Consider the integral $$I_{n}(\alpha) := \int_{\mathbb{R}^{n}}e^{-\alpha\Vert x \Vert_{2}^{2}}\operatorname{d} x,\quad \alpha>0,$$ onde $\Vert x \Vert_{2} = \sqrt{x_{1}^{2} + ... + x_{n}^{2}}$ denotes the euclidean norm of $x \in \mathbb{R}^{n}$, e $\operatorname{d} x = \operatorname{d} x_{1}\operatorname{d} x_{2}...\operatorname{d} x_{n}$.
(a) Show that $I_{2}(\alpha) = \pi/\alpha$ and conclude that $I_{1}(\alpha) = \sqrt{\pi/\alpha}$.
(b) Show that $I_{n}(\alpha) = (\pi/\alpha)^{\frac{n}{2}}$.
(c) Show that $$\frac{d}{d\alpha}I_{n}(\alpha) = \int_{\mathbb{R}^{n}}\Vert x \Vert_{2}^{2}e^{-\alpha \Vert x \Vert_{2}^{2}}\operatorname{d}x$$
For (a). Using polar coordinates, we have $x = (r\sin \theta, r \cos \theta)$ e então $$I_{2}(\alpha) = \int_{0}^{2\pi}\lim_{\epsilon \to \infty}\int_{0}^{\epsilon}e^{-\alpha r^{2}}r \operatorname{d}r \operatorname{d}\theta = \pi/\alpha.$$ Thus, $$I_{1}(\alpha) = \int_{\mathbb{R}}e^{-\alpha x_{1}^{2}}\operatorname{d} x_{1} = \left(\int_{\mathbb{R}}\int_{\mathbb{R}}e^{-\alpha(x_{1}^{2} + x_{2}^{2})}\operatorname{d} x_{1} \operatorname{d} x_{2}\right)^{\frac{1}{2}} = \left(\int_{\mathbb{R}^{2}}e^{-\alpha\Vert x \Vert_{2}^{2}}\operatorname{d} x\right)^{\frac{1}{2}} = \sqrt{I_{2}(\alpha)}.$$
For (b). I'm using induction but, I'm not sure about my proof. $$I_{n+1}(\alpha) = \int_{\mathbb{R}^{n+1}}e^{-\alpha\Vert x \Vert_{2}^{2}}\operatorname{d}x = \int_{\mathbb{R}}\int_{\mathbb{R}^{n}}e^{-\alpha\Vert \tilde{x} \Vert}e^{-\alpha \Vert x_{n+1}\Vert_{2}^{2}}\operatorname{d}\tilde{x}\operatorname{d}x_{n+1} = I_{n}(\alpha)\int_{\mathbb{R}}e^{-\alpha \Vert x_{n+1} \Vert_{2}^{2}}\operatorname{d}x_{n+1} = (\pi/\alpha)^{\frac{n}{2}}(\pi/\alpha)^{\frac{1}{2}} = (\pi/\alpha)^{\frac{n+1}{2}}.$$
For (c). I'm having trouble. I'm tried to write $\Vert x \Vert_{2}^{2} = u$, but I didn't get anything interesting.
Can someone help me?
Your (b) is fine. Alternatively, you can avoid induction by directly showing $I_n(\alpha) = (I_1(\alpha))^n$.
For (c), differentiate under the integral, and recall $\frac{d}{d\alpha} e^{c\alpha} = c e^{c \alpha}$. Also should the right-hand side be $- \int \|x\|^2 e^{-\alpha \|x\|^2}\, dx$?