@TedShifrin, the proof that the cohomology class defined by the curvature of the Chern connection is the so-called Atiyah class seems rather technical---do you think it's worthwhile to go through it in detail? Huybrechts provides a much shorter argument to show that the class does not depend on the Hermitian structure (but not that it's the Atiyah class).
That the cohomology class is independent of choice of connection is the famous Chern-Weil theorem (the proof of which leads to things like secondary characteristic classes).
A prof briefly mentioned that one can think about a group as a set with $3$ operations, the "standard" binary one, an $1$-ary one that gives the inverses and a $0$-ary that selects the identity. Is there any motivation to do so?
@TedShifrin I like geometric interpretations when possible. But myself I also like to find physical examples, especially if it gives some intuition for what's going on.
For better or worse, though, that tends to be less and less helpful the more technical the math becomes
Hi everyone, [no intention to spam], I'm trying to figure out how I could evaluate an integral of the form $\int\log(ax+b)\exp(-0.5x^2)dx$. I have asked a question, for which I also started a bounty, [here](http://math.stackexchange.com/questions/1989341/on-the-evaluation-of-the-integral-int-fracba-frac1-ba-log-left). If you are interested, could you please take a look and lend a hand? Please be patient with me; I'm here to learn, not to spam :) Thank you very much!
@BalarkaSen @TedShifrin @Alessandro I guess there's that if you want to show that something is a Lie group or a topological group, you need that the inverse operator is continuous as well, not just the multiplication operator. But the "pick an identity element" operator is automatically continuous and smooth so I'm less inspired.
@nullgeppetto if you are interested in a numerical evaluation, what you can do is use that $$\int_0^1 x^n\ln(x) dx = -\frac1{(1+n)^2}$$ and expand the exponential in a power series
@s.harp, thanks for your response. The truth is that I would like to avoid going to a numerical solution (like using power series). I would very much like to have a form of solution, even if it's complicated.
@s.harp, I believe that some solution could be found, since there is an integral (here: en.wikipedia.org/wiki/…) where the Euler-Mascheroni constant appears and looks similar.
@nullgeppetto Mathematica can evaluate $\int_0^1 \log(x) \exp(-a x^2)dx$, the expression is -HypergeometricPFQ[{1/2, 1/2}, {3/2, 3/2}, -a], proofs involving such expressions are beyond me though.
Mathematica also can evaluate $\int_0^1 x^n \log(x)\exp(-a x^2)$, but it is more complicated.
It cannot handle $\int_0^1 \exp(-b x)\log(x)\exp(-ax^2)$ though
@Semiclassic: I like physical interpretations, too, when I can offer them. (Balarka just found one in my notes re geodesics on a surface of revolution.) But I don't know nearly enough physics for all the math that shows up.
@s.harp, I tried some similar forms in WolframAlpha, but still didn't find what I want. Besided, I'm really interested in the proof. Anyway, thank you very much for your responses :)
