*5.2. Second Step: Polar Decomposition*

Let the independent real variables *z*1, ... , *zp* be transformed to the general polar coordinates *r*, *θ*1, ... , *θp*−<sup>1</sup> as follows, where *r* > 0, −*π*/2 < *θ<sup>j</sup>* ≤ *π*/2, *j* = 1, ... , *p* − 2, −*π* < *θp*−<sup>1</sup> ≤ *π* [40],

$$z\_1 = r \sin \theta\_1 \tag{54}$$

$$z\_2 = r \cos \theta\_1 \sin \theta\_2 \tag{55}$$

$$z\_j = r \cos \theta\_1 \cos \theta\_2 \dots \cos \theta\_{j-1} \sin \theta\_{j}, \quad j = 2, 3, \dots, p - 1 \tag{56}$$

$$z\_P = r \cos \theta\_1 \cos \theta\_2 \dots \cos \theta\_{p-1}.\tag{57}$$

The Jacobian determinant according to theorem (1.24) in [40] is

$$\mathrm{d}z\_1\ldots\mathrm{d}z\_p = r^{p-1} \prod\_{j=1}^{p-1} |\cos\theta\_j|^{p-j-1} \mathrm{d}r \mathrm{d}\theta\_j. \tag{58}$$

It is clear that with the last transformations, we get ∑*<sup>p</sup> <sup>i</sup>*=<sup>1</sup> *<sup>z</sup>*<sup>2</sup> *<sup>i</sup>* = *<sup>r</sup>*<sup>2</sup> and the multiple integral in (53) is then given as follows

$$E\_{\mathbf{X}^1} \{ \ln[1 + \mathbf{X}^T \Sigma\_2^{-1} \mathbf{X}] \} = A \int\_0^{+\infty} \frac{r^{p-1}}{[1 + r^2]^{\frac{1+p}{2}}} \int\_{-\pi/2}^{\pi/2} \dots \int\_{-\pi}^{\pi} \left( \prod\_{j=1}^{p-1} |\cos \theta\_j|^{p-j-1} \right) \times$$

$$\ln \left[ 1 + r^2 (\lambda\_1 \sin^2 \theta\_1 + \dots + \lambda\_p \cos^2 \theta\_1 \dots \cos^2 \theta\_{p-1}) \right] \text{d}r \prod\_{j=1}^{p-1} \text{d}\theta\_j. \tag{59}$$

By replacing the expression of sin2 *<sup>θ</sup><sup>j</sup>* by 1 <sup>−</sup> cos2 *<sup>θ</sup>j*, for *<sup>j</sup>* <sup>=</sup> 1, ... , *<sup>p</sup>* <sup>−</sup> 1, we have the following expression

$$
\lambda\_1 \sin^2 \theta\_1 + \dots + \lambda\_p \cos^2 \theta\_1 \dots \cos^2 \theta\_{p-1} = \lambda\_1 + (\lambda\_2 - \lambda\_1) \cos^2 \theta\_1
$$

$$
+ \dots + (\lambda\_p - \lambda\_{p-1}) \cos^2 \theta\_1 \cos^2 \theta\_2 \dots \cos^2 \theta\_{p-1}.\tag{60}
$$

Let *xi* = cos2 *θ<sup>i</sup>* be a transformation to use where d*xi* = 2*x*1/2 *<sup>i</sup>* (<sup>1</sup> <sup>−</sup> *xi*)1/2d*θi*. Then the expectation given by the multiple integral over all *θj*, *j* = 1, . . . , *p* − 1 is as follows

$$2A\int\_0^{+\infty} \frac{r^{p-1}}{\left[1+r^2\right]^{\frac{1+p}{2}}} \int\_0^1 \dots \int\_0^1 \left(\prod\_{j=1}^{p-1} \mathbf{x}\_j^{\frac{p-j}{2}-1} (1-\mathbf{x}\_j)^{-\frac{1}{2}}\right) \ln\left[1+r^2\mathcal{B}\_p(\mathbf{x}\_1,\dots,\mathbf{x}\_{p-1})\right] \text{d}\mathbf{r} \mathbf{dx}\_1\dots\mathbf{dx}\_{p-1} \tag{61}$$

where *Bp*(*x*1,..., *xp*−1) = *λ*<sup>1</sup> + (*λ*<sup>2</sup> − *λ*1)*x*<sup>1</sup> + ... + (*λ<sup>p</sup>* − *λp*−1)*x*1*x*<sup>2</sup> ... *xp*−1, *p* ≥ 1 and *B*<sup>1</sup> = *λ*1. In the following, we use the notation *Bp* instead of *Bp*(*x*1, ... , *xp*−1) to alleviate writing equations.

