Let \(\mathbb{R}^d\) be the Euclidean \(d\)-dimensional space with the Euclidean norm \[|x|=\bigg(\sum_{j=1}^{d}|x_j|^2\bigg)^{1/2}=\sqrt{\langle x,x \rangle},\] which is induced by the standard inner product \[\langle x,y \rangle= x\cdot y= \bigg(\sum_{j=1}^{d}x_jy_j\bigg)^{1/2},\] for any \(x=(x_1,\ldots,x_d), y=(y_1,\ldots,y_d)\in\mathbb R^d\).
Jacobians If \(E \subseteq \mathbb{R}^d\) is open and \(f:E \to \mathbb{R}^d\) is differentiable at a point \(x \in E\), the determinant of the linear operator \(D_xf=f'(x)\) is called the Jacobian of \(f\) at \(x\in\mathbb R^d\) and denoted by \[J_f(x)=\det D_xf=\det f'(x).\] We shall also use the notation \[\frac{\partial (y_1,\ldots,y_d)}{\partial (x_1,\ldots,x_d)}=J_f(x) =\det \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \ldots & \frac{\partial f_1}{\partial x_d}\\ \vdots & \ddots & \vdots \\ \frac{\partial f_d}{\partial x_1} & \ldots & \frac{\partial f_d}{\partial x_d} \end{bmatrix}.\] if \((y_1,\ldots,y_d)=(f_1(x_1,\ldots,x_d), \ldots, f_d(x_1,\ldots,x_d))=f(x_1,\ldots,x_d)\).
In terms of Jacobians, the crucial hypothesis in the inverse function theorem is that \(J_f(x) \neq 0\).
We will investigate behavior of the Lebesgue integral under linear transformations. We will identify a linear map \(T:\mathbb{R}^d \to \mathbb{R}^d\) with the matrix \[(T_{ij})_{1\le i, j\le d}=(e_i \cdot Te_j)_{1\le i, j\le d},\] where \(\{e_j: 1\le j\le d\}\) is the standard basis in \(\mathbb{R}^d\).
We denote the determinant of this matrix by \(\det T\) and recall that \[\det(T \circ S)=(\det T)(\det S).\]
Furthermore, \(\text{GL}(d,\mathbb{R})\) denotes the general linear group of invertible linear transformations of \(\mathbb{R}^d\).
Recall that every \(T \in {\rm GL}(d,\mathbb{R})\) can be written as the product of finitely many transformations of three elementary types:
\[T_1(x_1,\ldots,x_j,\ldots,x_d)=(x_1,\ldots,cx_j,\ldots,x_d),\]
\[T_2(x_1,\ldots,x_j,\ldots,x_d)=(x_1,\ldots,x_j+cx_k,\ldots,x_d), \quad \text{ for } \quad k \neq j,\]
\[T_3(x_1,\ldots,x_j,\ldots,x_k,\ldots,x_d)=(x_1,\ldots,x_k,\ldots,x_j,\ldots,x_d).\]
That every \(T \in {\rm GL}(d,\mathbb{R})\) is a product of transformations of these three types is simply the fact that every invertible matrix can be row-reduced to the identity matrix.
From now on we shall consider \(\mathbb R^d\) with Lebesgue measure \(\lambda_d\) either on \(\mathcal L(\mathbb R^d)\) or on \({\rm Bor}(\mathbb R^d)=\mathcal B(\mathbb R^d)\). We shall also abbreviate \(d\lambda_d(x)\) to \(dx\).
Suppose \(T \in {\rm GL}(d,\mathbb{R})\).
If \(f\) is a Lebesgue measurable function on \(\mathbb{R}^d\), so is \(f \circ T\). If \(f \geq 0\) or \(f \in L^1(\mathbb R^d)\), then \[\qquad\qquad \int_{\mathbb{R}^d}f(x)dx=|\det T|\int_{\mathbb{R}^d}f \circ T(x)dx. \qquad\qquad {\color{purple}(*)}\]
If \(E \in \mathcal{L}(\mathbb R^d)\), then \(T[E] \in \mathcal{L}(\mathbb R^d)\) and \[\lambda_d(T[E])=|\det T|\lambda_d(E).\]
Proof. First suppose that \(f\) is Borel measurable, then \(f \circ T\) is Borel measurable since \(T\) is continuous.
