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Show that $$\int_{-\infty}^{\infty}\exp\bigg({-\bigg(\sqrt {4\pi^{2} t}\xi+\frac{i(y-x)}{2\sqrt t}~\bigg)^{2}}\bigg)~d\xi =\frac{1}{\sqrt {4\pi t}}~,$$ where $x,y\in{\bf R}~,t\in{\bf R}_{>0}$ are given .

Here is my working :

For $x,y\in{\bf R}~,t\in{\bf R}_{>0},$ we have \begin{align} \int_{-\infty}^{\infty}\exp\bigg({-\bigg(\sqrt {4\pi^{2} t}\xi+\frac{i(y-x)}{2\sqrt t}~\bigg)^{2}}\bigg)~d\xi&=\int_{-\infty}^{\infty}~~e^{-\big(\xi{\sqrt{4\pi^{{2}}t}}\big)^{2}}~d\xi~~~\color{blue}{-(1)}~\\ &=\frac{1}{\sqrt {4\pi t}}\int_{-\infty}^{\infty}e^{-({\sqrt{\pi}\xi})^{2}}~d\xi\\ &=\frac{1}{\sqrt {4\pi t}}\\ \end{align} ,where we use the fact that $\displaystyle\int_{\bf R}e^{-s^{2}}ds=\sqrt{\pi}$ and change of variable .

Here is a detail we use the contour integration to yield the equation $\color{blue}{(1)}$ :

Now , take $z=\sqrt {4\pi^{2} t}\xi+\frac{i(y-x)}{2\sqrt t}$ and note that $e^{-z^{2}}$ is an analytic function in the complex plane so its integral around the rectangle is $0$ by Cauchy's theorem , and hence for all $a>0$ we have \begin{align} 0&=\oint e^{-z^{2}}dz\\ &=\int_{-a}^{a}e^{-\big(\xi{\sqrt{4\pi^{2}t}\big)^{2}}}d\xi+\int_{0}^{\frac{y-x}{2\sqrt t}}e^{-\big(a\sqrt{4\pi^{2}t}+\eta i\big)^{2}}d\eta\\&+\int_{a}^{-a}e^{-\big(\xi\sqrt{4\pi^{2}t}+i\frac{y-x~~}{2\sqrt{t}}\big)^{2}}d\xi+\int_{\frac{y-x}{2\sqrt{t}}}^{0}e^{-\big(-a\sqrt{4\pi^{2}t}+\eta i\big)^{2}}d\eta \end{align}

Whence , \begin{align} &\bigg|\int_{-a}^{a}e^{-\big(\xi\sqrt{4\pi^{2}t}+i\frac{y-x~~}{2\sqrt{t}}\big)^{2}}d\xi-\int_{-a}^{a}e^{-\big(\xi\sqrt{4\pi^{2}t}\big)^{2}}d\xi~\bigg|\\=&~\bigg|\int_{0}^{\frac{y-x}{2\sqrt t}}e^{-\big(a\sqrt{4\pi^{2}t}+\eta i\big)^{2}}d\eta-\int_{0}^{\frac{y-x}{2\sqrt t}}e^{-\big(-a\sqrt{4\pi^{2}t}+\eta i\big)^{2}}d\eta\bigg|\\ =&~\bigg|\int_{0}^{\frac{y-x}{2\sqrt t}}e^{-\big(a\sqrt{4\pi^{2}t}+\eta i\big)^{2}}-e^{-\big(-a\sqrt{4\pi^{2}t}+\eta i\big)^{2}}d\eta\bigg|\\ \le&\int_{0}^{\frac{|y-x|}{2\sqrt{t}}}\bigg|e^{-\big(a\sqrt{4\pi^{2}t}+\eta i\big)^{2}}-e^{-\big(-a\sqrt{4\pi^{2}t}+\eta i\big)^{2}}~\bigg|~d\eta\\ =&~2e^{-4\pi^{2}ta^{2}}\int_{0}^{\frac{|y-x|}{2\sqrt{t}}} e^{\eta^{2}}~d\eta\\ \le&~2e^{-4\pi^{2}ta^{2}}\cdot\frac{|y-x|}{2\sqrt{t}} e^{\frac{(y-x)^{2}}{4t}}<\infty \end{align}

Therefore , we yield the equation $\color{blue}{(1)}$ provided that $a\longrightarrow \infty$ .

Can anyone check my proof for validity if you have the time, otherwise ignore it that it's okay . Any comment or valuable suggestion I will be grateful . Thanks for your patient reading .

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  • $\begingroup$ I've taken a brief skim of the proof the conceptionally it seems very accurate aside from some grammar errors to avoid confusion and maybe to make it easier on the behalf of the reader perhaps you could define your rectangular contour it might make things a bit more clearer $\endgroup$ – Zophikel Aug 29 '18 at 23:42

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