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bio website linkedin.com/pub/…
location Los Angeles, CA
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visits member for 3 years, 1 month
seen Nov 26 at 15:25

I am an enthusiastic math and physics student.


Oct
24
awarded  Popular Question
Sep
24
awarded  Autobiographer
Sep
12
awarded  Tumbleweed
Sep
5
revised Behavior of lower incomplete gamma function at complex infinity
added tag
Sep
5
asked Behavior of lower incomplete gamma function at complex infinity
Sep
1
accepted Show $\prod \limits_{\substack{k=0\\k\neq k_0}}^{K-1}\left(\exp\left[{\frac{2 \pi i (k-k_0)}{K}}\right]-1\right)=(-1)^{K-1} K$
Sep
1
revised Show $\prod \limits_{\substack{k=0\\k\neq k_0}}^{K-1}\left(\exp\left[{\frac{2 \pi i (k-k_0)}{K}}\right]-1\right)=(-1)^{K-1} K$
corrected wrong formula
Sep
1
comment Show $\prod \limits_{\substack{k=0\\k\neq k_0}}^{K-1}\left(\exp\left[{\frac{2 \pi i (k-k_0)}{K}}\right]-1\right)=(-1)^{K-1} K$
You are right of course, I made a mistake when typing up my notes. Could you please explain your expansion of $f(x)$? If it's a well--known result, maybe you can point me to a source?
Sep
1
asked Show $\prod \limits_{\substack{k=0\\k\neq k_0}}^{K-1}\left(\exp\left[{\frac{2 \pi i (k-k_0)}{K}}\right]-1\right)=(-1)^{K-1} K$
Jul
7
awarded  Nice Question
Jul
2
awarded  Curious
Apr
16
comment Complex contour integral with sign function:$-i \int \limits_{-\infty}^\infty \frac{{\rm sgn}(x)^2 ~x~ e^{i x}}{1+ax^2} dp$
For complex $z$, $\operatorname{sign}(z)=z/|z|$ (mathworld.wolfram.com/Sign.html). Hence, $\operatorname{sign}(i b)=i \operatorname{sign}(b)=i~,~b>0$
Apr
16
comment Complex contour integral with sign function:$-i \int \limits_{-\infty}^\infty \frac{{\rm sgn}(x)^2 ~x~ e^{i x}}{1+ax^2} dp$
Right, but my question was why do I get a different result when I include the $\operatorname{sign}^2(x)$ part, not how to compute it... The answer seems to be (see comments to the question) that $\operatorname{sign}(x)$ is not holomorphic. Maybe you have some insights about the n-dimensional case (see Note)?
Apr
14
comment Complex contour integral with sign function:$-i \int \limits_{-\infty}^\infty \frac{{\rm sgn}(x)^2 ~x~ e^{i x}}{1+ax^2} dp$
@DanielFischer I guess my intuition fails at this because I still don't see if one approach works, why don't they all. Where is the loss of information? But in any case, thank you very much, I really appreciate your help. (If you care about those things, I will this as an answer.)
Apr
14
comment Complex contour integral with sign function:$-i \int \limits_{-\infty}^\infty \frac{{\rm sgn}(x)^2 ~x~ e^{i x}}{1+ax^2} dp$
@DanielFischer Right $x^2=||{\bf x}||^2$. And, yes: this is essentially a Fourier transform with a modified measure. A colleague of mine got an answer (by writing it as $\int d^n {\bf x} \int\limits_0^\infty ds e^{-s(1+ax^2)}e^{-i{\bf x\cdot y}}$ and using en.wikipedia.org/wiki/…), so I don't see why I wouldn't be able to do it this way.
Apr
14
comment Complex contour integral with sign function:$-i \int \limits_{-\infty}^\infty \frac{{\rm sgn}(x)^2 ~x~ e^{i x}}{1+ax^2} dp$
@DanielFischer I see... The idea is to compute something like $\int e^{i{\bf x \cdot y}}/(1+ax^2) d^n {\bf x} \sim \int\limits_0^\infty \left(e^{ixy}-e^{-ixy}\right)/(1+ax^2) x^{n-1} d x$, where ${\bf x, y}$ are n-dimensional vectors. I turned it into an integral over the whole real line to apply the residue theorem, but I guess that doesn't work for even $n$. From what you say, I'm gathering there's no solution to this problem?
Apr
13
comment Complex contour integral with sign function:$-i \int \limits_{-\infty}^\infty \frac{{\rm sgn}(x)^2 ~x~ e^{i x}}{1+ax^2} dp$
@DanielFischer is that because of convergence? I am evaluating this as a way to compute: $\int \limits_0^\infty x^{n-2} \sin{(x)}/(1+ax^2) dx$. Did I go the wrong way? (maybe I should create a new question...)
Apr
13
comment Complex contour integral with sign function:$-i \int \limits_{-\infty}^\infty \frac{{\rm sgn}(x)^2 ~x~ e^{i x}}{1+ax^2} dp$
@DanielFischer I figured $\operatorname{sign}$ was causing trouble. Could you please explain what you mean by splitting the integral? any suggestions about the symmetries? If you could help me a bit more, I would take that as an answer.
Apr
13
comment Complex contour integral with sign function:$-i \int \limits_{-\infty}^\infty \frac{{\rm sgn}(x)^2 ~x~ e^{i x}}{1+ax^2} dp$
@DanielFischer I can forget that in the case $n=3$ (see the note), but not in general... so I want to know why I get different things with and without the sign function
Apr
13
comment Complex contour integral with sign function:$-i \int \limits_{-\infty}^\infty \frac{{\rm sgn}(x)^2 ~x~ e^{i x}}{1+ax^2} dp$
@Pragabhava typo, should be $dx$. Thanks.