The integral is zero! $\int \frac{\mathrm{d}^d k}{(2\pi)^d} = 0$ In using dimensional regularization in QFT calculations, one comes across integrals over propagators, they might look like $(d = \text{dimension of spacetime}, n = \text{a number})$ 
$$\tag{1}I(d,n)=\int \frac{\mathrm{d}^d k}{(2\pi)^d}\frac{1}{\big[k^2-\Delta\big]^n}$$
where one can consider the integral to be a function of the spacetime dimension $d$ which here need not be an integer. Now there's a formula for the integral $(1)$, which is given by (see e.g. Appendix of Peskin and Schroeder)
\begin{align}\tag{2}I(d,n)&=\int \frac{\mathrm{d}^d k}{(2\pi)^d}\frac{1}{\big[k^2-\Delta\big]^n} \\
&=\frac{(-1)^n\mathrm{i}}{(4\pi)^{d/2}}\frac{\Gamma(n-d/2)}{\Gamma(n)}\left(\frac{1}{\Delta}\right)^{n-d/2}. \end{align}
My question is why does the integral 
$$I(d,0) = 0.$$
Expanding for small $n$ we get 
$$\frac{i 2^{-d} \pi ^{-d/2} (-1)^n \left(\frac{1}{\Delta }\right)^{n-\frac{d}{2}}
   \Gamma \left(n-\frac{d}{2}\right)}{\Gamma (n)} = i 2^{-d} \pi ^{-d/2}  \left(\frac{1}{\Delta }\right)^{-d/2} \Gamma
   \left(-\frac{d}{2}\right)n+O\left(n^2\right)$$
Which is proportional to $n$ and which according to mathematica is equation to zero in the $n\rightarrow 0$ limit. So that 
$$\tag{3}\int \frac{\mathrm{d}^d k}{(2\pi)^d}  = 0~\text{(in dimreg)}.$$ 
This reminds me of that thing in string theory 
$$1+2+3+4+\cdots = -\frac{1}{12}.$$
So where/what in the derivation of the dimreg result equation $(2)$ does the assumption(s) come into the picture making equation (3) possible? Because $(3)$ looks like the volume of spacetime which should be infinite, all this is really strange. 
 A: As Prahar said in a comment, dim. reg. allows us to see that two functions are equal in a half-plane, and if one of them is analytic in the whole plane, the other function may be analytically continued as well.
So the volume of the spacetime is indeed zero in dimensional regularization. More generally, any power law divergence is set to zero in dimensional regularization. Only logarithmic divergences may be "genuine".
This result, zero, is totally compatible with the values assigned by the zeta regularization. Note that
$$ \sum_{n=1}^\infty n = \zeta(-1)= -\frac{1}{12} $$
but the simpler version you want here is
$$ \sum_{n=1}^\infty 1 = \zeta(0)= -\frac{1}{2} $$
The latter relationship is self-consistent because it implies
$$ \sum_{n=-\infty}^\infty 1 = 1 + 2 \sum_{n=1}^\infty 1 = 1+ 2\times (-1/2) = 0 $$
I've divided the sum over all integers to the sum over positive ones (which is $-1/2$), the negative ones (also $-1/2$), and an additional $1$ from $n=0$. The sum of these three parts cancels which is no coincidence.
The integral may be written as
$$ \int_{-\infty}^\infty dx\,1 = \lim_{\epsilon\to 0} \sum_{n=-\infty}^{\infty} \epsilon = 0\epsilon = 0 $$
where I used the substitution $x=n\epsilon$ and a Riemann definition of the integral. In fact, because the already vanishing result from the summation was multiplied by another $\epsilon$ from the $dx$, we got a "stronger zero" than needed. For that reason, the continuous "integral" version would be zero even if we had things like the integral over $x$ from zero, and even if the integrand were a positive power of $x$. The minus one twelfth would get multiplied by $\epsilon^2\to 0$. You may check it explicitly.
A: using zeta regularization and Euler maclaurin series the integral $ \int_{0}^{\infty}x^{m} $ is not zero but is related to $ \zeta (-m) $ see my paper http://vixra.org/abs/1009.0047 adn in particular $\int_{0}^{\infty}dx=1+\zeta (0) 4$ by euler maclaurin formula
