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 Dec 30 comment Finding Cauchy Principal Value for $\int_0^\infty \frac{\ln^2x }{(1-x)^2\sqrt{x}(x-4)} dx$ Thanks Ron. I look forward to your "messy but systematic presentation" complex method, and I will no longer worry about 'usurpation'. :) Dec 29 revised Evaluation of $\int \limits _0 ^{2 \pi} \frac {(r \cos \phi +x) \cos(n\phi)} {r^2+2xr \cos \phi +x^2} d\phi$ added more detail. Dec 29 revised Evaluation of $\int \limits _0 ^{2 \pi} \frac {(r \cos \phi +x) \cos(n\phi)} {r^2+2xr \cos \phi +x^2} d\phi$ typo and addendum Dec 29 revised Evaluation of $\int \limits _0 ^{2 \pi} \frac {(r \cos \phi +x) \cos(n\phi)} {r^2+2xr \cos \phi +x^2} d\phi$ added 103 characters in body Dec 29 revised Evaluation of $\int \limits _0 ^{2 \pi} \frac {(r \cos \phi +x) \cos(n\phi)} {r^2+2xr \cos \phi +x^2} d\phi$ add more thoughts on the topic. Dec 29 answered Evaluation of $\int \limits _0 ^{2 \pi} \frac {(r \cos \phi +x) \cos(n\phi)} {r^2+2xr \cos \phi +x^2} d\phi$ Dec 28 comment Finding Cauchy Principal Value for $\int_0^\infty \frac{\ln^2x }{(1-x)^2\sqrt{x}(x-4)} dx$ Hi RG. Wow, you did not get the same thing?. May I ask what you got?. I double checked and keep getting the same solution. But, not matching with you get makes me worry I done something wrong somewhere; which is probably more likely than you. I just noticed that Lucian posted a comment matching my answer. It was under a link that said, "3 more comments" and I hadn't noticed it at first. Dec 25 answered Finding Cauchy Principal Value for $\int_0^\infty \frac{\ln^2x }{(1-x)^2\sqrt{x}(x-4)} dx$ Dec 6 revised Closed form for Euler sum with $H_{2n}$?. fix typo Dec 6 revised Closed form for Euler sum with $H_{2n}$?. added more information Dec 6 awarded Custodian Dec 6 reviewed Approve Closed form for Euler sum with $H_{2n}$?. Dec 6 comment Closed form for Euler sum with $H_{2n}$?. I emended the heading to be more specific about what I was asking. Dec 6 revised Closed form for Euler sum with $H_{2n}$?. edited title Dec 6 comment Closed form for Euler sum with $H_{2n}$?. Yes, of course. Can anyone find a closed form for said Euler sum?. Dec 6 asked Closed form for Euler sum with $H_{2n}$?. Dec 3 comment another product of log integral Hey nospoon, here is another integral I found if you're interested. I think it can be done the way you done the last one. But, I found: $$\int_{0}^{1}\log(1-x^{3})\log(1+x^{3})dx=1/4\left(-72+2 \gamma^{2}+\pi^{2}+4 \gamma \psi(1/6)+2\psi(1/6)-2\psi_{1}(7/6)\right)-\frac{1}{2}\left(3\sqrt{3}\pi +\frac{\pi^{2}}{6}+9\log(3)+12\log(2)-36\right)-\sum_{n=1}^{\infty}\frac{H_{2n}}‌​{(n(6n+1))}$$. I got stuck on that last sum....so far. Dec 3 comment another product of log integral Wow, thanks nospoon. Very nice work. There are lots of polylog identities. Maybe reference to Lewin's book we can find a way to simplify them. I managed to discover a solution, but it is not as efficient as yours. As mentioned above, I used a bunch of Euler sums. Dec 3 accepted another product of log integral Nov 29 revised another product of log integral added another sum.