User talk:Xetrov

Welcome!

Hello,, and welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. Here are some pages that you might find helpful: I hope you enjoy editing here and being a Wikipedian! Please sign your name on talk pages using four tildes ( ~ ); this will automatically produce your name and the date. If you need help, check out Questions, ask me on my talk page, or place  on your talk page and ask your question there. Again, welcome! -- Jeff3000 20:45, 21 February 2007 (UTC)
 * The five pillars of Wikipedia
 * How to edit a page
 * Help pages
 * Tutorial
 * How to write a great article
 * Manual of Style


 * Greetings, Jeff3000. Thank you kindly for the resource-laden message. Xetrov 23:10, 21 February 2007 (UTC)

Gaussian_integral#Higher-order_polynomials
http://en.wikipedia.org/wiki/Gaussian_integral#Higher-order_polynomials

Hello,

Do you have any references for the formula given (or the general technique of using series approximations of Gaussian integrals)? Would you happen to know what aspect of quantum field theory involves these integrals? I searched a lot for these things, but the terms Gaussian and integral are so general that I'm not getting anywhere without more specific information.

Thanks in advance! --Xeṭrov 01:24, 24 January 2011 (UTC)


 * I don't know any references for the formula, but it follows from the definition of Γ, using $$ t \equiv a x^n, \quad {\rm d}t = {\rm d}x \, a n x^{n-1}, \quad z \equiv \frac{j + 1}{n} $$:
 * $$ \Gamma(z) \equiv \int_0^\infty {\rm d}t \, t^{z-1} e^{-t} $$
 * $$ \Gamma\left(\frac{j + 1}{n}\right) = \int_0^\infty {\rm d}x \, a^{\frac{j + 1}{n}} n x^j e^{-a x^n} $$
 * $$ \int_0^\infty {\rm d}x \, x^j e^{-a x^n} = \frac{1}{n} \frac{\Gamma\left(\frac{j + 1}{n}\right)}{a^{\frac{j + 1}{n}}} $$
 * and expanding $$ \int_0^\infty {\rm d}x \, e^{-a x^n + b x^{n - 1} + \dots + c x} $$ in terms of $$ b \dots c $$ gives
 * $$ \int_0^\infty {\rm d}x \, e^{-a x^n + b x^{n - 1} + \dots + c x} = \sum_{B \dots C = 0}^{\infty} \frac{b^B}{B!} \dots \frac{c^C}{C!} \int_0^\infty {\rm d}x \, x^{(n-1)B + \dots + C} e^{-a x^n} $$
 * $$ \int_0^\infty {\rm d}x \, e^{-a x^n + b x^{n - 1} + \dots + c x} = \frac{1}{n} \sum_{B \dots C = 0}^{\infty} \frac{b^B}{B!} \dots \frac{c^C}{C!} \frac{\Gamma\left(\frac{(n-1)B + \dots + C + 1}{n}\right)}{a^{\frac{(n-1)B + \dots + C + 1}{n}}} $$.
 * Path integrals are usually gaussian integrals or similar. Κσυπ Cyp  18:28, 25 January 2011 (UTC)


 * Thank you very much for explaining this. It looks like approximations are in order, but it's good to know that there is an exact formula. --Xeṭrov 22:47, 27 January 2011 (UTC)


 * What happens to $$x^j$$ after the "and expanding [...] in terms of" line? I assume the sums come from the infinite sum expansions of $$e^{[term]}$$. Is it possible to expand $$e^{-a x^n}$$ so each term is just a product of expansion terms? --Xetrov 01:54, 2 February 2011 (UTC)


 * Oh, sorry, that $$x^j$$ wasn't supposed to be there (if it was there, the j would be added to the $$\frac{(n-1)B + \dots + C + 1}{n} + j$$ terms), as in:
 * $$ \int_0^\infty {\rm d}x \, x^j e^{-a x^n + b x^{n - 1} + \dots + c x} = \frac{1}{n} \sum_{B \dots C = 0}^{\infty} \frac{b^B}{B!} \dots \frac{c^C}{C!} \frac{\Gamma\left(\frac{(n-1)B + \dots + C + 1}{n} + j\right)}{a^{\frac{(n-1)B + \dots + C + 1}{n} + j}} $$
 * If also expanding the $$e^{-a x^n}$$ term, then all terms would be of the form $$\int_0^\infty {\rm d}x \, x^A = \infty$$. If expanding the $$e^{-a x^n}$$ term but leaving the $$e^{-b x^m}$$ term (with $$m<n$$, then the sum would not converge, due to $$\Gamma\left(\frac{nA + \dots}{m}\right)$$ growing faster than $$A!$$. Κσυπ Cyp  02:04, 3 February 2011 (UTC)