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 .