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A delta function can be written as

$$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}dp\,e^{ipx}.$$

I have a very poor understanding of the Wick rotation technique used in quantum field theory. Does it make sense to write the following equation?

$$\delta(x)=\frac{-i}{2\pi}\int_{-i\infty}^{+i\infty}dE\,e^{Ex}$$

This question seems trivial. The reason why I ask this question is that I am confused by an equation from ''Gauge Fields and Strings'' by A. M. Polyakov. The following is taken from chapter 9 of Gauge Fields and Strings, page 153.

\begin{align} G\left(x,x'\right) \boldsymbol{=} &\int\dfrac{\mathscr{D}h(\tau)}{\mathscr{D}f(\tau)}\exp\left(\boldsymbol{-}m_{0}\int\limits_{0}^{1}(h(\tau))^{1/2}\mathrm d\tau\right) \nonumber\\ & \times\int\mathscr{D}x(\tau)\delta(\overset{\,\centerdot}{x}{}^{2}(\tau)\boldsymbol{-}h(\tau)) \tag{9.7}\label{9.7} \end{align} where we have introduced a $^{\prime\prime}$metric tensor$^{\prime\prime}$ $h(\tau)$ on the path and inserted a functional $\delta$-function into the integrand. Let us begin with computation of the second integral in \eqref{9.7} and use a Lagrange multiplier to define the $\delta$-function: \begin{align} \mathscr{K}(x &, x'\!,h(\tau) ) \boldsymbol{=} \nonumber\\ & \boldsymbol{\equiv} \int\mathscr{D}x(\tau)\delta(\overset{\,\centerdot}{x}{}^{2}(\tau)\boldsymbol{-}h(\tau)) \nonumber\\ & \boldsymbol{=}\int\limits_{\boldsymbol{c-\mathrm i\infty}}^{\boldsymbol{c+\mathrm i\infty}}\!\!\!\!\mathscr{D}\lambda(\tau)\exp\left(\int\limits_{0}^{1}\lambda(\tau) h(\tau)\mathrm d\tau\right)\int\mathscr{D}x(\tau)\exp\left(\boldsymbol{-}\int\limits_{0}^{1}\lambda(\tau)\overset{\,\centerdot}{x}{}^{2}(\tau)\mathrm d\tau\right) \tag{9.8}\label{9.8} \end{align}

My question is about the delta functional in equation \eqref{9.8}. In QFT, the delta functional is usually written as

$$\delta[\dot{x}^{2}(\tau)-h(\tau)]=\int\mathcal{D}\lambda(\tau)\exp\left(i\int_{0}^{1}d\tau\lambda(\tau)\left[h(\tau)-\dot{x}^{2}(\tau)\right]\right).$$

It seems that Polyakov is doing a "Wick Rotation" of $\lambda(\tau)$ to give it an imaginary part so that one can get rid of the factor $i$ in the exponential. Is that correct?

What does it mean to write the functional integral

$$\int_{c-i\infty}^{c+i\infty}\mathcal{D}\lambda(\tau)?$$

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  • $\begingroup$ Shouldn't it just be understood as a complex contour integral? As in: $\int_{c-i\infty}^{c+i\infty}\mathcal{D}\lambda(\tau) f\left[\lambda(\tau)\right] = i \int_{-\infty}^{\infty}\mathcal{D}\lambda(\tau) f\left[c+ i \lambda(\tau)\right]$ were on the right it's just an ordinary real scalar field being integrated. $\endgroup$ – Tal Sheaffer Jan 26 at 22:15
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    $\begingroup$ Yes, OP is right, it's defined via Wick rotation. Similar question on Math.SE: math.stackexchange.com/q/2829562/11127 $\endgroup$ – Qmechanic Jan 26 at 23:08

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