2,120 reputation
21446
bio website anhhuy.ch
location Switzerland
age 22
visits member for 4 years, 1 month
seen 13 hours ago

My name is Anh Huy Truong and I am currently studying mathematics at the ETH Zürich.


May
15
comment What do I know when the curl of a vector field equals 0?
We had a theorem: "Let $\Omega \subseteq \mathbb{R}^2$ be open and star-shaped. A $C^1$-vector field $K: \Omega \to \mathbb{R}^2$ is conservative iff $\mathrm{rot}~K = 0$." So this domain is not star-shaped, because the origin of the x-y-plane is missing. How do I proceed then?
May
15
comment What do I know when the curl of a vector field equals 0?
Which one did I miss?
May
15
asked What do I know when the curl of a vector field equals 0?
May
10
answered Finding double root. An easier way?
May
9
answered Mathematicians whose names are commonly mispronounced
May
8
awarded  Autobiographer
May
8
awarded  Enthusiast
May
6
comment A simple way to evaluate $\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx$?
There's one part I don't understand at this very moment: Why is $\int_{-\infty}^\infty \frac{x}{2t} (-2tx \exp(-tx^2)) \, \mathrm dx = \frac{-1}{2t} \int_{-\infty}^\infty \exp(-tx^2) \, \mathrm dx$?
May
6
comment A simple way to evaluate $\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx$?
So is there no way similar to this way which actually is correct? :/
May
6
comment A simple way to evaluate $\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx$?
Would I be allowed to solve $\int_{-\infty}^\infty e^{-tx} \, \mathrm dx$ (which is $\sqrt{\frac{\pi}{t}}$) and then replace $t$ by $i$? However I wonder why so many users upvote wrong answers?
May
6
accepted A simple way to evaluate $\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx$?
May
6
comment A simple way to evaluate $\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx$?
It worked like a harm. Thanks a lot. I accept this answer, as - opposed to the other two answers - it actually answers my original question.
May
6
comment A simple way to evaluate $\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx$?
We already know that integral equals &\sqrt{\pi}$, so I think it'll be okay. But there's something else I don't understand: You use a complex number in your substitution. How will the lower and upper bound of the integral change? I assumed they'd stay the same, but that doesn't seem clean to me as u is a complex number and I'd integrate in a totally different direction; or aren't I? If not, how so? If yes, why can I do this? This question arises because we've never done a complex substitition yet; as far as I remember.
May
5
comment A simple way to evaluate $\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx$?
Thank you very much for the quick answer, I wonder why our professor wants us to solve the integral the awkward way. Probably to actually solve that final integral.
May
5
comment A simple way to evaluate $\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx$?
This approach indeed does seem easier than wolframalpha's. I will try to solve the integral in this way as soon as I've had enough sleep.
May
5
comment A simple way to evaluate $\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx$?
This indeed is a very nice way to get it. However, how exactly do I solve $\int_{-\infty}^\infty e^{-i x^2} \, \mathrm dx$?
May
5
answered Proving a function is infinitely differentiable
May
5
comment A simple way to evaluate $\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx$?
Then, it's basically the way described by wolframalpha. I'm looking for a simpler way than that one. (if it's already the simplest way, I'd be quite disappointed :( )
May
5
comment A simple way to evaluate $\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx$?
$(x^2-x+1)(x^2+x+1) = x^4 + x^3 + x^2 - x^3 - x^2 - x + x^2 + x + 1 = x^4 + x^2 + 1$.
May
5
asked A simple way to evaluate $\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx$?