Hello,


I'm trying to solve the following integral :


$\int\_0^\infty \frac{1}{t^{d/2}}(e^{-\gamma t} - e^{-\delta t})dt$.


I know it equals 


$\Gamma(1-\frac{d}{2})[\gamma^{\frac{d}{2}-1}-\delta^{\frac{d}{2}-1}]$ for every $d<4$.


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.


Any advice on how to proceed? Which integration method would work here?


For the story, this integral is part of the cooperon correction to the conductivity which causes weak localization in mesoscopic systems.


Thanks in advance!



Undefined gamma function problem
A direct calculation for $d=2$ is also possible and interesting:


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.


Clearly $G(\lambda, \mu) = G(\lambda/\mu, 1)$ by a substitution, so we need only calculate $F(\lambda) = G(\lambda, 1) = \int\_0^\infty \frac{e^{- \lambda t} - e^{- t}}{t} dt$.


Now by differentiation under the integral sign,


$$ F'(\lambda) = -\int\_0^\infty e^{- \lambda t} dt = -1/ \lambda $$


Since clearly $F(1) = 0$, we have $F(\lambda) = -\log \lambda$, so (*as Anatoly Kochubei correctly says in another answer*), the final answer is $G(\gamma, \delta) = -\log(\gamma/\delta) = \log \delta - \log \gamma$.


(*Note that these calculations are all easy to justify rigorously, by simple estimates.*)


