I am trying to evaluate the following integral on $\mathbb{R}^{n-1}$
$$\int_{\mathbb{R}^{n-1}}\frac{1}{(1+|x|^2)^{\frac{n}{2}}}dx$$
I claim that this is equal to the half the surface area of the sphere in $\mathbb{R}^n$. One solution to this problem uses polar coordinates, I have written this out at the end of the question. However I am attempting to explain this a different way. Observe that the value $(1+|x|^2)^{n/2}$ is equal to $|(x,1)|^n$ where this is the norm in $n$-space. Therefore we can write this integral as
$$\int_H\frac{1}{|x|^n}d\sigma_H(x)$$
Where $H$ is the hyperplane $\{x\in \mathbb{R}^n:x_n=1\}$ and $\sigma_H$ is the surface measure on this hyperplane. Now to relate this to the sphere, I would love to consider the following map $x\mapsto x/|x|$, which takes this hyperplane to the upper hemisphere! But what happens to the above integral? Does it become the integral of the constant function 1? I would love it to, and it really looks like maybe it is, maybe the change of variables gives a factor of $|x|^n$? But I can't prove this, the matrix of derivatives is messy and it just doesn't seem to come out easily. My question is how do I fill the hole in the above argument? Can it be done?
Note that for context, this question originates from my reading of Mathematical Analysis and Numerical Methods for Science and Technology Volume 1 by Dautray and Lions, at the end of the proof of II.3.3.Prop 11 (top of page 302) an integral almost equal to what I wrote appears, except theirs is over $\mathbb{R}^n$! However I am sure this must be a mistake because all the integral leading up to this point are over $\mathbb{R}^{n-1}$. They do not prove the identity.
Here is the alternative proof:
\begin{align}
\int_{\mathbb{R}^{n-1}}\frac{1}{(1+|x|^2)^{\frac{n}{2}}}dx&=\omega_{n-1}\int_0^\infty\frac{r^{n-2}}{(1+r^2)^{n/2}}dr\\
&=\frac{2\pi^{(n-1)/2}}{\Gamma(1/2(n-1))}\sqrt{\pi} \frac{\Gamma(1/2 (n-1)))}{2 Γ(n/2)}\\
&=\frac{1}{2}\frac{2\pi^{n/2}}{\Gamma(n/2)} = \frac{1}{2}\omega_n
\end{align}
Where I denote by $\omega_n$ the surface area of the $n$-sphere.
Best Answer
Yes, fortunately, things work out nicely. The integral in question is exactly $\int_{\Sigma}1\,d\sigma$, where $\Sigma$ is the upper hemisphere of $S^{n-1}$ in $\Bbb{R}^n$. Hence, this evaluates to $\frac{1}{2}A_{n-1}\equiv\frac{1}{2}\omega_n=\frac{1}{2}\frac{2\pi^{n/2}}{\Gamma(n/2)}$.
To see this, consider the parametrization $\phi:\Bbb{R}^{n-1}\to\Sigma\subset\Bbb{R}^n$ given by $\phi(x)=\frac{(x,1)}{\sqrt{1+|x|^2}}$. This is pretty much what you wrote. Its $i^{th}$ partial derivative is \begin{align} \frac{\partial \phi}{\partial x_i}&=-\frac{1}{2}\frac{2x_i}{(1+|x|^2)^{3/2}}(x,1)+\frac{1}{\sqrt{1+|x|^2}}e_i=\frac{1}{\sqrt{1+|x|^2}}\left[-\frac{x_i}{1+|x|^2}(x,1)+e_i\right], \end{align} where $e_i=(0,\dots, 1,\dots, 0, 0)$ is the vector in $\Bbb{R}^n$ with $1$ in the $i^{th}$ slot and $0$ elsewhere. With this, we can compute the inner products: \begin{align} G_{ij}&:=\left\langle\frac{\partial \phi}{\partial x_i},\frac{\partial \phi}{\partial x_j}\right\rangle\\ &=\frac{1}{\left(\sqrt{1+|x|^2}\right)^2}\left[\frac{x_ix_j}{(1+|x|^2)^2}\cdot (1+|x|^2)-\frac{2x_ix_j}{1+|x|^2}+\delta_{ij}\right]\\ &=\frac{1}{1+|x|^2}\left[\delta_{ij}-\frac{x_ix_j}{1+|x|^2}\right]\\ &\equiv c(\delta_{ij}-a_ia_j), \end{align} where I have abbreviated $c=\frac{1}{1+|x|^2}$ and $a_i=\frac{x_i}{\sqrt{1+|x|^2}}$ for $1\leq i\leq n-1$. Using a standard matrix determinant identity, we have $\det(G_{ij})=c^{n-1}(1-\|a\|^2)=\frac{1}{(1+|x|^2)^n}$. Note the $c^{n-1}$ part is clear (by multilinearity), it is the $1-\|a\|^2$ part which is a perhaps a little less well known, so I’ll prove it below. Therefore, we have as desired, \begin{align} \int_{\Sigma}1\,d\sigma&=\int_{\phi(\Bbb{R}^{n-1})}1\,d\sigma=\int_{\Bbb{R}^{n-1}}(1\circ \phi)\cdot\sqrt{\det(G_{ij})}\,dx=\int_{\Bbb{R}^{n-1}}\frac{1}{(1+|x|^2)^{n/2}}\,dx. \end{align}
A curious determinant formula.
