Let $\tau=\inf\{t>0\mid B_t=1\}$ where $(B_t)$ is a Brownian motion. I need to prove that $\mathbb E[\tau]=\infty $. So, I tried as follow :
$$\mathbb P\{\tau>t\}=\mathbb P\{\sup_{s\in [0,t]}B_s<1\}=1-\mathbb P\{\sup_{s\in [0,t]}B_s\geq 1\}=1-2\mathbb P\{B_t>1\}=1-\sqrt{\frac{2}{\pi t}}\int_1^\infty e^{-x^2/2t}dx.$$ Now, since in the integral, there is a $t$ dependance, I'm not so sure if $\int_0^\infty \mathbb P\{\tau>t\}dt=\infty $ or not. Is there an easy way to see that it's really $\infty $ ?
Notice that if $\tau\in L^1$, then by Optional Sampling Theorem, $\mathbb E[B_\tau]=\mathbb E[B_0]=0$. However, $\mathbb E[B_\tau]=1$. Therefore, $\tau\notin L^1$.
To conclude using your technique, notice that $e^{-\frac{x^2}{2t}}\leq \frac{2t}{x^2}$. Therefore, $$g(t):=1-\frac{2}{\sqrt{2\pi t}}\int_1^\infty e^{-\frac{x^2}{2t}}dx\geq 1+\frac{4}{\sqrt{2\pi}}\sqrt t,$$ and thus $\int_0^\infty g=\infty $.