How find stochastic logarithm of $B^2(t)+1$.

Question

Find the stochastic logarithm of $B^2(t)+1$.
I know that for find stochastic logarithm According to Theorem we must use the The following formula $$X(t)=\mathcal L(U)(t)= \ln(\frac{U(t)}{U(0)})+\int_{0}^{t} \frac{d[U,U](t)}{2U^2(t)} $$

I start and if $U(t)=B^2(t)+1$ Then $dU(t)=dt+2B(t)dB(t)$ and $d[U,U](t)=4 B^2(t)dt$

and my problem is I cannot get a close form or I cannot Calculate the integral.

thanks for help

Answer 1

You might be expected to note that $$ X_t=\int_0^t\frac{\mathrm dU_s}{U_s}=2\int_0^t\frac{B_s}{B_s^2+1}\mathrm dB_s+\int_0^t\frac{\mathrm ds}{B_s^2+1}. $$

Answer 2

I happen to meet this problem as Exercise 5.8 on Fima C. Klebaner's Introduction to Stochastic Calculus with Applications (3ed).

Just want to make an observation: the two results -- computed directly by definition and by the formula mentioned in the question -- are indeed the same.

On one hand, we may directly compute by definition. That is Did's answer:

$$X(t)=\mathcal L(U)(t)=\int_0^t\frac{\mathrm dU(s)}{U(s)}=\int_0^t\frac{2B(s)}{B(s)^2+1}\mathrm dB(s)+\int_0^t\frac{1}{B(s)^2+1}\mathrm ds$$

On the other hand, we may use the formula mentioned in the question, which in this case yields

$$ X(t)=\mathcal L(U)(t)= \ln(B^2(t)+1) \ +\int_{0}^{t} \frac{2B^2(s)}{[B^2(s)+1]^2}\mathrm ds $$

The two results are indeed the same, which can be seen if we take the differential of $\ln(B^2(t)+1)$ by Ito's formula, which is

$$ \mathrm d\ln(B^2(t)+1) = \frac{2B(s)}{B^2(s)+1}\mathrm dB(s) + \frac{1-B^2(s)}{[B^2(s)+1]^2}\mathrm ds $$

Note that after combining the $\mathrm ds$ terms, $$ \frac{1-B^2(s)}{[B^2(s)+1]^2} + \frac{2B^2(s)}{[B^2(s)+1]^2} = \frac{B^2(s)+1}{[B^2(s)+1]^2} = \frac{1}{B^2(s)+1} $$

So both the terms on $\mathrm dB(s)$ and $\mathrm ds$ are the same for the two formulae.