If (*) is true for \(T,S \in {\rm GL}(d,\mathbb{R})\) it is also true for \(T \circ S\) since \[\begin{aligned} \int_{\mathbb{R}^d}f(x)dx=|\det T|\int_{\mathbb{R}^d}f(Tx)dx =|\det T||\det S|\int_{\mathbb{R}^d}f(TSx)dx. \end{aligned}\] It suffices to prove (*) when \(T\in\{T_1,T_2,T_3\}\) described above, which will be a simple consequence of the Fubini–Tonelli theorem.
Indeed, for \(T_3\) we interchange the order of integration in the variables \(x_j\) and \(x_k\), and for \(T_1\) and \(T_2\) we integrate first with respect to \(x_j\) and use the one-dimensional formulas \[\int_{\mathbb{R}}f(t)dt=|c|\int_{\mathbb{R}}f(xt)dt, \quad \text{ and } \quad \int_{\mathbb{R}}f(t+a)dt=\int_{\mathbb{R}}f(t)dt.\]
It is easily verified that \(\det T_1=c\), \(\det T_2=1\), \(\det T_3=-1\), so the formula (*) is proved.
Moreover, if \(E\) is a Borel set, then \(\lambda_d(T[E])=|\det T|\lambda_d(E)\) by taking \(f=\mathbf{1}_{{T[E]}}\) in (*) since \(T[E]\) is a Borel set as well. WHY? In particular, the class of Borel null sets is invariant under \(T\) and \(T^{-1}\).
Suppose \(E\in \mathcal{L}(\mathbb R^d)\), since \(\mathcal{L}(\mathbb R^d)\) is a completion of \(\mathcal{B}(\mathbb R^d)\), thus \(E=A\cup B\), where \(A\in \mathcal{L}(\mathbb R^d)\) and \(B\subseteq C\in \mathcal{B}(\mathbb R^d)\) and \(\lambda_d(C)=0\).
Then \(\lambda_d(T[B])\le \lambda_d(T[C])=|\det T|\lambda_d(C)=0\) and \(T[B]\in \mathcal{L}(\mathbb R^d)\), since \(\lambda_d\) is complete. Thus \[\lambda_d(T[E])=\lambda_d(T[A])=|\det T|\lambda_d(A)= |\det T|\lambda_d(E),\] since \(\lambda_d(T[E])=\lambda_d(T[A]\cup T[B])=\lambda_d(T[A])\) and \(\lambda_d(A)=\lambda_d(E)\).
By the previous argument (*) holds for \(f=\mathbf{1}_{{E}}\) if \(E\in \mathcal{L}(\mathbb R^d)\). By linearity (*) is true for step functions, by the is true for nonnegative \(\mathcal{L}(\mathbb R^d)\)-measurable functions. Consequently, (*) remains true for all \(\mathcal{L}(\mathbb R^d)\)-measurable functions by considering positive and negative parts of the real and imaginary parts of \(f\).$$\tag*{$\blacksquare$}$$
Lebesgue measure is invariant under rotations.
Proof. \(T \in {\rm GL}(d,\mathbb{R})\) is a rotation iff \(TT^*=I\), where \(T^{*}\) is the transpose of \(T\). Since \(\det T=\det T^*\) we obtain \(|\det T|=1\).$$\tag*{$\blacksquare$}$$
We now generalize previous theorem to differentiable maps. Let \(\Omega \subseteq \mathbb{R}^d\) be an open set and \(G=(g_1,\ldots,g_n):\Omega\to \mathbb{R}^d\) be a map with \(g_j\in C^1(\Omega)\).
