2
$\begingroup$

I need to show that $\int \limits_{\mathbb{R}^2}|x+ty|e^{-(x^2+y^2)/{2}}dxdy=\sqrt{8\pi (t^2+1)}$, for any given $t \in \mathbb{R}$.

I've tried the substitution $\left\{\begin{matrix} u=x+ty & \\ v= x-ty & \end{matrix}\right.$ that didn't work.

Also, polar coordinates yield: $ I=\left (\int_{0}^{\infty}r^2e^{-r^2/2}dr \right ) \left (\int_{0}^{2\pi}|\cos\theta + t\sin \theta| d\theta \right ) $ which is already a better expression but I still don't think that it can lead to an answer.

Any other ideas?

  • 0
    can't you restrict yourself by symmetry to a quadrant of the plane where the absolute value is not necessary? – 2017-01-24
  • 0
    I think that actually works, either. I'll try that too. – 2017-01-24

2 Answers 2

2

The substitution $$ u = \frac{x+t y}{\sqrt{1+t^2}},\qquad v = \frac{t x-y}{\sqrt{1+t^2}} $$ works far better, leading to

$$ \sqrt{1+t^2}\int_{\mathbb{R}^2} |u| e^{-\frac{u^2+v^2}{2}}\,du\,dv = \sqrt{8\pi(1+t^2)}\int_{0}^{+\infty}u e^{-u^2/2}\,du = \color{red}{\sqrt{8\pi (1+t^2)}} $$ as wanted, by Fubini's theorem and parity.

1

Kind of "cheating" by using properties of the Gaussian distribution, may be useful to others besides the OP:

If $X$ and $Y$ are independent standard normal, the density of $(X,Y)$ is $\frac{1}{2\pi} e^{-(x^2+y^2)/2}$, and $X+tY \sim N(0, 1+t^2)$. Then, if $Z$ is independent standard normal, $\sqrt{1+t^2}Z \overset{d}{=} X+tY$, so \begin{align} \frac{1}{2\pi}\iint |x+ty| e^{-(x^2+y^2)/2} \mathop{dx} \mathop{dy} &= \mathbb{E}|X+tY|\\ &= \sqrt{1+t^2}\mathbb{E} |Z|\\ &= \sqrt{\frac{2(1+t^2)}{\pi}}. \end{align} [The step $\mathbb{E} |Z|=\sqrt{2/\pi}$ can be obtained by integrating directly.] Multiplying both sides by $2\pi$ gives your answer.