I have a function $I(D)$ defined by the following integral representation
$$ I(D)=\int_0^\infty\mathrm{d}\alpha\,(1+2\alpha)^{-D/2} $$
which is clearly only sensible for $D>2$. The result of the computation for $D>2$ is $1/(D-2)$, itself valid for any $D\neq2$. So the result can be understood as the analytical continuation of $I(D)$.
My question: Can one derive (easily) an integral representation valid for $D<2$ starting from the original integral representation?
My motivation for this is the following: I have more complicated but similar looking integrals, again only convergent for $D>2$, that are rather complicated to calculate. I found a way to calculate them up to a factor, but I need to fix this factor by the knowledge of a specific value of $I(D)$, for which the integral is particularly simple. Now if I was able to easily derive an integral representation valid for $D<2$, I might be able to extract for example $I(D=0)$ without a difficult calculation.
Anyway it would be interesting to see if one can construct an analytical continuation - integral representation - starting from the original integral.
Note that if $D<2$ then $1<2/D$, and that the original integral representation was valid when $D/2>1$. Thus, $$ \int_0^{\infty} 2(1+2\alpha)^{-2/D}d\alpha = \frac{2}{\frac{4}{D}-2}=1+\frac{2}{D-2} $$ The integral representation for $D<2$ is now, $$ \int_0^{\infty}(1+2\alpha)^{-2/D}d\alpha -\frac{1}{2}. $$