Calculate $\mathbb{E}\left(\Phi\left(W_{t}\right)\right)$ where $W_{t}$ is a brownian motion and $\Phi$ the standard normal cumulative distribution.
Solution using integrals: $$ \begin{gathered} \mathbb{E}\left(\Phi\left(W_{t}\right)\right)=\int_{-\infty}^{+\infty}\left(\int_{-\infty}^{x} \frac{1}{\sqrt{2 \pi}} \exp \left(\frac{-u^{2}}{2}\right) d u\right) \frac{1}{\sqrt{2 \pi t}} \exp \left(\frac{-x^{2}}{2 t}\right) d t \\ \mathbb{E}\left(\Phi\left(W_{t}\right)\right)=\int_{-\infty}^{+\infty} \Phi(x) \Phi^{\prime}(x) d x=\left[\frac{\Phi^{2}(x)}{2}\right]_{-\infty}^{+\infty}=\frac{1}{2} \end{gathered} $$ I don't understand why the first integral is structured like that and also the inner part is $\Phi(x)$ but what is $\Phi^{\prime}(x)$ in the first integral?
The first integral is structured like that because, by definition, $$ \Phi(x)=\int_{-\infty}^{x} \frac{1}{\sqrt{2 \pi}} \exp \left(\frac{-u^{2}}{2}\right)\,du. $$ I hope this tells you what $\Phi'(x)$ is. Also note that, for any function $f(x)$, $$ \mathbb E(f(W_t))=\int_{-\infty}^{+\infty}f(x)\,\underbrace{\frac{1}{\sqrt{2 \pi t}} \exp \left(\frac{-x^{2}}{2t}\right)}_{(*)}\,dx. $$ because the pdf of $W_t$ is (*) which obviously equals $\Phi'(x/\sqrt{t})/\sqrt{t}\,.$ Your second equation looks wrong. It should be \begin{align} \mathbb E(\Phi(W_t))&=\int_{-\infty}^{+\infty}\Phi(x)\,\Phi'(x/\sqrt{t})\frac{dx}{\sqrt{t}}\,. \end{align} To solve this I prefer to write it as \begin{align}\tag{1} \mathbb E(\Phi(W_t))&=\int_{-\infty}^{+\infty}\Phi(ay+b)\,\Phi'(y)\,dy\,. \end{align} In your case $a=\sqrt{t},b=0,y=x/\sqrt{t}\,.$ Eq. (1) is $$\tag{2} \mathbb P\{X\le aY+b\} $$ where $X,Y$ are two independent standard normal random variables. Obviously, $$ Z:=\frac{X-aY}{\sqrt{1+a^2}} $$ is a standard normal RV. Therefore, (2) is $$ \mathbb P\{Z\le b\}=\Phi(b/\sqrt{1+a^2})\,. $$ In your case $b=0$. Therefore (1) and (2) equal $1/2\,.$