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I'm asked to calculate the Fourier transform of $f(x)=e^{-x^{2}-y^2-xy}$ and I have no idea how to do this. Can someone help me?

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  • $\begingroup$ Make a linear transformation to turn it into $e^{-u^2-v^2}$. You should work out the details yourself. $\endgroup$ – WhatsUp Oct 17 at 13:23
  • $\begingroup$ Do you know how to diagonalize a quadratic form? $\endgroup$ – md2perpe Oct 17 at 14:31
  • $\begingroup$ Did you mean the 2 variable Fourier transform of $f(x,y)$ ? If so, either we can do as the other users said, or we can evaluate the double integral directly, in $x$ then in $y$ (knowing the bilateral Laplace transform of a Gaussian, which follows from a bit of complex analysis) $\endgroup$ – reuns Oct 17 at 16:06
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Correcting error. If you mean 2 dim. Fourier transform as mentioned in comment by reuns and in the format of $F(s,w)=\frac 1{2\pi}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y)\exp(\!-\!isx\!-\!iwy)\:dx\:dy\;$ so $\;f(x,y)=\frac1{2\pi}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} F(s,w)\exp(isx\!+\!iwy)\:ds\:dw\;$ and you really meant $f(x,y)=e^{(-x^2-y^2\!-xy)}\;$ then the answer should be $F(s,w)\!=\!\frac1{\sqrt{3}}\exp({\frac{sw-s^2-w^2}3})\;$ but if you really meant only a function of the one variable x with y as a cst or parameter in the format $F(s)\!=\!\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)\exp(\!-\!isx)\:dx\;$ so $\;f(x)=\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}F(s)\exp(isx)\:ds\;$ then $F(s)\!=\!\ \frac1{\sqrt{2}}\exp({\frac{2isy-s^2-3y^2}4})\;$ Very briefly to get the answer as mentioned in comments you first complete a square to rid the linear term in the exponent so it's only a function of one quadratic variable but in the completing of the square there will also be another factor considered constant {since it is not the integration variable atleast not in the first integration} function, (...) generally complex, of form $e^{(...)}$ times a gaussian integral of form $\int_{-\infty}^{\infty}e^{-ax^2}\:dx\;$ which would assume you can do. Or in a little more detail in this particular case use $\exp{(ax^2\!+\!bx)}=\exp{(-\frac{b^2}{4a})}\exp{\left (a(x\!+\!\frac b{2a})^2\right )}$, and substitute say $z=x\!+\!\frac b{2a}$, noting $dx\!=\!dz$ , so then take $\exp{(-\frac{b^2}{4a})}\int_{-\infty}^{\infty}\exp{(az^2)}\:dz\;$ but there is a little more to it such as also multiplying by a phase factor and perhaps another factor which result from details to consider when doing contour integration since at first integration $b\!=\!-y\!-\!is$ is complex so will also need to do some complex analysis which I won't go into here.

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