Proving that $T_t := S_t -\left| x \right| -\frac {n-1}{2} \int _0 ^t \frac {1}{S_u}~du$ is a brownian motion Consider $B=(B_t)_{t\geq 0}$ $\mathcal F_t$ - brownian motion in $\mathbb R ^n, \  (n\geq 2)$ starting at zero, in a probability space $(\Omega, \mathcal F, (\mathcal F_t)_{t\geq 0}, \mathbb P)$. Then, consider $X_t ^x = x + B_t $ and  $S_t = \left|X^x_t\right|$.
I was trying to show that $(T_t)_{t\geq 0}$ defined by
$$T_t := S_t -\left| x \right| -\frac {n-1}{2} \int _0 ^t \frac {1}{S_u}~du$$
is a real $\mathcal F_t$ - brownian motion.
It's not difficult to see that $W_t = S_t -\left| x \right|$ is a real $\mathcal F_t$ - brownian motion.
Inded, Ito's lemma implies that
$$ W_t = \int _0 ^t \frac{X^x_t}{\left| X^x_t \right| }~ dB_s $$
wich is well defined since $\mathbb P (\exists t>0 :  S_t =\left| X^x_t \right| =0) =0$. Then by Pauls-Lévy Theorem we have that $(W_t)$ is a real $\mathcal F_t$ - brownian motion.
So, $$T_t = W_t  -\frac {n-1}{2} \int _0 ^t \frac {1}{S_u}~du.$$
Have anybody a smarter idea that compute its mean and covariance in order to apply again Paul-Lévy theorem ? If not, any smart advices to simplify the calculation?
Thank's in advance.
Edit: Maybe this other result could helpful to crack this proof.
I could show that $\mathbb E \left\{ \int _0 ^t \frac {1}{S_u}~du\right\} =0$  since $M_t := \int _0 ^t \frac {1}{S_u}~du$ is a bounded local martingale.
Then for the covariation we have
$$ \mathbb E \left \{T_s T_t  \right\}=t \wedge s -c (\mathbb E \left \{W_s M_t  \right\} +\mathbb E \left \{M_s W_t  \right\}) + \mathbb E \left \{M_s M_t  \right\}$$
New Edit:
Could someone please check the question at the end of my solution try at the answer ?
 A: I'd like to show that $M^\lambda = (M_t^\lambda)_{t\geq0}$ 
$$ M_t^\lambda := \exp\left(i\lambda T_t-\frac{\lambda^2}{2}t \right) $$
 is a complex martingale so $T$ is a brownian motion.
Indeed, by Ito's lemma we have that 
$$dM_t ^\lambda = \frac{\lambda^2}{2}M_t ^\lambda dt + i\lambda M_t ^\lambda dT_t - \frac{\lambda^2}{2} d\langle T \rangle_t$$
but also we have that 
$$ \langle T \rangle_t = \langle W \rangle_t + \langle \int _0 ^ . \frac{1}{S_u}~du \rangle_t = t$$ 
so $$dM_t ^\lambda = i\lambda M_t ^\lambda dT_t$$
and  $$\mathbb E \left\{  \int _0 ^t  \lambda^2 (M_t^ \lambda)^2 d\langle T \rangle_t\right\}\leq \exp(\lambda^2 t)< +\infty$$
However, we must remember $T$ is a local martingale. Then even if $\phi_t := i\lambda M_t ^\lambda \in \mathbb H ^2(T)$ can we conclude that $M^\lambda$ is a martingale ?
A: First, your sentence "$W_t=S_t-|x|$ is a Brownian motion" is wrong, or Brownian motion is bounded below by |x| almost surely. The aim of the integral is precisely to compensate for the absolute value making this not to be a Brownian motion. 
Have you checked out Revuz-Yor or Chaumont-Yor on this ? (I think your exercise in the latter book)
