"Time"-Laplace transforms of Brownian motion transition and first hitting time laws

Question

I was having a look at Girsanov's theorem and some of its applications. One of them is the possibility of affirming the following theorem.

Given a probability space $(\Omega, \mathcal{F},P)$ equipped with a filtration $\{\mathcal{F_t} \}_{t\geq1}$, let $B^\mu$ be a Brownian motion with drift applied to said filtration. Also, let $T_\beta^\mu = \inf\{ t : B^\mu(t) = \beta\}$ be a random variable expressing the motion's first hitting time of a level $\beta$. Then

$$P(\max_{0<s<t}B^\mu(s)\geq\beta) = P(T_\beta^\mu\leq t)=\int_0^t\frac{|\beta|}{\sqrt{2\pi s^3}}\exp\left\{{-\frac{(\beta -\mu s)^2}{2s}}\right\}ds$$

I was reading an excerpt of a proof of this fact. The proof strategy consists of using the continuity theorem of Laplace transforms to show that the two distributions are almost certainly identical.

Two identities that were used and that I could not understand were the following.

$$(1) \quad \int_0^\infty \frac{e^{-\gamma t}}{\sqrt{2\pi t}} \exp\left\{{\frac{-x^2}{2t}}\right\}dt = \frac{e^{-|x|\sqrt{2\gamma}}}{\sqrt{2\gamma}}$$

$$(2) \quad \int_0^\infty \frac{e^{-\gamma t}}{\sqrt{2\pi t^3}} |\beta|\exp\left\{{\frac{-\beta^2}{2t}}\right\}dt = e^{-|\beta|\sqrt{2\gamma}}$$

How would I go about prooving these results? I have tried substitutions and integration by parts, but it seems to bring nowhere. Does anybody have a hint as to what direction/method would be most appropriate?