\(G:\Omega\to \mathbb{R}^d\) is called a diffeomorphism if \(G\) is injective and \(D_xG\) is invertible for all \(x \in \Omega\). In this case, the inverse function theorem guarantees that \(G^{-1}:G[\Omega] \to \Omega\) is also a \(C^1\) diffeomorphism and that \[D_x(G^{-1})=(D_{G^{-1}(x)}G)^{-1}\quad \text{ for all } \quad x \in G[\Omega].\]
Let \(\Omega\subseteq \mathbb{R}^d\) be open and let \(G:\Omega \to \mathbb{R}^d\) be a \(C^1(\Omega)\) diffeomorphism.
If \(f\) is a Lebesgue measurable function on \(G[\Omega]\) then \(f \circ G\) is Lebesgue measurable on \(\Omega\). If \(f \geq 0\) or \(f \in L^1(G[\Omega],\lambda_d)\), then \[\int_{G[\Omega]}f(x)dx=\int_{\Omega}f \circ G(x)|\det D_x G|dx.\]
If \(E \subseteq \Omega\) and \(E \in \mathcal{L}(\mathbb R^d)\) then \(G[E] \in \mathcal{L}(\mathbb R^d)\) and \[\lambda_d(G[E])=\int_E |\det D_x G|dx.\]
For \(x \in \mathbb{R}^d\) and \(T=(T_{ij})_{1\le i, j\le d}\in {\rm GL}(d,\mathbb{R})\) we set \[|x|_{\infty}=\max_{1 \leq j \leq d}|x_j|, \qquad \|T\|=\max_{1 \leq i \leq d}\sum_{j=1}^d |T_{ij}|.\] Then it is easy to see that \[|Tx|_{\infty} \leq \|T\| |x|_{\infty}\] and \[a+[-r, r]^d=\{x \in \mathbb{R}^d: |x-a|_{\infty} \leq r\},\] is the cube of side length \(2r\) centered at \(a \in \mathbb{R}^d\).
Proof. It suffices to consider Borel measurable functions and sets. Since \(G\) and \(G^{-1}\) are both continuous, there are no problems with measurability.
Let \(Q=\{x \in \mathbb{R}^d: |x-a|_\infty \leq r\}\subseteq \Omega\). By the mean value theorem: \[g_j(x)-g_j(a)=\sum_{i=1}^d(x_i-a_i)\frac{\partial g_j}{\partial x_i}(y)\] for some \(y\) lying on the the line segment joining \(x\) and \(a\), so that \[|G(x)-G(a)|_\infty \leq r\big(\sup_{y \in Q}\|D_y G\|\big)\quad \text{ for } \quad x \in Q.\]
In other words, \(G[Q]\) is contained in the cube of side length \(\sup_{y \in Q}\|D_y G\|\) times that of \(Q\), so by the previous theorem, \[\lambda_d(G[Q]) \leq \big(\sup_{y \in Q}\|D_yG\|\big)^d\lambda_d(Q).\]
If \(T \in {\rm GL}(d,\mathbb{R})\) we can apply this formula with \(G\) replaced by \(T^{-1} \circ G\) together with the previous theorem to obtain \[\begin{aligned} \qquad \lambda_d(G[Q])&=|\det T|\lambda_d(T^{-1}(G[Q]))\\ &\leq |\det T|\Big(\sup_{y \in Q}\|T^{-1}D_y G\|\Big)^d\lambda_d(Q). \qquad {\color{purple}(**)} \end{aligned}\]
Since \(y \mapsto D_y G\) is continuous, for any \(\varepsilon>0\) we choose \(\delta>0\) so that \[\|(D_z G)^{-1}(D_y G)\|^d \leq 1+\varepsilon\] if \(y,z \in Q\) and \(|y-z|_{\infty} \leq \delta.\)
Let us subdivide \(Q\) into subcubes \(Q_1,\ldots,Q_N\) centered respectively at \(x_1,\ldots,x_N\), whose interiors are disjoint, and whose side lengths are at most \(\delta\).
