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I have the following integral: $$ I_1= \int_{-\infty}^{\infty} \frac{d\tau}{2\pi i} \int_{-\infty}^{\infty} \frac{d\tau'}{2\pi i} \frac{1}{(\tau - i \epsilon)(\tau' - i\epsilon')}. M(\tau, \tau')$$ where

$$M(\tau, \tau') = exp\Bigg[-\tau^2 -\tau'^2 - \tau \tau' \Bigg[2-AB.\sqrt{2\pi}. e^{-\frac{A^2B^2}{2}}. \Big[\frac{erf\Big[{\frac{iAB}{\sqrt{2}}}\Big]}{i} + i\Big]\Bigg]\Bigg].$$

Here $erf$ is the error function. Also $\epsilon$ and $\epsilon'$ are small positive constants (primed variables don't mean a derivative, they are independent variables) and $A$ is a real variable and $B$ is a real constant ($B \neq 0$). I solved the above integral in the limit $A \to 0$, where I get the following simplification for the coefficient of $\tau \tau'$:

$$\lim_{A \to 0} \Bigg[2-AB.\sqrt{2\pi}. e^{-\frac{A^2B^2}{2}}. \Big[\frac{erf\Big[{\frac{iAB}{\sqrt{2}}}\Big]}{i} + i\Big]\Bigg]\Bigg] = 2,$$

thereby making the integral $I_1$ to be of the following form

$$I_1= \int_{-\infty}^{\infty} \frac{d\tau}{2\pi i} \int_{-\infty}^{\infty} \frac{d\tau'}{2\pi i} \frac{e^{-(\tau + \tau')^2}}{(\tau - i \epsilon)(\tau' - i\epsilon')}.$$

The above integral can be evaluated to give $I_1 = \frac{1}{2}$. Also in the limit $A \to \infty$,

$$\lim_{A \to \infty} \Bigg[2-AB.\sqrt{2\pi}. e^{-\frac{A^2B^2}{2}}. \Big[\frac{erf\Big[{\frac{iAB}{\sqrt{2}}}\Big]}{i} + i\Big]\Bigg]\Bigg] = 0,$$

so therefore we get

$$I_1= \int_{-\infty}^{\infty} \frac{d\tau}{2\pi i} \int_{-\infty}^{\infty} \frac{d\tau'}{2\pi i} \frac{e^{-\tau^2 - \tau'^2}}{(\tau - i \epsilon)(\tau' - i\epsilon')}.$$

The above integral can be evaluated to give $I_1 = \frac{1}{4}$.

So my question is this: Does there exist any value of $A \neq 0$ for which $I_1 = \frac{1}{2}$. Can it be proved that the integral $I_1 = \frac{1}{2}$ iff $A=0$?

EDIT: An useful thing might be to see that the following function is

$$\Bigg[2-AB.\sqrt{2\pi}. e^{-\frac{A^2B^2}{2}}. \Big[\frac{erf\Big[{\frac{iAB}{\sqrt{2}}}\Big]}{i} + i\Big]\Bigg]\Bigg] = 2$$

iff $A=0$.

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  • $\begingroup$ The integrals depend on $\epsilon$ and $\epsilon'$. If you mean the limits, take the closed form computed here and prove that the only possible value is $a = 1$. $\operatorname{erf}(i x)/i$ is real, therefore $i A B = 0$. $\endgroup$ – Maxim Oct 28 '18 at 7:21

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