quantum field theory - Problem understanding sign of volume integral in Minkowski space

My professor told me that a 4-dimensional Minkowski - Space Integral I was working on can be written as the product of a metric tensor and a scalar:

$\int d^4 k \frac{k^\mu k^\nu}{(k^2-m_1^2)(k^2-m_2^2)} = g^{\mu \nu} B_2$

where $B_2$ is a scalar.

I am surprised that apparently for $\mu = \nu = 0$ , the sign of the integral is opposite to the sign for, say, $\mu = \nu = 1$. This seems wrong to me, and in an Eucledian Space it certainly would be. If I had been asked to guess, I would have written the above equation with $g^\mu_\nu B_2$ on the right hand side instead..

Can someone bring clarity?

Answer

The key insight is that the integral you wrote down is a Lorentz-invariant tensor, so whatever it evaluates to must also be Lorentz-invariant.

To illustrate this, let's be a bit more general and consider any integral of the form $$\begin{align} I^{\mu\nu}[f] = \int d^4k\, f(k^2)k^\mu k^\nu \end{align}$$ for any admissible function $f$. Then notice that for any Lorentz transformation $\Lambda = (\Lambda^\mu_{\phantom\mu\nu})$ we have $$\begin{align} \Lambda^\alpha_{\phantom\alpha\mu}\Lambda^\beta_{\phantom\beta\nu}\,I^{\mu\nu}[f] &= \int d^4k\, f(k^2)(\Lambda^\alpha_{\phantom\alpha\mu}k^\mu)( \Lambda^\beta_{\phantom\beta\nu}k^\nu)\\ &= \int d^4 u f(u^2) u^\alpha u^\beta \\ &= I^{\alpha\beta}[f] \end{align}$$ where in the second equality we made the change of variables $u^\alpha = \Lambda^\alpha_{\phantom\alpha\mu}k^\mu$, and we noted that the term $f(k^2)$ becomes $f(u^2)$ because $k^2$ is Lorentz-invariant, and the measure $d^4 k$ is Lorentz-invariant.

Addendum. (Following comments on answer v1)

The above argument demonstrates that provided $f$ is such that $I^{\mu\nu}[f]$ is well-defined, the integral is a Lorentz-invariant two-tensor and therefore is of the form $g^{\mu\nu}B[f]$ for some scalar functional $B$ as you wrote.

However, as is it is written, the integral you wrote down is divergent both because the integrand is singular (it has poles at $k^2 = m_1^2$ and $k^2 = m_2^2$), and by power counting which shows that the integrand scales linearly with $k$ for large $k$ so it is UV divergent. This ill-definedness leads to all sorts of apparent "paradoxes." For example, as pointed out by user10001, if we were to set $m_1 = m_2$, then the integrand is manifestly positive unless $k^\mu = 0$, so how could $I^{00}$ and $I^{ii}$ have different signs?

The resolution is to note that this integral comes from QFT where one uses the so-called $i\epsilon$ prescription (for good physics reasons) which removes the poles from the integrand by shifting them away from the real axis. Moreover, one fixes the UV divergence by regularizing the integral (using e.g. dimensional regularization). The object one then needs to compute is $$\begin{align} I^{\mu\nu}(d) = \lim_{\epsilon\to 0}\int_{\mathbb R^4} d^dk \frac{k^\mu k^\nu}{(k^2 - m_1^2 + i\epsilon)(k^2-m_2^2+i\epsilon)}, \tag{$\star$} \end{align}$$ for $d\neq 4$ and then analytically continue this to a function $I^{\mu\nu}(z)$ on the complex plain (minus some isolated singular points) which allows one to parametrize the divergence near $d=4$ by plugging in $z = 4-\epsilon$.

Note that $(\star)$ is not plagued by any of the apparent paradoxes that concerned us in the comments.