By (**) with \(Q_j\) in place of \(Q\) and \(T=D_xG\) we see that
\[\begin{aligned} \lambda_d(G[Q]) &\leq \sum_{j=1}^N \lambda_d(G[Q_j])\\ &\leq \sum_{j=1}^{N}|\det D_{x_j}G|\Big(\sup_{y \in Q_j}\|(D_{x_j}G)^{-1}D_yG\|\Big)^d \lambda_d(Q_j)\\ &\leq (1+\varepsilon)\sum_{j=1}^N |\det D_{x_j}G|\lambda_d(Q_j). \end{aligned}\]
This last sum is the integral of \[\sum_{j=1}^{N}|\det D_{x_j}G|\mathbf{1}_{{Q_j}}(x).\]
Then we see that \[\sum_{j=1}^N |\det D_{x_j}G|\mathbf{1}_{{Q_j}} \ _{\overrightarrow{\delta\to 0}} \ |\det D_x G|\mathbf{1}_{{Q}}\] converges uniformly, since \(D_x G\) is continuous.
Thus letting \(\delta \to 0\) and \(\varepsilon \to 0\) we find that \[\qquad \qquad \lambda_d(G[Q]) \leq \int_Q |\det D_x G|dx. \qquad \qquad {\color{purple}(***)}\]
We claim that (***) holds with \(Q\) replaced by any Borel set in \(\Omega\). Indeed, if \(U \subseteq \Omega\) is open we can write \[U=\bigcup_{j=1}^{\infty}Q_j,\] where \(Q_j\)’s are cubes with disjoint interiors.
Since the boundaries of the cubes have Lebesgue measure zero, hence by (***) we have \[\lambda_d(G[U])=\sum_{j=1}^{\infty}\lambda_d(G[Q_j]) \leq \sum_{j=1}^{\infty}\int_{Q_j}|\det D_x G|dx=\int_{U}|\det D_x G|dx.\]
Next, let \[W_k=\{x \in \Omega: |x|_{\infty}<k \text{ and } |\det D_x G|<k\},\] and note that \(W_k\subseteq W_{k+1}\) for \(k\in\mathbb N\).
If \(E\) is a Borel subset of \(W_k\) there is a decreasing sequence of open sets \(U_j \subseteq W_{k+1}\) such that \[E \subseteq \bigcap_{j=1}^\infty U_j \quad \text{ and } \quad \lambda_d\Big(\bigcap_{j=1}^\infty U_j \setminus E\Big)=0.\]
By the proceeding estimate and the dominated convergence theorem we have
\[\begin{aligned} \lambda_d(G[E])& \leq \lambda_d\bigg(G\Big[\bigcap_{j=1}^{\infty}U_j\Big]\bigg) =\lim_{j \to \infty}\lambda_d(G[U_j])\\ &\leq \lim_{j \to \infty}\int_{U_j}|\det D_x G|dx=\int_{E}|\det D_xG|dx. \end{aligned}\]
Finally, if \(E\) is any Borel subset of \(\Omega\) we apply this argument to \(E \cap W_k\) let \(k \to \infty\) and conclude via the monotone convergence theorem that \[\lambda_d(G[E]) \leq \int_E |\det D_x G|dx.\]
If \(f=\sum_{j=1}^M a_j\mathbf{1}_{{A_j}}\) is a nonnegative simple function on \(G[\Omega]\) we therefore have \[\begin{gathered} \int_{G[\Omega]}f(x)dx=\sum_{j=1}^M a_j\lambda_d(A_k) \leq \sum_{j=1}^M a_j\int_{G[A_j]}|\det D_x G|dx\\ =\int_{\Omega}f \circ G(x) |\det D_x G|dx. \end{gathered}\]
By the monotone convergence we conclude that the same is true for any nonnegative measurable function \(f\), i.e. \[\begin{gathered} \int_{G[\Omega]}f(x)dx\le \int_{\Omega}f \circ G(x) |\det D_x G|dx. \end{gathered}\]
The same reasoning applied with \(G\) replaced by \(G^{-1}\) and \(f\) replaced by \(f \circ G\), yields that
\[\begin{gathered} \int_{\Omega}f \circ G(x)|\det D_x G|dx \\ \leq \int_{G[\Omega]}f \circ G \circ G^{-1}|\det D_{G^{-1}(x)}G||\det D_x G^{-1}|dx\\ =\int_{G[\Omega]}f(x)dx. \end{gathered}\]
This establishes Theorem (a) for all \(f \geq 0\).
