Solution
= Solution
For $t>0$, $B_t$ has the <normal distribution> $N(0,t)$. Evaluate its <Gaussian integral>:
$$
L(t,\alpha,0)
=\frac1{\sqrt{2\pi t}}\int_{\mathbb R}
\exp\!\left[-\left(\alpha+\frac1{2t}\right)x^2\right]dx
=\frac1{\sqrt{1+2\alpha t}}.
$$
At $t=0$ the defining random variable is one, consistent with the same formula. Thus
$$
\boxed{L(t,\alpha,0)=(1+2\alpha t)^{-1/2}\qquad(t,\alpha\geq0).}
$$