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Alexandre Eremenko
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Fourier transform of $f(x)=e^{-tx^2}$ is $\hat{f}(\xi)=ce^{-\xi^2/(4t)}$. Multiplication of $f$ by a polynomial results isin applying a differential operator with constant coefficients to $\hat{f}$, and for our $\hat{f}$ this is equivalent to multiplication on some polynomial. These polynomials play no role in convergence (unless they are both zero), since the expression under the exponent in the product $f(x)\hat{f}(\xi)e^{2|x\xi|}$ is $$-tx^2+2|x\xi|-\xi^2/(4t).$$ Since this quadratic expression is sometimes positive in the first quadrant, the integral diverges, unless $P=0$.

Fourier transform of $f(x)=e^{-tx^2}$ is $\hat{f}(\xi)=ce^{-\xi^2/(4t)}$. Multiplication of $f$ by a polynomial results is applying a differential operator with constant coefficients to $\hat{f}$, and for our $\hat{f}$ this is equivalent to multiplication on some polynomial. These polynomials play no role in convergence (unless they are both zero), since the expression under the exponent in the product $f(x)\hat{f}(\xi)e^{2|x\xi|}$ is $$-tx^2+2|x\xi|-\xi^2/(4t).$$ Since this quadratic expression is sometimes positive in the first quadrant, the integral diverges, unless $P=0$.

Fourier transform of $f(x)=e^{-tx^2}$ is $\hat{f}(\xi)=ce^{-\xi^2/(4t)}$. Multiplication of $f$ by a polynomial results in applying a differential operator with constant coefficients to $\hat{f}$, and for our $\hat{f}$ this is equivalent to multiplication on some polynomial. These polynomials play no role in convergence (unless they are both zero), since the expression under the exponent in the product $f(x)\hat{f}(\xi)e^{2|x\xi|}$ is $$-tx^2+2|x\xi|-\xi^2/(4t).$$ Since this quadratic expression is sometimes positive in the first quadrant, the integral diverges, unless $P=0$.

Source Link
Alexandre Eremenko
  • 91.8k
  • 9
  • 260
  • 431

Fourier transform of $f(x)=e^{-tx^2}$ is $\hat{f}(\xi)=ce^{-\xi^2/(4t)}$. Multiplication of $f$ by a polynomial results is applying a differential operator with constant coefficients to $\hat{f}$, and for our $\hat{f}$ this is equivalent to multiplication on some polynomial. These polynomials play no role in convergence (unless they are both zero), since the expression under the exponent in the product $f(x)\hat{f}(\xi)e^{2|x\xi|}$ is $$-tx^2+2|x\xi|-\xi^2/(4t).$$ Since this quadratic expression is sometimes positive in the first quadrant, the integral diverges, unless $P=0$.