Timeline for An integral with Gamma functions
Current License: CC BY-SA 3.0
20 events
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| Jun 26, 2013 at 17:44 | history | edited | Igor Khavkine | CC BY-SA 3.0 |
deleted 2 characters in body
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| Apr 30, 2013 at 19:22 | comment | added | Anirbit | @Igor Khavkine Hmm..I did find a copy in my library..hope to read up the details! Thanks for the help! | |
| Apr 26, 2013 at 18:04 | comment | added | Igor Khavkine | @Anirbit, probably the best thing for you to do is get a copy of Gelfand and Shilov and read carefully through their treatment of these distributions. | |
| Apr 26, 2013 at 16:58 | comment | added | Anirbit | @Igor Khavkine You say that any $\nu_1$ and $\nu_2$ is allowed but isn't something uncomfortable happening when either or both are equal to $0$? | |
| Apr 22, 2013 at 15:33 | comment | added | Anirbit | @Igor Khavkine Ah..sorry..I guess the point is that the paper uses the notation of $k = \vert \vec{k} \vert$ and hence the flip of the sign doesn't matter and you get that effect by writing it as $\vec{k}.\vec{k}$. And I guess by thinking of $(\vec{k}+\vec{q})^2$ as $((-\vec{k}) - \vec{q})^2$ you are getting across the $(-1)^d$ issue that the OP had already mentioned. (...this is quite a queer integral I would say - very hard to find anywhere and softwares like Mathematica won't know either..) | |
| Apr 22, 2013 at 6:11 | comment | added | Igor Khavkine | @Anirbit: If that is what you are worried about, you should perhaps try an example, say, compute $k^3$ and $(k^2)^{1.5}$ for $k=(1,2,3)$. Think carefully about what the notation means. | |
| Apr 21, 2013 at 23:20 | comment | added | Anirbit | @Igor Khavkine My point about $k+q$ is this - if you see the linked reference then the answer is framed as "~k^$d - 2(\nu_1 + \nu_2)}$" In this form $k$ and $-k$ would give different answers. But you frame the answer as "$~(k^2)^(d/2 - (\nu_1+\nu_2))$" in which $k$ and $-k$ would give the same answers. | |
| Apr 21, 2013 at 21:26 | comment | added | Igor Khavkine | @Anirbit: $(k+q)^2=[(-k)-q]^2$. The $\nu$s could be any complex number, excluding the poles of the final answer. | |
| Apr 21, 2013 at 20:44 | comment | added | Anirbit | @IgorKhavkine Thanks! And I wanted to clarify if anything here changes if in the denominator instead of $k-q$ it were $k+q$? Also is this applicable for any value of $\nu_1$ and $\nu_2$? | |
| Mar 27, 2013 at 19:41 | history | edited | Igor Khavkine | CC BY-SA 3.0 |
Fixed typo in final exponent.
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| Mar 25, 2013 at 22:19 | comment | added | Igor Khavkine | Yes, it's legitimate; no subtleties beyond consistently carrying out multiplication of series to a given order. Moreover, the $\Gamma$ function never vanishes on the real line, so the denominator does not contribute any poles. | |
| Mar 25, 2013 at 17:05 | comment | added | Anirbit | @Igor My question is this - the 3 $\Gamma$ functions in the numerator and one in the denominator are potentially going to have poles. Now is it legitimate to just ordinarily multiply the 3 Laurent series in the numerator and divide by the one in the denominator and then look for the pole of the result? Is such multiplication and division with Laurent series legal? Aren't there anything subtle? | |
| Mar 22, 2013 at 23:38 | comment | added | Igor Khavkine | @curious,Anirbit: It's very simple to work through the details yourself, which is what I recommend. The steps are straightforward as long as you know some basic properties of Fourier transforms and convolutions. These you can look up on Wikipedia or in an elementary textbook. Same advice for figuring out the poles. The poles of a product are easy to get once you know the poles of the factors: just take products of the Laurent expansions. Even if you don't have access to Gelfand & Shilov, you can take my Fourier transform formulas for granted and check the rest of the arithmetic. | |
| Mar 22, 2013 at 21:19 | comment | added | Anirbit | @Igor Khavkine I guess you are using the convolution theorem somewhere (mathworld.wolfram.com/ConvolutionTheorem.html). Seems you are doing the "inverse" of it - as in you are calculating the given convolution by taking the inverse Fourier transform of the product of the Fourier transforms of each. right? But I don't understand the technique of finding the poles when all the 4 functions few in denominator and few in the numerator) all can have poles! | |
| Mar 22, 2013 at 21:03 | comment | added | curious | @Igor I don't have this book you refer to with me right now. Can you kindly write down the Fourier transform and the convolution steps? At least the steps and the results. (..I didn't get why you needed to Fourier transform to do the convolution..) | |
| Mar 22, 2013 at 21:00 | comment | added | curious | @Igor Khavkine Thanks for the reply. The issue about the poles that I am confused about is this - for generic values of $\nu_1$ and $\nu_2$ all the $4$ Gamma functions in the value of the integral which have a $d$ in the argument are going to get poles. So individually each of the $4$ Gamma functions has a pole. Now how does one multiply and divide such things? | |
| Mar 22, 2013 at 9:31 | comment | added | Igor Khavkine | Hmm, any comments to go along with the downvote? | |
| Mar 22, 2013 at 4:25 | comment | added | Igor Khavkine | Alas, not my notation. | |
| Mar 22, 2013 at 3:50 | comment | added | Noam D. Elkies | "$d^d q$"? Ouch! | |
| Mar 22, 2013 at 3:35 | history | answered | Igor Khavkine | CC BY-SA 3.0 |