In a massless theory we often get integrals of the form $$\int\frac{d^Dk}{(2\pi)^D} \frac{1}{k^{2n}} \tag{*}$$ where $D=4-2\varepsilon$. I have tried to calculate this in two ways in Minkowski space and in Euclidean space. With the former I would be integrating $$\int\frac{d^Dk}{(2\pi)^D} \frac{1}{(k^2+i\varepsilon')^n}$$ but in this case I appear to get a factor of $1/(\varepsilon')^\epsilon$ in my final answer (for the case of $n=2$) in which I can't take $\varepsilon'\rightarrow 0$. In the Euclidean case I get an intergral of the form: $$\int^\infty_0 t^{\varepsilon/2-1}$$ which as far as I can tell diverges for $\varepsilon\gt 0$ (which it is). Thus my question is as follows: What is the standard and easiest way to calculate (*) ideally based only on Feynman parametrization, Schwinger parametrization and guassian integrals?
Integrals of the form $\int\frac{d^Dk}{(2\pi)^D} \frac{1}{k^{2n}}$ in $D=4-2\varepsilon$ dimensions?
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2$\begingroup$ In dim reg, the value of this integral is just zero. (Related question here.) $\endgroup$– knzhouDec 27, 2017 at 8:39
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1$\begingroup$ related question by OP: Renormalization of a Feynman diagram with zero bare mass. $\endgroup$– AccidentalFourierTransformDec 27, 2017 at 10:33
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$\begingroup$ See Collins "Renormalization" for a discussion of dimensional regularization, including this particular integral. $\endgroup$– BlazejDec 27, 2017 at 12:51
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