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I struggle with finalizing my argument and making sure all my steps are correct.

I realize it might be good to look at the integral this way: $\iint\limits_{\mathbb{R^2}} |\langle \begin{pmatrix} 1 \\ t\end{pmatrix} , \vec{x}\rangle| e^{-\langle \frac{1}{2}I\vec{x}, \vec{x}\rangle}dx$.

Then, use substitution in a way that $x' = A^{-1} x, A^{-1} = \begin{pmatrix} 1 & t \\ -t & 1\end{pmatrix}$, for the sake of simplifying the dot product. (Resulting with $A = \frac{1}{1+t^2} \begin{pmatrix} 1 & -t \\ t & 1\end{pmatrix}$)

In that way, I can get: $\iint\limits_{\mathbb{R^2}} |x_1| e^{-\langle \frac{1}{2}IA\vec{x}, A\vec{x}\rangle}|det{A}|dx_1dx_2$, which due to symmetry of 1st-4th quarters and 2nd-3rd quarters, results with

2$\iint\limits_{\mathbb{R^2\ right\ plane}} x_1 e^{-\langle A^t\frac{1}{2}IA\vec{x}, \vec{x}\rangle}dx_1dx_2$.

I am then stuck with making sure of my steps, as it looks like an integration by part might help, but it is abit tricky to me, and it looks like I want to take advantage of the fact that $\iint\limits_{\mathbb{R^n}} e^{-\langle B\vec{x}, \vec{x}\rangle}dx = \frac{\pi^{\frac{n}{2}}}{\sqrt{|detB|}}$ for positive definite matrices.

The solution should result with $\sqrt{8\pi(1+t^2)}$.

Thanks in advance to any helpers!

Eyal Skop
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    Hi, welcome to Math SE. Hint: you should find this transformation rewrites your integral as $\iint_{\Bbb R^2}\frac{1}{t^2+1}|x_1|\exp\left(-\frac{x_1^2+x_2^2}{2(t^2+1)}\right)dx_1dx_2$. Now switch to polar coordinates viz. $x_1=r\cos\theta,,x_2=r\sin\theta$. If you wish to use that determinant identity in the polar step, the relevant special case has $B$ proportional to the identity matrix, so you get $\int_{\Bbb R}e^{-Br^2}dr=\sqrt{\pi/B}$ for real $B>0$; now take $-\partial_B$ to get the $r^2$ factor. – J.G. May 29 '23 at 14:02

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Following your approach we have that: $$\begin{cases} u=x+ty\\ v=-tx+y \end{cases} \implies \begin{cases}x=\frac{u-tv}{1+t^2}\\ y=\frac{tu+v}{1+t^2} \end{cases}.$$ Therefore $$\begin{align} \iint_{\mathbb{R}^2}{|x+ty|e^{-\frac{x^2+y^2}{2}}}dxdy &=\iint_{\mathbb{R}^2}{|u|e^{-\frac{(u-tv)^2+(tu+v)^2}{2(1+t^2)^2}}}\frac{dudv}{1+t^2} \\&=\frac{1}{1+t^2}\iint_{\mathbb{R}^2}{|u|e^{-\frac{u^2+v^2}{2(1+t^2)}}}dudv\\ &=\frac{1}{1+t^2}\int_{0}^{\infty}\int_{0}^{2\pi}{r|\cos(t)|e^{-\frac{r^2}{2(1+t^2)}}}rdrdt\\ &=\frac{4}{1+t^2}\int_{0}^{\infty}{r^2e^{-\frac{r^2}{2(1+t^2)}}}dr\\ &=4\sqrt{1+t^2}\int_{0}^{\infty}s^2e^{-s^2/2}ds\\ &=4\sqrt{1+t^2}\sqrt{\frac{\pi}{2}}=\sqrt{8\pi(1+t^2)} \end{align}$$ where $s=\frac{r}{\sqrt{1+t^2}}$ and we applied Integrate $x^2 e^{-x^2/2}$ .

Robert Z
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