The case \(f \in L^1(\mathbb R^d)\) follows immediately.
Theorem (b) is a special case of Theorem (a) where \(f=\mathbf{1}_{{G[E]}}\).
The case if \(f\) is Lebesgue measurable can be handled in much the same ways as the previous theorem, we omit the details. $$\tag*{$\blacksquare$}$$
We show that \[\int_{\mathbb{R}}e^{-x^2}dx=\sqrt{\pi}.\]
Consider \[\left(\int_{\mathbb{R}}e^{-x^2}dx\right)^2 =\int_{\mathbb{R}}\int_{\mathbb{R}}e^{-x^2-y^2}dxdy\] and a mapping \(G:(0,\infty)\times [0,2\pi) \to \mathbb{R}^2 \setminus [0,+\infty)\) given by \[G(r,\theta)=(r\cos(\theta),r\sin(\theta))\] Then we have \[D_{r,\theta}G= \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -r\sin(\theta) & r\cos(\theta) \end{bmatrix}, \quad \det D_{(r,\theta)}=r\cos(\theta)^2+r\sin(\theta)^2=r.\]
Thus \[\begin{aligned} \int_{\mathbb{R}}\int_{\mathbb{R}}e^{-x^2-y^2}dxdy&=\int_0^\infty \int_0^{2\pi}e^{-r^2}r d\theta dr\\ &=2\pi \int_0^{\infty}re^{-r^2}dr\\ &=\pi \int_0^{\infty}2re^{-r^2}dr\\ &=\pi \int_0^\infty e^{-t}dt=\pi. \end{aligned}\] Thus \[\qquad {\color{red}\int_{\mathbb{R}}e^{-x^2}dx =\left(\int_{\mathbb{R}}\int_{\mathbb{R}}e^{-(x^2+y^2)}dxdy\right)^{1/2}=\sqrt{\pi}.}\qquad\tag*{$\blacksquare$}\]
The most important nonlinear coordinate system in \(\mathbb{R}^2\) and \(\mathbb{R}^3\) are polar coordinates \[\begin{gathered} {\color{red}(r,\theta) \longmapsto (r\cos(\theta),r\sin(\theta))},\\ {\color{blue}(r,\phi,\theta) \longmapsto (r\sin(\phi)\cos(\theta),r\sin(\phi)\sin(\theta),r\cos(\theta))}. \end{gathered}\]
The multi-dimensional change of variables formula yields \[\begin{gathered} {\color{red}dx\,dy=r\,dr\,d\theta},\\ {\color{blue}dx\,dy\,dz=r^2\sin(\phi)\,dr\,d\theta\,d\phi}. \end{gathered}\]
Similar coordinate systems exist in higher dimensions, but they become more complicated as the dimension increases.
For most purposes, however, it is sufficient to know that Lebesgue measure is effectively the product of the measure \({\color{purple}r^{n-1}dr}\) on \(\mathbb R_+=(0,\infty)\) and a certain "surface measure \(\sigma\)" on the unit sphere (\(d\sigma(\theta)=d\theta\) for \(d=2\); and \(d\sigma(\theta, \phi)=\sin(\phi)\,d\theta\,d\phi\) for \(d=3\)).
Our construction of this measure is motivated by a familiar fact from the plane geometry. Namely, if \(S_{\theta}\) is a sector of a disc of radius \(r>0\) with an angle \(\theta\), the area \(\lambda_2(S_{\theta})\) is proportional to \(\theta\). In fact, we have \[\lambda_2(S_{\theta})=\frac{1}{2}r^2\theta.\]
The equation above can be solved for \(\theta\) and used to define the angular measure \(\theta\) in terms of the area of \(\lambda_2(S_{\theta})\). The same works in \(\mathbb{R}^d\) for \(d \geq 2\).
