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Try to calculate this integral: $$\int_{-\infty}^\infty \exp(z_0t- t^2)d t,$$ where $z_0=x_0+i y_0$ is a fixed complex number.

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\begin{align} I & = \int_{-\infty}^{\infty} \exp(z_0 t - t^2)dt = \int_{-\infty}^{\infty} \exp(-(t-z_0/2)^2+z_0^2/4)dt\\ & = \exp(z_0^2/4)\int_{-\infty}^{\infty} \exp(-(t-z_0/2)^2)dt = \exp(z_0^2/4)\int_{-\infty-z_0/2}^{\infty-z_0/2} \exp(-y^2)dy \end{align} Since the integrand $\exp(-y^2)$ is analytic, we can move the contour $$\Gamma_{z_0} = -\infty-z_0/2 \text{ to } \infty-z_0/2$$ to the contour $$\Gamma_{0} = -\infty \text{ to } \infty$$ and the value of the integral remains the same. This is due the fact that $$\int_{-T-z_0/2}^{T-z_0/2} f(z) dz + \int_{T-z_0/2}^{T-x_0/2} f(z) dz + \int_{T-x_0/2}^{-T-x_0/2} f(z) dz + \int_{-T-x_0/2}^{-T-z_0/2} f(z) dz = 0$$ whenever $f(z)$ is analytic. It is now easy to show that $\displaystyle \int_{T-z_0/2}^{T-x_0/2} f(z) dz \to 0$ and $\displaystyle \int_{-T-x_0/2}^{-T-z_0/2} f(z) dz \to 0$ as $T \to \infty$ since the integrand itself $\to 0 $ as $T \to \infty$.

Hence, $$\int_{\Gamma_{z_0}} f(z) dz = \int_{\Gamma_{0}} f(z) dz$$ From this we can conclude that $$I = \exp(z_0^2/4)\int_{-\infty-z_0/2}^{\infty-z_0/2} \exp(-y^2)dy = \exp(z_0^2/4)\int_{-\infty}^{\infty} \exp(-y^2)dy = \exp(z_0^2/4)\sqrt{\pi}$$

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Sorry, two point I have lost:1. why $y\to t-z0/2$ holds; 2. why the move of the contour hold? your explanation is not so clear, maybe give the exactly contour will help me figuring it out? –  van abel Nov 21 '12 at 4:45
    
good point, I almost figured it out! –  van abel Nov 21 '12 at 4:58

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