Mysterious path integral divergence after Hubbard-Stratonovich transformation Let us consider common gaussian path integral over some complex random field $\displaystyle \Psi (\mathbf{r})$:
\begin{equation*}
N=\int D\Psi ^{*} D\Psi \ \exp\left( -\int d^{n} r\ \Psi ^{*}\hat{K} \Psi \right)
\end{equation*}
Here $\displaystyle \hat{K}$ is some operator, which posses $\displaystyle \hat{K} \langle \Psi ^{*}(\mathbf{r}_{1}) \Psi (\mathbf{r}_{2}) \rangle =\delta (\mathbf{r}_{1} -\mathbf{r}_{2})$, so $\displaystyle \hat{K}^{-1}(\mathbf{r}_{1} ,\mathbf{r}_{2}) =\langle \Psi ^{*}(\mathbf{r}_{1}) \Psi (\mathbf{r}_{2}) \rangle $ and we work with usuall gaussian statistics. Now I want to calculate average ≪intensity≫ of my field in $\displaystyle \mathbf{r} =\mathbf{y}$. So I write:
\begin{equation*}
\langle \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y}) \rangle =\dfrac{1}{N}\int D\Psi ^{*} D\Psi \ \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y}) \ \exp\left( -\int d^{n} r\ \Psi ^{*}\hat{K} \Psi \right)
\end{equation*}
This integral is common and we can easily calculate it introducing source terms:
\begin{gather*}
\langle \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y}) \rangle =\dfrac{1}{N} \partial _{\lambda } \partial _{\eta }\int D\Psi ^{*} D\Psi \ \exp\left( -\int d^{n} r\ \Psi ^{*}\hat{K} \Psi +\eta \Psi ^{*}(\mathbf{y}) +\lambda \Psi (\mathbf{y})\right)\biggl|_{\lambda ,\eta =0} =\\
=\partial _{\lambda } \partial _{\eta }\exp\left(\int d^{n} r\int d^{n} r'\ \eta \delta (\mathbf{r} -\mathbf{y})\hat{K}^{-1}(\mathbf{r} ,\mathbf{r} ') \lambda \delta (\mathbf{r} '-\mathbf{y})\right)\biggl|_{\lambda ,\eta =0} =\langle \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y}) \rangle 
\end{gather*}
So we see, that the calculation is self-consistent. Let us now perform it in an other way, using HS transformation:
\begin{gather*}
\langle \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y}) \rangle =\dfrac{1}{N}\int D\Psi ^{*} D\Psi \ \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y}) \ \exp\left( -\int d^{n} r\ \Psi ^{*}\hat{K} \Psi \right) =\\
\dfrac{1}{N}\int D\Psi ^{*} D\Psi \int\limits ^{\infty }_{-\infty }\dfrac{d\xi }{\sqrt{\pi }}\left( \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y})\right)^{2}\exp\left( -\int d^{n} r\ \Psi ^{*}\hat{K} \Psi -\left( \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y})\right)^{2} \xi ^{2}\right) =\\
=\dfrac{1}{N}\int D\Psi ^{*} D\Psi \int\limits ^{\infty }_{-\infty }\dfrac{d\xi }{\sqrt{\pi }}\dfrac{1}{2\xi }\dfrac{\partial }{\partial \xi }\exp\left( -\int d^{n} r\ \Psi ^{*}\hat{K} \Psi -\left( \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y})\right)^{2} \xi ^{2}\right) =\\
=\dfrac{1}{2\sqrt{\pi } N}\int D\Psi ^{*} D\Psi \int\limits ^{\infty }_{-\infty } d\xi \dfrac{1}{\xi }\dfrac{\partial }{\partial \xi }\int \dfrac{d\alpha ^{*} \land d\alpha }{2\pi i}\dfrac{1}{\xi ^{2}}\exp\left( -\int d^{n} r\ \Psi ^{*}\hat{K} \Psi -\dfrac{\alpha ^{*} \alpha }{\xi ^{2}} +\left( \alpha ^{*} -\alpha \right) \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y})\right) =\\
\\
=\dfrac{1}{2\sqrt{\pi } N}\int D\Psi ^{*} D\Psi \int\limits ^{\infty }_{-\infty } d\xi \dfrac{1}{\xi }\dfrac{\partial }{\partial \xi }\int \dfrac{d\alpha ^{*} \land d\alpha }{2\pi i}\dfrac{1}{\xi ^{2}}\int \dfrac{d\beta ^{*} \land d\beta }{2\pi i\left( \alpha ^{*} -\alpha \right)}\exp\left( -\int d^{n} r\ \Psi ^{*}\hat{K} \Psi -\dfrac{\alpha ^{*} \alpha }{\xi ^{2}} -\dfrac{\beta ^{*} \beta }{\alpha ^{*} -\alpha } +\beta \Psi ^{*}(\mathbf{y}) +\beta ^{*} \Psi (\mathbf{y})\right) =