Let *t* = *r*<sup>2</sup> be transformation to use. Then, one can write

$$=A\int\_0^{+\infty} \frac{t^{\frac{p}{2}-1}}{[1+t]^{\frac{1+p}{2}}} \int\_0^1 \dots \int\_0^1 \left(\prod\_{j=1}^{p-1} x\_j^{\frac{p-j}{2}-1} (1-x\_j)^{-\frac{1}{2}}\right) \ln[1+tB\_p] \text{d}t \text{d}x\_1 \dots \text{d}x\_{p-1}.\tag{62}$$

In order to solve the integral in (62), we consider the following property given by log(*x*)*f*(*x*) <sup>d</sup>*<sup>x</sup>* <sup>=</sup> <sup>−</sup> *<sup>∂</sup> ∂a x*−*<sup>a</sup> f*(*x*)d*x* + + *<sup>a</sup>*=<sup>0</sup> and the following equation given as follows

$$\left(1 + B\_p t\right)^{-a} = \frac{1}{\Gamma(a)} \int\_0^{+\infty} y^{a-1} e^{-(1+B\_p t)y} dy. \tag{63}$$

Making use of the above equation, we obtain a new expression of (62) given as follows

$$\begin{split} &E\_{\mathbf{X}^{\prime}}\{\ln[1+\mathbf{X}^{\prime}\Sigma\_{2}^{-1}\mathbf{X}]\} \\ &= -\frac{\partial}{\partial a}\Big\{\frac{A}{\Gamma(a)}\int\_{0}^{+\infty} \frac{t^{\frac{p}{2}-1}}{[1+t]^{\frac{1+p}{2}}} \int\_{0}^{+\infty} y^{a-1} e^{-(1+B\_{\mathbf{f}}t)y} \int\_{0}^{1} \dots \int\_{0}^{1} \prod\_{j=1}^{p-1} \mathbf{x}\_{j}^{\frac{p-j}{2}-1} (1-\mathbf{x}\_{j})^{-\frac{1}{2}} \mathbf{dx}\_{j} \mathbf{d}y \mathbf{d}t \Big\} \Big|\_{\mathbf{x}=0} \\ &= -\frac{\partial}{\partial a} \Big\{\frac{A}{\Gamma(a)} \int\_{0}^{+\infty} \frac{t^{\frac{p}{2}-1}}{[1+t]^{\frac{1}{2}}} \int\_{0}^{+\infty} y^{a-1} e^{-y} \mathbf{H}(t,y) \mathbf{d}y \mathbf{d}t \Big\} \Big|\_{\mathbf{x}=0} \end{split} \tag{64}$$

$$\mathcal{I} = -\frac{\partial}{\partial a} \left\{ \frac{A}{\Gamma(a)} \int\_0^{+\infty} \frac{t^{\frac{p-q}{2}}}{[1+t]^{\frac{1+p}{2}}} \int\_0^{+\infty} y^{\rho-1} e^{-y} \mathbf{H}(t,y) dy dt \right\} \Big|\_{a=0}$$

where **H**(*t*, *y*) is defined as

$$\mathbf{H}(t,y) = \int\_0^1 \dots \int\_0^1 e^{-B\_p ty} \prod\_{j=1}^{p-1} \mathbf{x}\_j^{\frac{p-j}{2}-1} (1-\mathbf{x}\_j)^{-\frac{1}{2}} d\mathbf{x}\_j. \tag{66}$$

*5.3. Third Step: Expression for H(t,y) by Humbert and Beta Functions*

Let *x <sup>i</sup>* = 1 − *xi*, *i* = 1, . . . , *p* − 1 be transformations to use. Then

> . .

.

$$(\lambda\_2 - \lambda\_1)\mathbf{x}\_1 = (\lambda\_2 - \lambda\_1)(1 - \mathbf{x}\_1')\tag{67}$$

$$(\lambda\_3 - \lambda\_2)\mathbf{x}\_1\mathbf{x}\_2 = (\lambda\_3 - \lambda\_2)(1 - \mathbf{x}\_1')[1 - \mathbf{x}\_2'] \tag{68}$$

$$(\lambda\_4 - \lambda\_3)\mathbf{x}\_1\mathbf{x}\_2\mathbf{x}\_3 = (\lambda\_4 - \lambda\_3)(1 - \mathbf{x}\_1')(1 - \mathbf{x}\_2')[1 - \mathbf{x}\_3'] \tag{69}$$

$$\dot{\mathbf{i}} = \dot{\mathbf{i}}$$

$$(\lambda\_p - \lambda\_{p-1}) \prod\_{i=1}^{p-1} \mathbf{x}\_i = (\lambda\_p - \lambda\_{p-1}) \prod\_{i=1}^{p-1} (1 - \mathbf{x}\_i'). \tag{70}$$

