Prove this equation: $$\mathbb{E}[(X_3-c)^{+}]=\frac{1}{\sqrt{2\pi}}\cdot \mathrm{e}^{\frac{-c^2}{2}}-c(1-\Phi(c) \quad (1)$$ with $X_3 \sim N(0,1)$, $$y^{+}:= \left\{\begin{array}{ll} y, & y\ge 0\\ 0, & y<0\end{array}\right. .$$
So these are my calculations: $$ \begin{align*} \mathbb{E}[(X_3-c)^{+}] &= \frac{1}{\sqrt{2\pi}}\int\limits_{0}^{\infty}(x-c)\cdot \mathrm{e}^{-\frac{1}{2}(x-c)^2}dx \\ &=\frac{1}{\sqrt{2\pi}} \int\limits_{0}^{\infty} x\cdot \mathrm{e}^{-\frac{1}{2}(x-c)^2}dx -c\cdot \int\limits_{0}^{\infty} \mathrm{e}^{-\frac{1}{2}(x-c)^2}dx \end{align*}$$
I think it's the right beginning, but the limits don't fit. I know that $\Phi(1-c)=\Phi(-c)=\int\limits_{-\infty}^c \frac{1}{\sqrt{2\pi}}\mathrm{e}^{\frac{-c^2}{2}}$ and the first summand of (1) ist the PDF of the standard normal distribution $f_{X_3}(c)$.
How I get these things together? Thank you all!
You made a mistake by using the formular for the expectation.
It holds for any measurable function $f$ that $$E[f(X_3)] = \int_{-\infty}^\infty f(x)f_{X_3}(x) dx$$ where $f_{X_3}$ is the density of $X_3$.
So in your case you have $f(x) = (x-c)^+$ and get
$$E[(X_3 - c)^+] = \frac{1}{\sqrt{2\pi}}\int\limits_{c}^{\infty}(x-c)\cdot \mathrm{e}^{-\frac{1}{2}x^2}dx$$