Let \(\mathbb S^{d-1}=\{x \in \mathbb{R}^d:|x|=1\}\) be the unit sphere in \(\mathbb{R}^d\). If \(x \in \mathbb{R}^d \setminus \{0\}\) the polar coordinates of \(x\) are \[r=|x| \in (0,\infty) \quad \text{ and } \quad x'=\frac{x}{|x|} \in \mathbb S^{d-1}.\]
The map \(\Phi(x)=(r,x')\) is a continuous bijection from \(\mathbb{R}^d \setminus \{0\}\) to \((0,\infty) \times \mathbb S^{d-1}\), whose continuous inverse is \[\Phi^{-1}(r,x')=rx'.\]
We denote by \(m_{d}\) the Borel measure on \((0,\infty) \times \mathbb S^{d-1}\) induced by \(\Phi\) from Lebesgue measure on \(\mathbb{R}^d\), that is \[m_d(E)=\lambda_d(\Phi^{-1}[E]).\]
Moreover, we define the measure \(\rho=\rho_d\) on \((0,\infty)\) by \[{\color{red}\rho(E)=\int_Er^{d-1}dr}.\]
There is a unique Borel measure \(\sigma=\sigma_{d-1}\) on \(\mathbb S^{d-1}\) such that \(\lambda_d=\rho \times \sigma\). If \(f\) is Borel measurable on \(\mathbb{R}^d\) and \(f \geq 0\) or \(f \in L^1(\mathbb R^d)\) then \[\qquad \qquad \int_{\mathbb{R}^d}f(x)dx =\int_0^{\infty}\int_{\mathbb S^{d-1}}f(rx')r^{d-1}d\sigma(x')dr. \qquad \qquad{\color{purple}(*)}\]
Remark. This theorem can be extended to Lebesgue measurable functions by considering the completion of the measure \(\sigma\).
Proof. Equation (*) when \(f\) is a characteristic function of a set is a restatement of the equation \(m_{d}=\rho \times \sigma\) and it follows for general \(f\) by the usual linearity and approximation arguments. Hence we need only to construct \(\sigma\).
If \(E\) is a Borel set in \(\mathbb S^{d-1}\) for \(a>0\) let \[E_{a}=\Phi^{-1}((0,a] \times E)=\{rx': (r, x')\in (0,a] \times E\}.\] If (*) is to hold with \(f=\mathbf{1}_{{E}}\), we must have \[\lambda_d(E_1)=\int_0^1 \int_E r^{d-1}d\sigma(x')dr =\sigma(E)\int_0^1 r^{d-1}dr =\frac{\sigma(E)}{d}.\]
We therefore define
\[{\color{red}\sigma(E)=d \cdot \lambda_d(E_1)}.\]
Since the map \(E \mapsto E_1\) takes Borel sets to Borel sets and commutes with unions, intersections, and complements, it is clear that \(\sigma\) is a Borel measure on \(\mathbb S^{d-1}\). Also, since \(E_a\) is the image of \(E_1\) under the map \(x \mapsto ax\), it follows that \(\lambda_d(E_a)=a^d\lambda_d(E_1)\).
Fix \(E \in \mathcal{B}(\mathbb S^{d-1})\) and let \(\mathcal{A}_E\) be a the collection of finite disjoint unions of sets of the form \((a,b] \times E\). We know that \(\mathcal{A}_E\) is an algebra on \(\mathbb R_+ \times E\) that generates the \(\sigma\)-algebra \(\mathcal{M}_E=\{A \times E: A \in \mathcal{B}(\mathbb R_+)\}\).
We show that \(m_d=\rho \times \sigma\) on \(\mathcal{A}_E\). If \(0<a<b\), then we have \[\begin{aligned} m_d((a,b] \times E)&=\lambda_d(E_b \setminus E_a)=\frac{b^d-a^d}{d}\sigma(E)\\ &=\sigma(E)\int_a^b r^{d-1}dr={\color{blue}\rho \times \sigma((a,b] \times E).} \end{aligned}\]
Since \(m_d=\rho \times \sigma\) on \(\mathcal{A}_E\), hence by the uniqueness assertion of Caratheodory’s extension theorem we have \(m_d=\rho \times \sigma\) on \(\mathcal{M}_E\).