\end{gather*}
Now we deal with common gaussian path integral with outer sources, and we can integrate $\displaystyle \Psi $ out, so:
\begin{equation*}
=\dfrac{1}{2\sqrt{\pi }}\int\limits ^{\infty }_{-\infty } d\xi \dfrac{1}{\xi }\dfrac{\partial }{\partial \xi }\int \dfrac{d\alpha ^{*} \land d\alpha }{2\pi i}\dfrac{1}{\xi ^{2}}\int \dfrac{d\beta ^{*} \land d\beta }{2\pi i\left( \alpha ^{*} -\alpha \right)}\exp\left( -\dfrac{\alpha ^{*} \alpha }{\xi ^{2}} -\dfrac{\beta ^{*} \beta }{\alpha ^{*} -\alpha }\right) \cdot \exp\left( \beta ^{*} \beta \hat{K}^{-1}(\mathbf{y} ,\mathbf{y})\right) =
\end{equation*}
And now very weird thing. The integral over $\displaystyle \xi $ is simple, we can evaluate it:
\begin{gather*}
=\dfrac{1}{2\sqrt{\pi }}\int\limits ^{\infty }_{-\infty } d\xi \dfrac{1}{\xi }\dfrac{\partial }{\partial \xi }\int \dfrac{d\alpha ^{*} \land d\alpha }{2\pi i}\dfrac{1}{\xi ^{2}}\int \dfrac{d\beta ^{*} \land d\beta }{2\pi i\left( \alpha ^{*} -\alpha \right)}\exp\left( -\dfrac{\alpha ^{*} \alpha }{\xi ^{2}} -\dfrac{\beta ^{*} \beta }{\alpha ^{*} -\alpha }\right) \cdot \exp\left( \beta ^{*} \beta \hat{K}^{-1}(\mathbf{y} ,\mathbf{y})\right) =\\
=\dfrac{1}{4}\int \dfrac{d\alpha ^{*} \land d\alpha }{2\pi i}\int \dfrac{d\beta ^{*} \land d\beta }{2\pi i}\dfrac{\left( \alpha ^{*} \alpha \right)^{-3/2}}{\alpha ^{*} -\alpha }\exp\left( -\dfrac{\beta ^{*} \beta }{\alpha ^{*} -\alpha } +\beta ^{*} \beta \hat{K}^{-1}(\mathbf{y} ,\mathbf{y})\right)
\end{gather*}
And here we easily see, that integral over $\displaystyle \alpha $ is suddenly divergent at $\displaystyle \alpha \rightarrow 0$! Why?! How?! I understand, that in this case Hubbard-Stratonovich transformation is not the most rational way to obtain the answer, but I ask this question due to a little bit more complicated problem, where you can't just introduce sources and get the answer. Thanks everybody in advance!
 A: I suppose the case is related to some vague mathematical thing like changing of limits etc. So, we can think about it in such a way: okay, we have been playing with integrals and the integrals have won, so let us regularize the answer to kill the divergency related to our vague tricks. Let $\displaystyle \alpha =x+iy,\ \alpha ^{*} =x-iy$, then we write:
\begin{array}{l}
=\dfrac{1}{8i\pi }\int \dfrac{d\beta ^{*} \land d\beta }{2\pi i}\int \dfrac{dxdy}{y\left( x^{2} +y^{2}\right)^{3/2}}\exp\left( i\ \dfrac{\beta ^{*} \beta }{2y} +\beta ^{*} \beta \hat{K}^{-1}(\mathbf{y} ,\mathbf{y})\right) =\\
=\dfrac{1}{4i\pi }\int \dfrac{d\beta ^{*} \land d\beta }{2\pi i}\int \dfrac{dy}{y^{3}}\exp\left( i\ \dfrac{\beta ^{*} \beta }{2y} +\beta ^{*} \beta \hat{K}^{-1}(\mathbf{y} ,\mathbf{y})\right) =\\
=\dfrac{1}{4i\pi }\int \dfrac{d\beta ^{*} \land d\beta }{2\pi i}\int d\eta \ \eta \ \exp\left( i\ \dfrac{\beta ^{*} \beta }{2} \eta +\beta ^{*} \beta \hat{K}^{-1}(\mathbf{y} ,\mathbf{y})\right) =\\
\dfrac{1}{2i\pi }\int \dfrac{d\beta ^{*} \land d\beta }{2\pi i} \delta '\left( \beta ^{*} \beta \right)\exp\left( \beta ^{*} \beta \hat{K}^{-1}(\mathbf{y} ,\mathbf{y})\right) \sim \hat{K}^{-1}(\mathbf{y} ,\mathbf{y})
\end{array}
It is definitely not a rigorous calculation and I still do not understand the nature of this divergence. However, it is at least some way to obtain the answer.
