# About $\displaystyle\int_{\mathbb{R}_y^3}\int_{\mathbb{S}^2}e^{-\frac{1}{2}|x-[(x-y)\cdot\omega]\omega|^2}d\omega dy$

An integral has been pushed me over the edge for several weeks. It reads as: $$\displaystyle\int_{\mathbb{R}_y^3}\int_{\mathbb{S}^2}e^{-\frac{1}{2}|x-[(x-y)\cdot\omega]\omega|^2}d\omega dy$$

I tried to calculate the surface integral inside using spherical coordinates, but it seems that I couldn't do any further calculation since the integrand function is something like $$e^{-\big(k_1(\varphi)\sin^2\theta+k_2(\varphi)\cos^2\theta+k_3(\varphi)\sin\theta\cos\theta\big)}\sin\theta .$$ Then I tried to use variable substitution to compute, similarly, I didn't get anything useful. I was also trying to use Maple to compute, but it didn't work at all. My original intention is to prove that the formula $$\displaystyle e^{-\frac{1}{2}|x|^2}\int_{\mathbb{R}_y^3}\int_{\mathbb{S}^2}e^{-\frac{1}{2}|x-[(x-y)\cdot\omega]\omega|^2}d\omega dy$$ is bounded.

I would be grateful if you could give me a definite result.

## 1 Answer

Without loss of generality you can orient the axes so that the vector $$\mathbf{x}$$ is along the $$x_3$$ axis. Using also that $$|\mathbf{\omega}|=1$$, one has $$|\mathbf{x}-[(\mathbf{x}-\mathbf{y})\cdot\mathbf{\omega}]\mathbf{\omega}|^2=(\mathbf{\omega}\cdot\mathbf{y})^2-x_3^2(\omega_3^2-1).$$ Subtitution into the integral $$I=\displaystyle\int_{\mathbb{R}_y^3}\int_{\mathbb{S}^2}e^{-\frac{1}{2}|\mathbf{x}-[(\mathbf{x}-\mathbf{y})\cdot\omega]\omega|^2}d\omega d\mathbf{y}$$ gives $$I=\displaystyle\int_{\mathbb{R}_y^3}\int_{\mathbb{S}^2}\exp\left[-\tfrac{1}{2}(\omega\cdot \mathbf{y})^2+\tfrac{1}{2}\left(\omega_3^2-1\right) x_3^2\right]\,d\omega d\mathbf{y}$$ $$=(2\pi)^{3/2}\int_{-\infty}^\infty dy_2\int_{-\infty}^\infty dy_3\int_0^\pi d\theta\,\frac{\sin\theta}{|\cos\theta|}e^{-\tfrac{1}{2}x_3^2\sin^2\theta}.$$ All three integrations are divergent.

• I can't quite understand what you mean. Could you be more specific? – Linxiaodiu Mar 25 at 13:17
• added some more intermediate steps; did I misinterpret your question or make a mistake? – Carlo Beenakker Mar 25 at 13:45
• @Carlo: I agree with Carlo, the $\mathbf y$ integral is already log-divergent in $\mathbb R_y^1$. – Fred Hucht Mar 26 at 14:59
• @ Carlo Beenakker: Thank you for your patience.I just don't get the idea ‘Without loss of generality you can orient the axes so that the vector $\mathbf{x}$ is along the $x_3$ axis’. – Linxiaodiu Mar 27 at 8:38
• @ Carlo Beenakker: With your help, I made the following calculations: $$\int_{\mathbb{R}_y^3}\int_{\mathbb{S}^2}e^{-\frac{1}{2}|x-[(x-y)\cdot\omega]\omega|^2}d\omega dy \overset{\text{(1)}}{=}\int_{\mathbb{S}^2}\int_{\mathbb{R}_y^3}e^{-\frac{1}{2}|x-[(x-y)\cdot\omega]\omega|^2} dy d\omega\\ =\int_{\mathbb{S}^{2}} e^{-\frac{1}{2}\left(\left|x\right|^{2}-\left|x \cdot \omega\right|^{2}\right)}\left(\int_{\mathbb{R}_{y}^{3}} e^{-\frac{1}{2}\left|y \cdot \omega\right|^{2}} d y\right) d \omega$$ $\int_{\mathbb{R}_{y}^{3}} e^{-\frac{1}{2}\left|y \cdot \omega\right|^{2}} dy$ diverges. – Linxiaodiu Mar 27 at 8:42