But \(\bigcup_{E \in \mathcal{B}(\mathbb S^{d-1})}\mathcal{M}_E\) is precisely the set of Borel rectangles in \(\mathbb R_+ \times \mathbb S^{d-1}\) so another application of the uniqueness theorem shows that \(m_d=\rho \times \sigma\) on all Borel sets.$$\tag*{$\blacksquare$}$$
If \(f:\mathbb{R}^d\to \mathbb C\) is a measurable function nonnegative or integrable, such that \(f(x)=g(|x|)\) for some measurable function \(g:\mathbb{R}_+\to \mathbb C\), then \[\int_{\mathbb{R}^d}f(x)dx =\sigma(\mathbb S^{d-1})\int_0^{\infty}g(r)r^{d-1}dr.\]
Proof. By the polar decomposition \[\begin{aligned} \int_{\mathbb{R}^d}f(x)dx&=\int_0^{\infty} \int_{\mathbb S^{d-1}}f(rx')d\sigma(x')dr\\ &=\int_0^\infty \int_{\mathbb S^{d-1}}g(r|x'|)r^{d-1}d\sigma(x')dr\\ &=\sigma(\mathbb S^{d-1})\int_{0}^{\infty}g(r)r^{d-1}dr. \qquad\qquad {\blacksquare} \end{aligned}\]
Proposition. For \(a>0\) we have \[I_d=\int_{\mathbb{R}^d}e^{-a|x|^2}dx=\left(\frac{\pi}{a}\right)^{d/2}.\]
Proof. For \(d=2\) we obtain \[I_2=2\pi\int_0^{\infty}re^{-ar^2}dr=\frac{\pi}{a},\] since \(\sigma(\mathbb S^1)=2\pi\). Observe now that \(e^{-a|x|^2}=\prod_{j=1}^de^{-ax_j^2}.\) Thus, by Tonelli’s theorem, \(I_d=(I_1)^d.\) In particular, \(I_1=I_2^{1/2}\), so \[\qquad\qquad I_d=(I_2)^{d/2}={\color{red}\left(\frac{\pi}{a}\right)^{d/2}}.\qquad\qquad \tag*{$\blacksquare$}\]
Proposition. Recall that \(\Gamma(x)=\int_0^{\infty}t^{x-1}e^{-t}dt\). Then one has \[\sigma(\mathbb S^{d-1})=\frac{2\pi^{d/2}}{\Gamma(d/2)}.\]
Proof. By the previous proposition we have \[\begin{aligned} \pi^{d/2}&=\int_{\mathbb{R}^d}e^{-|x|^2}dx=\sigma(\mathbb S^{d-1})\int_0^\infty e^{-r^2}r^{d-1}dr\\ &=\frac{1}{2} \sigma(\mathbb S^{d-1}) \int_0^{\infty}s^{d/2-1}e^{-s}ds =\frac{1}{2}\sigma(\mathbb S^{d-1})\Gamma(d/2). \end{aligned}\] Thus \[\qquad\qquad{\color{blue}2\pi^{d/2}\Gamma(d/2)^{-1}=\sigma(\mathbb S^{d-1})}.\qquad\qquad\tag*{$\blacksquare$}\]
Proposition. One has \[\lambda_d(B(0, 1))=\frac{\pi^{d/2}}{\Gamma(d/2+1)},\] where \(B(0, 1)=\{x \in \mathbb{R}^d: |x|<1\}.\)
Proof. Since \(x\Gamma(x)=\Gamma(x+1)\) we have \[\begin{aligned} \qquad\qquad\lambda_d(B(0, 1))& =\int_{\mathbb{R}^d}\mathbf{1}_{{B(0, 1)}}(x)dx\\ &=\sigma(\mathbb S^{d-1})\int_0^1 r^{d-1}dr\\ &=\frac{1}{d}\sigma(\mathbb S^{d-1})\\ &=\frac{2\pi^{d/2}}{d\Gamma(d/2)} =\frac{\pi^{d/2}}{\Gamma(d/2+1)}.\qquad\qquad{\blacksquare} \end{aligned}\]