35,840 reputation
378136
bio website sos440.tistory.com
location Los Angeles, CA
age 27
visits member for 3 years, 8 months
seen 4 hours ago

I am currently a UCLA student, particularly interested in probability theory.

I love mathematics, especially calculating some sort of integrals and summations with inner aesthetic appealing.

Language fluency: (G:green, A:amber, R:red)

  • Korean: Native
  • English: (Reading: G, Listening : A, Speaking : R, Writing : A)
  • Japanese: (Reading : A, Listening : A, Speaking : A, Writing : R)

19h
awarded  Necromancer
2d
answered How to construct a smooth function with compact support satisfying $f(x)+f(x^{-1})=1$
2d
awarded  Popular Question
Dec
17
awarded  closed-form
Dec
17
awarded  Self-Learner
Dec
16
revised Integral $\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right) \ \mathrm dx$
added 615 characters in body
Dec
16
answered Integral $\int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} - sx + 1} \right) \, \mathrm dx$
Dec
16
comment Closed form of $\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$
@Venus, thank you! I tried many ways until the hat fits her perfectly. Your compliment rewards my effort :)
Dec
16
revised Closed form of $\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$
added 118 characters in body
Dec
16
answered Closed form of $\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$
Dec
16
awarded  real-analysis
Dec
15
revised Evaluation of $\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$
added 570 characters in body
Dec
15
answered Evaluation of $\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$
Dec
13
comment Need help to find $\int_0^\infty \frac{e^{-x^2}\sin x^2}{\ln(1+x^2)}dx$
I cannot understand why you copy someone's comment into your question.
Dec
12
comment Economically computing $d\beta$
Is it allowed to use deRham cohomology?
Dec
12
answered Convergence of $ \sum_{k=1}^{\infty} \sqrt{a} \prod_{i=1}^{k}\frac{1}{1+i a}$ as $a \rightarrow 0$
Dec
11
answered Pie Integral $\int_0^1 \log\frac{(x+\sqrt{1-x^2})^2}{(x-\sqrt{1-x^2})^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.$
Dec
10
answered Finding the pdf of $(X+Y)^2/(X^2+Y^2)$ where $X$ and $Y$ are independent and normal
Dec
10
comment Pie Integral $\int_0^1 \log\frac{(x+\sqrt{1-x^2})^2}{(x-\sqrt{1-x^2})^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.$
I would like to point out that the integrand of your $I'(a)$ has a pole at $x = -a/\sqrt{1+a^2}$ if $a$ is negative. So this solution also shares the same issue with @gar's answer. Circumventin this problem, though, seems not hard. Oh, and +1 upvote, of course.
Dec
10
comment Pie Integral $\int_0^1 \log\frac{(x+\sqrt{1-x^2})^2}{(x-\sqrt{1-x^2})^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.$
A clever approach, but I have a doubt since $x - a\sqrt{1-x^2}$ can be negative. (So in your solution we have log of negative value in (1) and also a simple pole in the next line. I guess this solution can be modified (by using principal value) to yield a rigorous solution.