feb17 comment Conjecture $_2F_1\left(\frac14,\frac34;\,\frac23;\,\frac13\right)=\frac1{\sqrt{\sqrt{\frac4{\sqrt{2-\sqrt[3]4}}+\sqrt[3]{4}+4}-\sqrt{2-\sqrt[3]4}-2}}$ Dear Raymond - Outstanding. Beautifully resolves the "Update" conjecture. I was wondering if it also addresses the initial conjecture(4)? I.e., does the update come out of (4), or are they distinct? Thanks and best regards, Feb16 comment Evaluating $\sum_{n=1}^\infty\frac{\zeta(3n)}{2^{3n}}$ Nice generalization above - thanks very much. Best, Feb15 comment Evaluating $\sum_{n=1}^\infty\frac{\zeta(3n)}{2^{3n}}$ Dear Raymond, Maybe you would please elaborate a bit how to get from the line above "As for the multiplication theorem" to the last line. Thanks and all the best, Feb12 comment How to prove Gauss’s Multiplication Formula? @JacobMayle Me too, a big help. Feb10 revised Information about Riemann Zeta function added 236 characters in body Feb1 awarded Notable Question Jan31 answered Information about Riemann Zeta function Dec26 awarded Nice Question Dec17 comment Basic probability problem! What's with the exclamation point in the header! Dec17 revised Evaluating $\sum_{n=1}^\infty\frac{\zeta(3n)}{2^{3n}}$ edited tags Dec16 awarded Necromancer Dec12 revised Best Sets of Lecture Notes and Articles added 265 characters in body Dec11 comment Quickest way to teach myself college algebra @DepeHb Yes, I think you are right. That's why I stuck in the "Gelfand" suggestion.