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I try to find the following

$$\int_{\mathbb{R}}^{}\int_{\mathbb{R}}^{} e^{-y(x+z)-(x^2+z^2)} dxdz$$ and I change variables $x=r\cos(\theta)$ and $z=r\sin(\theta)$ and the integral becomes:

$$\int_{0}^{2\pi}\int_{0}^{+\infty} e^{r(-y\cos(\theta)-ysin(\theta)-1)}r drd\theta$$

Then I find the following

$$\int_{0}^{+\infty}{e^{-ax}x} dx = \left[-\frac{1}{a}e^{-ax}x-\frac{1}{a^2}e^{-ax}\right]_0^{+\infty} = \left[0-0-\left(0-\frac{1}{a^2}\right)\right]=\frac{1}{a^2} $$

now we can find the inner integral of the last double integral if we set $a=y\cos(\theta)+y\sin(\theta)+1$, so

$$\int_{0}^{+\infty} e^{r(-y\cos(\theta)-ysin(\theta)-1)}r dr = \frac{1}{(y(\cos(\theta)+sin(\theta))+1)^2}$$

and replace it to

$$\int_{0}^{2\pi}\int_{0}^{+\infty} e^{r(-y\cos(\theta)-ysin(\theta)-1)}r drd\theta=\int_{0}^{2\pi} \frac{1}{(y(\cos(\theta)+\sin(\theta))+1)^2} d\theta$$ after that I don't know what can I do. Any idea ?

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    $\begingroup$ Hint: Pull out the $y$ as it is a constant and rewrite $\cos{\theta}+\sin{\theta}=\sqrt2\cos({\theta+\pi/4)}$ $\endgroup$ – Shrey Aryan Jan 19 '16 at 19:33
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    $\begingroup$ $x^2+z^2=r^2$ not $r$ $\endgroup$ – Jack's wasted life Jan 19 '16 at 19:52
  • $\begingroup$ Where is the connection to Fourier here? $\endgroup$ – Ron Gordon Jan 19 '16 at 20:08
  • $\begingroup$ i try to calculate a fourier transform , then i see i forgot something , but the problem still to solve the above integral , sorry for the mention about fourier transform . i wrote the heading first and forgot to edit ... $\endgroup$ – chaviaras michalis Jan 19 '16 at 20:44
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Why do you need polar coordinates? It seems that the FT is separable:

$$\int_{-\infty}^{\infty} dx \, e^{-x^2-xy} \, \int_{-\infty}^{\infty} dz \, e^{-z^2-zy} = \left (\int_{-\infty}^{\infty} dx \, e^{-x^2-xy} \right )^2$$

Now,

$$\int_{-\infty}^{\infty} dx \, e^{-x^2-xy} = e^{y^2/4}\int_{-\infty}^{\infty} dx \, e^{-(x+y/2)^2} = \sqrt{\pi} \, e^{y^2/4}$$

Thus, the double integral is equal to $\pi \, e^{y^2/2}$.

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