Use Plancherel theorem to prove the identity

Question

Plancherel theorem states $\int \hat f(x) \hat g(x) \, dx=\int f(x)g(x) \, dx$, where $\hat f$ denotes the Fourier transformation of $f$. It is required to prove this: $$\int_0^\infty\frac{\sin(x\xi)}\xi(1-e^{-c^2\xi^2t})d\xi=\sqrt\pi\int_{\frac{x}{2c\sqrt t}}^\infty e^{-\xi^2}d\xi$$ I have obtained this $f(x)=\begin{cases}1,&|x|<1\\0,&|x|>1\\\end{cases}$ and $\hat f(\xi)=\frac{\sin(\xi)}{\sqrt\pi\xi}$, but I don't know how to handle the other part $1-e^{-c^2\xi^2t}$. Can somebody help?

Answer 1

We assume that $x>0$, $t>0$, and $c>0$.

First, note that since $\frac{\sin(x\xi)}\xi(1-e^{-c^2\xi^2t})$ is an even function of $\xi$, we can write

$$\begin{align} \int_0^\infty\frac{\sin(x\xi)}\xi(1-e^{-c^2\xi^2t})\,d\xi&=\frac12\int_{-\infty}^\infty\frac{\sin(x\xi)}\xi(1-e^{-c^2\xi^2t})\,d\xi \tag 1\\\\ &=\frac{\pi}{2}-\frac12 \int_{-\infty}^\infty\frac{\sin(x\xi)}\xi e^{-c^2\xi^2t}\,d\xi \end{align}$$

Next, note that we can write

$$\begin{align} \sqrt\pi\int_{\frac{x}{2c\sqrt t}}^\infty e^{-k^2}\,dk&=\frac{\sqrt\pi}{2} \int_{-\infty}^\infty \left(1-\text{rect}\left(\frac{k}{x/c\sqrt{t}}\right)\right)e^{-k^2}\,dk \\\\ &=\frac{\pi}{2}-\frac{\sqrt\pi}{2} \int_{-\infty}^\infty \text{rect}\left(\frac{k}{x/c\sqrt{t}}\right) e^{-k^2}\,dk \tag 2 \end{align}$$

where $\text{rect}(x)$ is the Rectangular Function

$$\text{rect}(x)=\begin{cases}1&,|x|<1/2\\\\1/2&,x=1/2\\\\0&,|x|>1/2\end{cases}$$

Enforcing the substitution $k\to k/2c\sqrt{t} $ in $(2)$, we obtain

$$\begin{align} \sqrt\pi\int_{\frac{x}{2c\sqrt t}}^\infty e^{-k^2}\,dk&=\frac{\pi}{2}-\frac{\sqrt\pi}{4c\sqrt{t}} \int_{-\infty}^\infty \text{rect}\left(\frac{k}{2x}\right) e^{-k^2/4c^2t}\,dk \tag 3 \end{align}$$

Finally, the Fourier Transform of $\text{rect}\left(\frac{k}{2x}\right)$ is

$$\int_{-\infty}^\infty \text{rect}\left(\frac{k}{2x}\right) e^{ik\xi}\,d\xi=2\frac{\sin(x\xi)}{\xi} \tag 4$$

while the Fourier Transform of $e^{-k^2/4c^2t}$ is

$$\int_{-\infty}^\infty e^{-k^2/4c^2t}e^{ik\xi}\,d\xi=2c\sqrt{\pi}\sqrt{t} e^{-c^2t\xi^2} \tag 5$$

Using Parseval's Theorem along with $(1)$, $(3)$, $(4)$ and $(5)$, we conclude that for $x>0$, $t>0$ and $c>0$

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty\frac{\sin(x\xi)}\xi(1-e^{-c^2\xi^2t})\,d\xi=\sqrt\pi\int_{\frac{x}{2c\sqrt t}}^\infty e^{-k^2}\,dk}$$