Undefined gamma function problem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T20:59:27Z http://mathoverflow.net/feeds/question/57794 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57794/undefined-gamma-function-problem Undefined gamma function problem Nigu 2011-03-08T08:40:37Z 2011-03-08T09:39:43Z <p>Hello,</p> <p>I'm trying to solve the following integral :</p> <p>$\int_0^\infty \frac{1}{t^{d/2}}(e^{-\gamma t} - e^{-\delta t})dt$.</p> <p>I know it equals </p> <p>$\Gamma(1-\frac{d}{2})[\gamma^{\frac{d}{2}-1}-\delta^{\frac{d}{2}-1}]$ for every $d&lt;4$.</p> <p>However, this does not work for $d=2$ as the gamma function is not defined in zero. According to some reference (E. Akkermans and G. Montambaux, Mesoscopic Physics of Electrons and Photons, ISBN 978-0521855129 (2007)), it is possible to solve it and the resulting law they obtain is logarithmic. However, they do not give any details in between.</p> <p>Any advice on how to proceed? Which integration method would work here?</p> <p>For the story, this integral is part of the cooperon correction to the conductivity which causes weak localization in mesoscopic systems.</p> <p>Thanks in advance!</p> http://mathoverflow.net/questions/57794/undefined-gamma-function-problem/57795#57795 Answer by Julien Puydt for Undefined gamma function problem Julien Puydt 2011-03-08T08:55:01Z 2011-03-08T08:55:01Z <p>The usual way such things are done is : if you can prove equality everywhere but at $d=2$, then as long as both expressions are holomorphic (in $d$) and admit an holomorphic prolongation to that point, then there's no problem.</p> http://mathoverflow.net/questions/57794/undefined-gamma-function-problem/57797#57797 Answer by Anatoly Kochubei for Undefined gamma function problem Anatoly Kochubei 2011-03-08T09:02:56Z 2011-03-08T09:02:56Z <p>The gamma function is not non-defined at zero, it has a pole there. You should expand the function in brackets by the Taylor formula (near $d=2$), use the identity $\Gamma (1+x)=x\Gamma (x)$, and pass to the limit, as $d\to 2$. If I calculated correctly, the result will be $\log \delta -\log \gamma$.</p> http://mathoverflow.net/questions/57794/undefined-gamma-function-problem/57800#57800 Answer by Zen Harper for Undefined gamma function problem Zen Harper 2011-03-08T09:33:44Z 2011-03-08T09:39:43Z <p>A direct calculation for $d=2$ is also possible and interesting:</p> <p>Let $G(\lambda, \mu) = \int_0^\infty \frac{e^{- \lambda t} - e^{- \mu t}}{t} dt$, for $\lambda, \mu > 0$. There are no problems at $t=0$ because $e^{- \lambda t} - e^{- \mu t} = (\mu - \lambda)t + O(t^2)$ near zero.</p> <p>Clearly $G(\lambda, \mu) = G(\lambda/\mu, 1)$ by a substitution, so we need only calculate $F(\lambda) = G(\lambda, 1) = \int_0^\infty \frac{e^{- \lambda t} - e^{- t}}{t} dt$.</p> <p>Now by differentiation under the integral sign,</p> <p>$$F'(\lambda) = -\int_0^\infty e^{- \lambda t} dt = -1/ \lambda$$</p> <p>Since clearly $F(1) = 0$, we have $F(\lambda) = -\log \lambda$, so (<em>as Anatoly Kochubei correctly says in another answer</em>), the final answer is $G(\gamma, \delta) = -\log(\gamma/\delta) = \log \delta - \log \gamma$.</p> <p>(<em>Note that these calculations are all easy to justify rigorously, by simple estimates.</em>)</p>