Fourier Transform of $v(x)=\frac{e^{ix}}{1+9x^2}$

Question

For a function $u(t)$, its Fourier transform is $\widehat{u}(x) = \int_{\mathbb{R}} u(t) e^{-itx} dt$.

I have to calculate a Fourier Transform of $v(x)=\dfrac{e^{ix}}{1+9x^2}$.

I know that the Fourier transform of $\dfrac{1}{1+x^2}$ is $\pi e^{-|x|}$. But it doesn't help me to calculate the Fourier transform of $v$. I really don't know how to do it. Could someone help me ? Thank you in advance.

Answer 1

Let $\hat f$ be the Fourier transform of $f$.

$$f(x)=\frac{1}{1+x^2},~~~\hat f(\omega)=\pi e^{-|\omega|}$$ $$g(x)=f(3x)=\frac{1}{1+9x^2},~~~\hat g(\omega)=\frac1{|3|}\hat f\left(\frac\omega 3\right)=\frac{\pi e^{-|\omega|/3}}{3}$$ $$h(x)=g(x)e^{ix}=\frac{e^{ix}}{1+9x^2},~~~\hat h(\omega)=g(\omega-1)=\frac{\pi e^{-|\omega-1|/3}}{3}$$

This result is what also Mathematica says:

Integrate[Exp[I x]/(1 + 9 x^2) Exp[-I t x], {x, -\[Infinity], \[Infinity]}]
ConditionalExpression[1/3 E^(-(1/3) Abs[-1 + t]) \[Pi], t \[Element] Reals]