Let $B$ be a standard Brownian motion. Its characteristic exponent (or Fourier transform) is easily calculated to be
$$ \mathbb E [e^{ixB_t}] = e^{-\frac 12 x^2 t}. $$
Now I want to apply a Girsanov transformation with density $\exp(\int_0^t A_s d B_s - \int_0^t A_s^2 ds)$ for some predictable process $A_s$ satisfying Novikov's condition. It is well-known that $B_t-\int_0^t A_sds$ is Brownian motion under the new measure. However, I am more interested in the question what the characteristic exponent of the <i>old</i> Brownian motion under the new measure is. Is there an explicit way to calculate this? Same question, but differently phrased: Is there an elegant way to calculate
$$ \mathbb E [ e^{\int_0^t A_s dB_s} e^{ixB_t}]? $$