# Analytic continuation of function given as integral

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.

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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, $$\int_0^{\infty}(1+2\alpha)^{-2/D}d\alpha -\frac{1}{2}.$$

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There seems to be a small mistake in your last equality. Another problem is that one has to know the result beforehand to construct a subtraction. My aim is to be able to insert a specific value of D that simplifies the integral drastically. Maybe I can work with the limit D->oo, where it vanishes. –  Tobias Apr 7 '13 at 21:47