Consider an $n\times n$ matrix $G=G_n$ whose $(i,j)$ entry is $\delta_{ij}-a_ia_j$. So, we have the recursive block format \begin{align} G_n&= \left( \begin{array}{ccc|c} &&&-a_1a_n\\ &G_{n-1}&&\vdots\\ &&&-a_{n-1}a_n\\ \hline -a_na_1 &\dots & -a_na_{n-1}&1-a_n^2 \end{array} \right). \end{align} Now, let us take the determinant of this matrix. In doing so, we shall use linearity in the last column to split this as a sum of determinants: \begin{align} \det G_n&= \det \left( \begin{array}{ccc|c} &&&0\\ &G_{n-1}&&\vdots\\ &&&0\\ \hline -a_na_1 &\dots & -a_na_{n-1}&1 \end{array} \right) + \det \left( \begin{array}{ccc|c} &&&-a_1a_n\\ &G_{n-1}&&\vdots\\ &&&-a_{n-1}a_n\\ \hline -a_na_1 &\dots & -a_na_{n-1}&-a_n^2 \end{array} \right). \end{align} For the first term, expanding along the last column gives $\det G_{n-1}$. For the second term, factor out $-a_n$ from the last row then another $-a_n$ from the last column. This gives \begin{align} \det G_n&=\det G_{n-1} + (-a_n)^2\det \left( \begin{array}{ccc|c} &&&a_1\\ &G_{n-1}&&\vdots\\ &&&a_{n-1}\\ \hline a_1 &\dots & a_{n-1}&-1 \end{array} \right). \end{align} Now, add $a_1$ times the last column to the first column. Then, add $a_2$ times the last column to the second column, and so on up to the $(n-1)^{th}$ column. These column operations do not change the determinant. We thus get \begin{align} \det G_n&=\det G_{n-1} + a_n^2\det \left( \begin{array}{ccc|c} &&&a_1\\ &I_{n-1}&&\vdots\\ &&&a_{n-1}\\ \hline 0 &\dots & 0&-1 \end{array} \right)\\ &=\det G_{n-1}+a_n^2(-1), \end{align} where we used the fact that the matrix is upper triangular, so the determinant is the product of the diagonals. We can easily solve this recurrence relation to get $\det G_n=1-a_1^2-\dots-a_n^2=1-\|a\|^2$, which is exactly what I claimed above.
By the way if the $(i,j)$ entry was instead $\delta_{ij}+a_ia_j$, then going through the proof again (or simply replacing $a_i,a_j$ above with $\sqrt{-1}a_i,\sqrt{-1}a_j$) shows that the determinant in that case will turn out to be $1+(a_1^2+\dots+a_n^2)=1+\|a\|^2$. The reason we can use this trick of multiplying by the imaginary $\sqrt{-1}$ is because my derivation above made no assumptions about the field of scalars which comprise the matrix entries.
Extra FYI
The matrix determinant formula actually shows up in other places. Most notably, if you have a $C^1$ function $f:A\subset\Bbb{R}^n\to\Bbb{R}$ (where $A$ is some open subset), then you get a parameterization of its graph via $\phi:A\to\Bbb{R}^{n+1}$, $\phi(x)=(x,f(x))$. In this case, the inner product is \begin{align} \left\langle\frac{\partial\phi}{\partial x^i},\frac{\partial\phi}{\partial x^j}\right\rangle&=\delta_{ij}+\frac{\partial f}{\partial x^i}\frac{\partial f}{\partial x^j}, \end{align} so the determinant of the matrix is $1+\|df\|^2\equiv 1+\|\text{grad}(f)\|^2$. This is why the volume of an $n$-dimensional graph of a function is given by the familiar formula $\int_A\sqrt{1+\|\text{grad}(f)\|^2}\,dx$. When $n=1$, this is the arclength, when $n=2$, it’s the surface area etc.