I am deriving a formula for a volume of Wiener sausage in one dimension.
$\mathbb{E}[\operatorname{vol}(W(t))] = 2r+\sqrt{\frac{8t}{\pi}}$ where $W(t) = \bigcup_{s\leq t}\mathcal{B}(\epsilon(s),r)$. $\mathcal{B}$ - ball (just a segment in this case). $\epsilon(s)$ - standard Brownian motion. Vol for volume (length in 1d).
I follow the trick my lecturer used for higher dimensions. Let $\tau_A$ stand for a hitting time of $\epsilon$ for $A$:
$\mathbb{E}[\operatorname{vol}(W(t))]=\int_{\mathbb{R}}\mathbb{P}(y \in \bigcup_{s\leq t}\mathcal{B}(\epsilon(s),r))\;dy=\int_{\mathbb{R}}(\tau_{\mathcal{B}(y,r)}\leq t))\;dy=\operatorname{vol}(\mathcal{B}(0,r))+$ ${}+\int_{\mathbb{R}\backslash \mathcal{B}(0,r)}(\tau_{\mathcal{B}(y,r)}\leq t))\;dy=2r+\int_{-\infty}^{-r}\mathbb{P}(\tau_{y+r}\leq t)\;dy+\int_{r}^{\infty}\mathbb{P}(\tau_{y-r}\leq t)\;dy$
I use $\mathbb{P(\tau_{|a|} \leq t)=\frac{2}{\sqrt{2\pi t}}\int_{|a|}^{\infty}}e^{-\frac{x^2}{2t}}dx$ and the latter 2 integrals evalute to $\infty$ each (after use of Fubini).
Thank you for any suggestions.
Edit: $\int_{-\infty}^{-r}\mathbb{P}(\tau_{y+r}\leq t)\;dy+\int_{r}^{\infty}\mathbb{P}(\tau_{y-r}\leq t)\;dy = 2\int_0^\infty\mathbb{P}(\tau_y\leq t)\;dy=$=$2\int_0^\infty\int_y^\infty\frac{2}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}\;dx\;dy=\frac{4}{\sqrt{2\pi t}}\int_0^\infty\int_0^xe^{-\frac{x^2}{2t}}\;dy\;dx=\frac{4}{\sqrt{2\pi t}}\int_0^\infty xe^{-\frac{x^2}{2t}}\;dx=\sqrt{\frac{8t}{\pi}}$