Prove that inverse fourier cosine transform of $\exp(-tkw^2)=\frac{1}{\sqrt{2kt}}\exp(-\frac{x^{2}}{4kt}) $

Question

In the process of solution of a PDE via Fourier cosine transform the author assumes at one step $$F_{c}^{-1}\exp(-tkw^2)=\frac{1}{\sqrt{2kt}}\exp(-\frac{x^{2}}{4kt}) $$ where Fc^{-1} is fourier cosine inverse operator. I know this is derived from formula of inverse fourier cosine transform which is $$F_{c}^{-1}[F_{c}(w)]=f(x)=\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}cos(wx)F_{c}(w)dw$$ It means I have to prove $$\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}cos(wx)\exp(-tkw^2)dw=\frac{1}{\sqrt{2kt}}\exp(-\frac{x^{2}}{4kt})$$ There are similar questions like inverse fourier transform but I am unable to comprehend analogy between them and my question.

Answer 1

You can use the same differentiation trick to find the transform: \begin{align} \frac{d}{dx}\int_{0}^{\infty}\cos(wx)e^{-tkw^2}dw & = -\int_{0}^{\infty}w\sin(wx)e^{-tkw^2}dw \\ & = \int_{0}^{\infty}\sin(wx)\frac{1}{2kt}\frac{d}{dw}e^{-tkw^2}dw \\ & = \frac{1}{2kt}\left[\left.\sin(wx)e^{-tkw^2}\right|_{w=0}^{\infty}-x\int_{0}^{\infty}\cos(wx)e^{-tkw^2}dw\right] \\ & = -\frac{x}{2kt}\int_{0}^{\infty}\cos(wx)e^{-tkw^2}dw. \end{align} That gives a first order differential equation for the Fourier cosine transform, resulting in a constant $C$ that does not depend on $x$ (but may depend on $k$, $t$) such that $$ \int_{0}^{\infty}\cos(wx)e^{-tkw^2}dw=Ce^{-x^2/4kt} $$ Setting $x=0$ gives the constant $C(k,t)$: \begin{align} C(k,t)&=\int_{0}^{\infty}e^{-tkw^2}dw \\ &= \frac{1}{\sqrt{tk}}\int_{0}^{\infty}e^{-(w\sqrt{tk})^2}d(w\sqrt{tk}) \\ & = \frac{1}{\sqrt{tk}}\int_{0}^{\infty}e^{-u^2}du \\ & = \frac{1}{\sqrt{tk}}\frac{\sqrt{\pi}}{2} \end{align} Therefore the Fourier cosine transform is \begin{align} \sqrt{\frac{2}{\pi}}\int_{0}^{\infty}\cos(wx)e^{-tkw^2}dw & = \sqrt{\frac{2}{\pi}}\frac{1}{\sqrt{kt}}\frac{\sqrt{\pi}}{2}e^{-x^2/4kt} \\ & = \frac{1}{\sqrt{2kt}}e^{-x^2/4kt}. \end{align}