Adding equations from (67) to (70), we can state that the new expression of the function *Bp* becomes

$$B\_p = \lambda\_p - (\lambda\_p - \lambda\_1)\mathbf{x}\_1' - (\lambda\_p - \lambda\_2)(1 - \mathbf{x}\_1')\mathbf{x}\_2' - (\lambda\_p - \lambda\_3)(1 - \mathbf{x}\_1')(1 - \mathbf{x}\_2')\mathbf{x}\_3'$$

$$-\dots - (\lambda\_p - \lambda\_{p-1})(1 - \mathbf{x}\_1')\dots(1 - \mathbf{x}\_{p-2}')\mathbf{x}\_{p-1}'.\tag{71}$$

Then, the multiple integral **H**(*t*, *y*) given by (66) can be written otherwise

$$\mathbf{H}(t,y) = \int\_0^1 \dots \int\_0^1 e^{-B\_p t y} \prod\_{j=1}^{p-1} (1 - \mathbf{x}\_j')^{\frac{p-j}{2} - 1} \mathbf{x}\_j'^{-\frac{1}{2}} \mathbf{dx}\_1' \dots \mathbf{dx}\_{p-1}'.\tag{72}$$

Let the real variables *x* <sup>1</sup>, *x* <sup>2</sup>, ... , *x <sup>p</sup>*−<sup>1</sup> be transformed to the real variables *<sup>u</sup>*1, *<sup>u</sup>*2, ... , *up*−<sup>1</sup> as follows

$$
\mu\_1 = \mathfrak{x}\_1'\tag{73}
$$

$$
\mu\_2 = (1 - \mathbf{x}\_1')\mathbf{x}\_2' = (1 - \mu\_1)\mathbf{x}\_2' \tag{74}
$$

$$
\mu\_3 = (1 - \mathbf{x}\_1')(1 - \mathbf{x}\_2')\mathbf{x}\_3' = (1 - \mu\_1 - \mu\_2)\mathbf{x}\_3' \tag{75}
$$

$$
\mu\_{p-1} = \prod\_{i=1}^{p-2} (1 - \mathbf{x}\_i') \mathbf{x}\_{p-1}' = (1 - \sum\_{i=1}^{p-2} u\_i) \mathbf{x}\_{p-1}'.\tag{76}
$$

The Jacobian determinant is given by

$$\mathbf{d}u\_1 \dots \mathbf{d}u\_{p-1} = \prod\_{j=1}^{p-1} \left( 1 - \sum\_{i=1}^{j-1} u\_i \right) \mathbf{d}x'\_1 \dots \mathbf{d}x'\_{p-1}.\tag{77}$$

Accordingly, the new expression of *Bp* becomes

. .

$$B\_p = \lambda\_p - \sum\_{i=1}^{p-1} (\lambda\_p - \lambda\_i) u\_i. \tag{78}$$

As a consequence, the new domain of the multiple integral (72) is Δ = {(*u*1, *u*2, ... , *up*−1) ∈ <sup>R</sup>*p*−1; 0 <sup>≤</sup> *<sup>u</sup>*<sup>1</sup> <sup>≤</sup> 1, 0 <sup>≤</sup> *<sup>u</sup>*<sup>2</sup> <sup>≤</sup> <sup>1</sup> <sup>−</sup> *<sup>u</sup>*1, 0 <sup>≤</sup> *<sup>u</sup>*<sup>3</sup> <sup>≤</sup> <sup>1</sup> <sup>−</sup> *<sup>u</sup>*<sup>1</sup> <sup>−</sup> *<sup>u</sup>*2, ... , and <sup>0</sup> <sup>≤</sup> *up*−<sup>1</sup> <sup>≤</sup> <sup>1</sup> <sup>−</sup> *<sup>u</sup>*<sup>1</sup> <sup>−</sup> *u*<sup>2</sup> ... − *up*−2}, and the expression of **H**(*t*, *y*) is given as follows

$$\mathbf{H}(t,y) = \int \dots \int\_{\Delta} e^{-B\_{p}t} \prod\_{j=1}^{p-1} \left(1 - \sum\_{i=1}^{j-1} u\_{i}\right)^{-1} \left(\frac{u\_{j}}{1 - \sum\_{i=1}^{j-1} u\_{i}}\right)^{-\frac{1}{2}} \prod\_{j=1}^{p-1} \left(1 - \frac{u\_{j}}{1 - \sum\_{i=1}^{j-1} u\_{i}}\right)^{\frac{p-1}{2}-1} \mathbf{d}u\_{j} \tag{79}$$
 
