The Fourier transform of conjugate Poisson Kernel.

Question

From the very early begining we knew that $P_t(x)=\frac{t}{\pi (t^2+x^2)}$ is the poisson kernel for upper half plane. Now we knew that \begin{equation}\hat{P_t}(\xi)=e^{-t|\xi|}\label{e}.\end{equation} Now the conjugate Poisson kernel is defined by $Q_t(x)=\frac{x}{\pi(x^2+t^2)}$. Now to prove that $\hat{Q_t}(\xi)=(-i)sgn(\xi) e^{-t\lvert\xi\rvert}$ using that formula $\hat{P_t}(\xi)=e^{-t|\xi|}$. I could not make any progress and also the existing stack answer could not help me in getting that. Can anyone give me some hint.

Answer 1

METHODOLOGY $1$: COMPLEX ANALYSIS

Let $Q_t(x)=\frac{x}{\pi^2(x^2+t^2)}$. Note that the Fourier Transform of $Q_t(x)$ is given by

$$\begin{align} \hat Q_t(\xi)&=\int_{-\infty}^\infty \frac{x}{\pi^2(x^2+t^2)}e^{i\xi x}\,dx\\\\ &=\begin{cases} 2\pi i \text{Res}\left(\frac{z}{\pi^2(z^2+t^2)}, z=it\right), &\xi>0\\\\ -2\pi i \text{Res}\left(\frac{z}{\pi^2(z^2+t^2)}, z=-it\right), &\xi<0 \end{cases}\\\\ &=\frac{i}{\pi}\text{sgn}(\xi)e^{-|\xi t| } \end{align}$$

where we used contour integration and the Residue Theorem to arrive at the coveted result.

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METHODOLOGY $2$: REAL ANALYSIS

Alternatively, we can evaluate the Fourier Transform using real analysis. Proceeding, we see that

$$\begin{align} \hat Q_t(\xi)&=\int_{-\infty}^\infty \frac{x}{\pi^2(x^2+t^2)}e^{i\xi x}\,dx\\\\ &=\int_{-\infty}^\infty \frac{(x^2+t^2)-t^2}{\pi^2x(x^2+t^2)}e^{i\xi x}\,dx\\\\ &=\frac i \pi \text{sgn}(\xi)-\frac{t^2}{\pi^2}\int_{-\infty}^\infty \frac{1}{x(x^2+t^2)}e^{i\xi x}\,dx \end{align}$$

Differenting twice, we find that

$$\begin{align} \hat Q_t''(\xi)&=\frac{t^2}{\pi^2}\int_{-\infty}^\infty \frac{x}{x^2+t^2}e^{i\xi x}\,dx\\\\ &=t^2\hat Q_t(\xi) \end{align}$$

Solving the ODE, $\hat Q_t(\xi)=C_1e^{|t|\xi}+C_2e^{-|t|\xi}$.

Now, as $t\to \infty$, $\hat Q_t(\xi)\to 0$. Hence, we must have $\hat Q_t(\xi)=C^+e^{-|t|\xi}$ for $\xi>0$ and $\hat Q_t(\xi)=C^- e^{|t|\xi}$ for $\xi<0$.

We also see that $\hat Q_t'(0)=-i|t|/\pi$ and $\hat Q_t'(0)=- \text{sgn}(\xi) |t|C^\pm$. Therefore, $C^\pm=\frac{i}\pi \text{sgn}(\xi)$.

Putting it all together, we have

$$\hat Q_t(\xi)=\frac{i}{\pi}\text{sgn}(\xi)e^{-|t\xi|}$$

as expected!