I have this integral $$u(x,t)=\int _{-\infty}^{\infty} f(\eta)\left[\frac{1}{2\pi}\int _{-\infty}^{\infty}e^{iw(x-\eta)-w^2t}\ dw\right]\ d\eta=\int _{-\infty}^{\infty}k(x-\eta,t)f(\eta)\ d\eta$$ I want to prove $$k(x,t)=\frac{1}{\sqrt{4\pi t} }\ e^{\Large \frac{-x^2}{4t}} $$
Thanks for helping me out.
Completing square gives \begin{align*} &= \int_{-\infty}^\infty \exp ( i\omega (x-\eta ) - \omega^2t)d\omega \\ &= \int_{-\infty}^\infty \exp \left( - \left( i (x-\eta ) \cdot \frac{1}{2\sqrt t}\right )^2+2 \omega \sqrt{t} i (x-\eta ) \cdot \frac{1}{2\sqrt t} - \omega^2t + \left( i (x-\eta ) \cdot \frac{1}{2\sqrt t}\right )^2\right)d\omega \\ &= \exp\left(- \frac{(x-\eta)^2}{4t} \right )\int_{-\infty}^\infty \exp \left(- \left(\omega \sqrt t - \frac {i (x-\eta)} {2 \sqrt t} \right )^2 \right )d\omega\\ &= \exp\left(- \frac{(x-\eta)^2}{4t} \right) \frac 1 {\sqrt t} \int_{-\infty}^\infty \exp \left(- \left(\omega \sqrt t - \frac {i (x-\eta)} {2 \sqrt t} \right )^2 \right )d(\omega \sqrt t) \\ &= \exp\left(- \frac{(x-\eta)^2}{4t} \right) \sqrt{\frac \pi t} \end{align*}
In other words, we want to prove $$ k(x-\eta,t)=\frac1{2\pi}\int_{-\infty}^\infty e^{\large i(x-\eta)\omega-t\omega^2}\ d\omega $$ or $$ k(x,t)=\frac1{2\pi}\int_{-\infty}^\infty e^{\large ix\omega-t\omega^2}\ d\omega=\frac1{\pi}\int_{0}^\infty e^{\large -(t\omega^2+ix\omega)}\ d\omega $$ In general (click the link below for the complete proof) $$ \int_{x=0}^\infty e^{-(ax^2+bx)}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}\exp\left(\frac{b^2}{4a}\right). $$ We have $a=t$ and $b=ix$, therefore \begin{align} k(x,t)&=\frac1{\pi}\frac{1}{2}\sqrt{\frac{\pi}{t}}\exp\left(\frac{(ix)^2}{4t}\right)=\large\color{blue}{\frac1{\sqrt{4\pi t}}\ e^{\Large-\frac{x^2}{4t}}}.\qquad\qquad\blacksquare \end{align}