$$\varepsilon\_{\varepsilon} \quad \varepsilon\_{\varepsilon} \quad \text{ $p-1$ } \quad \varepsilon\_{1} \quad \text{ $\{$ j $}$ } \quad \text{ $\{\varepsilon\}$ } \quad \text{ $\{\varepsilon\}$ }^{\frac{p-1}{2}-1} \left(\frac{u\_{j}}{1 - \sum\_{i=1}^{j-1} u\_{i}}\right)^{\frac{1}{2} - \frac{p-1}{2}} \text{ d}u\_{j} \tag{70}$$

$$\mathbf{u}\_{p} = \int\_{\cdot \cdot} \dots \int\_{\Delta} e^{-B\_{p}ty} \prod\_{j=1}^{p-1} u\_{j}^{-\frac{1}{2}} \left( 1 - \sum\_{i=1}^{j} u\_{i} \right)^{\frac{p-1}{2}-1} \left( 1 - \sum\_{i=1}^{j-1} u\_{i} \right)^{\frac{1}{2} - \frac{p-1}{2}} \, \mathrm{d}u\_{1} \dots \mathrm{d}u\_{p-1} \tag{80}$$

$$\hat{\rho} = \int \dots \int\_{\Delta} e^{-B\_p t y} \left( 1 - \sum\_{i=1}^{p-1} u\_i \right)^{\frac{p}{2} - \frac{p-1}{2} - 1} \prod\_{j=1}^{p-1} u\_j^{-\frac{1}{2}} \text{d}u\_j \tag{81}$$

$$\mathbf{u} = e^{-\lambda\_p ty} \int \dots \int\_{\Delta} \left( 1 - \sum\_{i=1}^{p-1} u\_i \right)^{-\frac{1}{2}} \prod\_{i=1}^{p-1} u\_i^{-\frac{1}{2}} e^{(\lambda\_p - \lambda\_i)u\_i ty} \mathbf{d}u\_i. \tag{82}$$

Using Proposition 1, we subsequently find that

$$\mathbf{H}(t,y) = e^{-\lambda\_{\mathcal{P}}ty} \mathbf{B}\left(\underbrace{\frac{1}{2}, \dots, \frac{1}{2}}\_{p}\right) \Phi\_2^{(p-1)}\left(\underbrace{\frac{1}{2}, \dots, \frac{1}{2}}\_{p-1}; \frac{p}{2}; (\lambda\_{\mathcal{P}} - \lambda\_1)ty, (\lambda\_{\mathcal{P}} - \lambda\_2)ty, \dots, (\lambda\_{\mathcal{P}} - \lambda\_{p-1})ty\right). \tag{83}$$

where Φ(*p*−1) <sup>2</sup> (.) is the Humbert series of *<sup>p</sup>* <sup>−</sup> 1 variables and *<sup>B</sup>*( <sup>1</sup> <sup>2</sup> , ... , <sup>1</sup> <sup>2</sup> ) is the multivariate beta function. Applying the following successive two transformations *r* = *ty* (d*r* = *t*d*y*) and *<sup>u</sup>* <sup>=</sup> 1/*<sup>t</sup>* (d*<sup>u</sup>* <sup>=</sup> <sup>−</sup>*u*2d*t*), the new expression of the expectation given by (65) is written as follows

$$\begin{split} &E\_{\mathbf{X}^{\prime}}\{\ln[1+\mathbf{X}^{T}\Sigma\_{2}^{-1}\mathbf{X}]\} = -\frac{\partial}{\partial a}\left\{\frac{A}{\Gamma(a)}B\left(\underbrace{\frac{1}{2},\dots,\frac{1}{2}}\_{p}\right)\int\_{0}^{+\infty}r^{a-1}e^{-\lambda\_{P}r} \\ &\times\Phi\_{2}^{(p-1)}\left(\underbrace{\frac{1}{2},\dots,\frac{1}{2}}\_{p-1};\frac{p}{2};(\lambda\_{p}-\lambda\_{1})r,\dots,(\lambda\_{p}-\lambda\_{p-1})r\right)\left(\int\_{0}^{+\infty}u^{a-\frac{1}{2}}(1+u)^{-\frac{1+p}{2}}e^{-ru}du\right)\text{d}r\right\}\Big|\_{a=0}.\end{split} \tag{84}$$
