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!

  • 1
    $\begingroup$ Use as $\displaystyle{\large\Gamma\left(2 - {d \over 2}\right)\,{\gamma^{d/2 - 1} - \delta^{d/2 - 1}\over 1 - d/2}}$ and take the limits $\displaystyle{\large d \to 2}$ with L'Hopital. $\endgroup$ Jan 19, 2014 at 22:51

3 Answers 3


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.)

  • $\begingroup$ @Zen. Thank you! Very clear and understandable explanation... $\endgroup$
    – Nigu
    Mar 8, 2011 at 10:55
  • $\begingroup$ You'll notice that if you compute using the technique I mentioned, you'll get the same result pretty fast too : putting $s=1-\frac{d}2$ and making $s$ go to zero, you'll find the second expression is $\Gamma(s)\left(\exp(-s\ln(\gamma))-\exp(-s\ln(\delta))\right)$, which is an easy limit. $\endgroup$ Mar 8, 2011 at 13:01

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.


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 $.


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