Calculating the expecation of the supremum of absolute value of a Brownian motion I got a Brownian motion $B(t)$ that starts in $0$ and want to calculate the expectated value of the supremum on the interval $[0,1]$ of the absolute value of it, i.e.
$E \left (\sup \limits_{t \in [0,1]}  |B(t) | \right )$.
I found some information on this question in this thread.
Here user3371583 posted in the comment chain that the expected value in the case of the interval $[0,1]$ should be $\sqrt{\frac{\pi}{8}}$ and he explained how he got to this result in the same post, but I cant seem to get it done.
What I got so far is this:
$E \left (\sup \limits_{t \in [0,1]}  |B(t) | \right ) = \int_0^\infty P(\sup \limits_{t \in [0,1]}  |B(t) | \geq y )  dy $
$ =  \int_0^\infty \sum_{k=-\infty}^\infty (-1)^k \text{sign}((2k+1) y) \text{Erfc} \left ( \frac{\vert (2k+1) y \vert }{\sqrt{2}}  \right ) dy $.
Here Erfc is the error function: 
$\text{Erfc}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \, dt$
Now the post mentions splitting the series up into a series over negative $k$ and one over positive (nonnegative I guess) $k$, so that I get
$= \int_0^\infty \sum_{k=-\infty}^{-1} (-1)^k \text{sign}((2k+1) y) \text{Erfc} \left ( \frac{\vert (2k+1) y \vert }{\sqrt{2}}  \right ) dy $
$ + \int_0^\infty \sum_{k=0}^\infty (-1)^k \text{sign}((2k+1) y) \text{Erfc} \left ( \frac{\vert (2k+1) y \vert }{\sqrt{2}}  \right ) dy  $
and now the next step seems to be using a change of variable to see that both terms are actually equal, but I cannot verify this. The first term is
$\int_0^\infty \sum_{k=1}^{\infty} (-1)^{-k} \text{sign}((2(-k)+1) y) \text{Erfc} \left ( \frac{\vert ((2(-k)+1) y \vert }{\sqrt{2}}  \right ) dy $
$= \int_0^\infty \sum_{k=1}^{\infty} (-1)^{k} \text{sign}((2k+1) y) (-1)\text{Erfc} \left ( \frac{\vert ((2(-k)+1) y \vert }{\sqrt{2}}  \right ) dy $ 
and this is where I'm stuck already. I know that the Errorfunction is an odd function, but I cant seem to bring it into a form where I can combine both sums into one.
Also, I don't understand how to use integration by parts to evaluate both of the integrals after combining the series.
Can anyone help me out with this? Thanks!
 A: First of all, note that Erfc does not denote the Error function but its "complement", i.e.
$$\DeclareMathOperator{\erfc}{Erfc}\DeclareMathOperator{\erf}{Erf}   \erfc(x) := \frac{2}{\sqrt{\pi}} \int_x^{\infty} \exp (-y^2) \, dy = 1- \erf(x)$$
where 
$$\erf(x) = \frac{2}{\sqrt{\pi}} \int_0^x \exp (-y^2) \, dy$$
is the Error function.

Since
$$|2(-k)+1| = |2k-1|$$
we have
\begin{align*} \DeclareMathOperator{\sign}{sgn}  \DeclareMathOperator{\erf}{Erfc} & - \int_{(0,\infty)} \sum_{k \geq 1} (-1)^k \underbrace{\sign((2k+1)y)}_{=1} \erf \left( \frac{|(2(-k)+1)y|}{\sqrt{2}} \right) \, dy \\ &= \int_{(0,\infty)} \sum_{k \geq 1} (-1)^{k+1} \erf \left( \frac{|(2k-1)y|}{\sqrt{2}} \right) \, dy \\
&\stackrel{k \to j+1}{=} \int_{(0,\infty)} \sum_{j \geq 0} (-1)^{j} \underbrace{(-1)^2}_{=1} \erf \left( \frac{|(2j+1)y|}{\sqrt{2}} \right) \, dy.
\end{align*}
Hence, by the computations from your question,
\begin{align*} \mathbb{E} \left( \sup_{t \in [0,1]} |B_t| \right) = 2 \int_{(0,\infty)} \sum_{j \geq 0} (-1)^j  \erf \left( \frac{|(2j+1)y|}{\sqrt{2}} \right) \, dy. \end{align*}
Interchanging sum and integration and applying the integration by parts formula we get
\begin{align*} \mathbb{E} \left( \sup_{t \in [0,1]} |B_t| \right) &= 2 \sum_{j \geq 0} (-1)^j \int_{(0,\infty)} \erf \left( \frac{(2j+1)y}{\sqrt{2}} \right) \, dy \\ &=  \frac{4}{\sqrt{2\pi}} \sum_{j \geq 0} (2j+1) (-1)^j \int_{(0,\infty)} y \exp \left(- \frac{(2j+1)^2}{2} y^2 \right) \, dy \\ &= \frac{4}{\sqrt{2\pi}} \sum_{j \geq 0} (-1)^j  \frac{1}{2j+1} \end{align*}
As $$\arctan(x) = \sum_{j \geq 0} (-1)^j \frac{x^{2j+1}}{2j+1}$$ we conclude that
$$\mathbb{E} \left( \sup_{t \in [0,1]} |B_t| \right) = \frac{4}{\sqrt{2\pi}} \arctan(1) =  \frac{4}{\sqrt{2\pi}}  \frac{\pi}{4} = \sqrt{\frac{\pi}{2}} $$
Remark I: Feel free to check the constants appearing in my computations. In the thread, which you linked, it is claimed that
$$\mathbb{E} \left( \sup_{t \in [0,1]} |B_t| \right) = \sqrt{\frac{\pi}{8}};$$
this is, however, wrong. It follows from the reflection principle that $\sup_{t \in [0,1]} B_t \sim |B_1|$, and therefore
\begin{align*} \mathbb{E} \left( \sup_{t \in [0,1]} |B_t| \right) \geq \mathbb{E} \left( \sup_{t \in [0,1]} B_t \right) &= \mathbb{E}(|B_1|) \\ &= \sqrt{\frac{2}{\pi}} \approx 0.797 \\ &> \sqrt{\frac{\pi}{8}} \approx 0.626 \end{align*}
Remark II: Let me provide a reference for the formula for $\mathbb{P}(\sup_{s \leq 1} |B_s| \geq r)$ which you used in your computations (just in case that somebody wants to look it up). In the Handbook of Brownian Motion by Borodin & Salminen it is stated on p. 339 (Part II, Chapter 3, Formula 1.1.4) that $$\mathbb{P}^x \left( \sup_{s \leq t} |B_s| \geq r \right) = \tilde{\text{cc}}_t(x,r)$$ where $$\tilde{\text{cc}}_t(x,r) := \sum_{k \in \mathbb{Z}} (-1)^k \sign(x+(2k+1)r) \erfc \left( \frac{|x+(2k+1)r|}{\sqrt{2t}} \right),$$ see p. 651 (Appendix II, Section 13).
