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I should evaluate Gaussian integrals in the 1+1 Minkowski space, which read

$$ I_{1}= \int d^{2}k \, {\rm Tr}\big[ \gamma^{5} \gamma^{\eta} \gamma^{\kappa} e^{\alpha k^{\mu}k_{\mu} + \beta \gamma^{\mu} \gamma^{\nu} M^{* \sigma}_{\mu} M^{\iota}_{\nu} k_{\sigma} k_{\iota}} \big],\\ I_{2}= \int d^{2}k \, {\rm Tr}\big[ \gamma^{5} e^{\alpha k^{\mu}k_{\mu} + \beta \gamma^{\mu} \gamma^{\nu} M^{* \kappa}_{\mu} M^{\iota}_{\nu} k_{\kappa} k_{\iota}} \big],\\ I_{3}= \int d^{2}k \, {\rm Tr}\big[ \gamma^{5} e^{\alpha k^{\mu}k_{\mu} + \beta \gamma^{\mu} \gamma^{\nu}\gamma^{5} M^{* \kappa}_{\mu} M^{\iota}_{\nu} k_{\kappa} k_{\iota}} \big], $$ where $\alpha$ and $\beta$ are real constants and the diagonal matrix $M$ has nonzero complex elements~($M_{\mu}^{\mu}=v_{\mu}$). Using the Euclidean convention $g^{\mu \nu} =- \delta^{\mu \nu}$ at $\beta=0$ results in $$ I_{1}= \int d^{2}k \, {\rm Tr}\big[ \gamma^{5} \gamma^{\eta} \gamma^{\kappa} e^{-\alpha k_{\mu}k_{\mu} } \big] =-2 \frac{\pi}{\alpha} \varepsilon^{\eta \kappa} , \\ I_{2}=0,\\ I_{3}=0, $$ where I have used ${\rm Tr}[\gamma^{5}]=0$, and ${\rm Tr}[\gamma^{5} \gamma^{\eta} \gamma^{\kappa}] = -2 \varepsilon^{\eta \kappa}$. How can I calculate $I_{1}$, $I_{2}$, and $I_{3}$ with nonzero $\beta$?

EDIT: A simplified question is how the following integral should be calculated $$ I_{0}= \int d^{2}k \, e^{ \gamma^{\mu} \gamma^{\nu} M^{* \sigma}_{\mu} M^{\iota}_{\nu} k_{\sigma} k_{\iota}} . $$

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    $\begingroup$ There's a problem of indices in your exponent, $\mu$ and $\nu$ appear three times in a product. $\endgroup$ Commented Feb 27, 2021 at 11:33
  • $\begingroup$ @JeanbaptisteRoux Corrected! $\endgroup$
    – Shasa
    Commented Feb 27, 2021 at 15:59
  • $\begingroup$ Have you tried calculating $\int dxe^{-Ax^2}$ where $A$ is some matrix? If you can find what properties $A$ must have, the generalization should be about as straightforward as the usual generalization from the 1d Gaussian integral to more complicated versions. I think it should be pretty lax...might need some weak form of invertibility like a pseudoinverse. $\endgroup$ Commented Feb 27, 2021 at 20:38
  • $\begingroup$ In this way, the matrix $A$ comprises $MM^{*}$ and $\gamma$ matrices. This Gaussian integral after the integration then generates a factor with the determinate of A in the denominator. This seems very weird to me. I think I should treat gamma matrices differently. $\endgroup$
    – Shasa
    Commented Mar 6, 2021 at 23:39

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You need to expand out the exponential using identites such a $(v^\mu \gamma_\mu)^2 = |v|^2{\mathbb I}$. It may well help introduce a new variable $\omega$ and us a Hubbard Stratanovich transform to replace $ (v^\mu \gamma_\mu)(v^{*\nu} \gamma_\nu) $ by something like $i\omega (v^\mu \gamma_\mu) + \frac 12 \omega^2 $ (signs not checked) befoe expanding.

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