- #1

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[tex]\int_{-\infty}^{\infty} e^{-Bx^2} e^{-iAx} dx[/tex]

Splitting it in two using Euler's identity show that the imaginary part is 0 (cuz integrand is odd). Remains the real part

[tex]2 \int_0^{\infty} cos(-Ax) e^{-Bx^2} dx[/tex]

for which integration by parts leads nowhere.