Finding $P(X$X \sim \operatorname{Exp}(1)$, $Y \sim N(0,1)$, $X$ and $Y$ are independent. 
I know the following fact:
$$P(X < Y+1) = \int_{-\infty}^\infty P(X < y+1 \mid Y = y)  f_Y(y)\, dy$$
Using this, I try to show the (known) fact that: 
$$P(X < Y+1) = Φ(1) - \frac{1}{2\sqrt e} = 0.5377346701 $$
I do the following:
$$P(X < Y+1) = \int_{-\infty}^\infty P(X < y+1)  f_Y(y)\, dy\,,$$ due to independence of $X$ and $Y$.
$$P(X < Y+1) = \frac{1}{\sqrt{2π}} \int_{-\infty}^\infty (1-e^{-(y+1)})(e^{-\frac{y^2}{2}})\, dy\,,$$ by using the pdf of $Y$ and the probability of $X$.
Which gives me the result (when put into an integral calculator), of:
$$ \frac{\sqrt{e}-1}{\sqrt{e}}  = 0.3934693403 $$
I thought I had done every step correctly yet I am quite lost as to where I could have gone wrong. I feel that it is related to my limits.
 A: Up to this point:

$P(X < Y+1) = \int_{-\infty}^\infty P(X < y+1)  f_Y(y) dy$

you are completely correct. The problem is in replacing $P(X<y+1)$ with $1-e^{-(y+1)}$, since this only holds for $y \geq -1$. To see why, recall that 
$$P(X  \le x) = \int_{-\infty}^xf_X(x) \ dx$$
where $f_X(x)$ is the pdf of $X$. If $X \sim \text{Exp}(1)$, then $f_X(x)=\begin{cases} e^{-x} & x \geq 0 \\ 0 & x<0 \end{cases}$. So we have:
$$P(X \leq x) = \int_{-\infty}^x f_X(x) \ dx = \begin{cases} \int_0^x e^{-x} \ dx & x \geq 0 \\ 0 & x<0 \end{cases} = \begin{cases} 1-e^{-x} & x \geq 0 \\ 0 & x<0 \end{cases}.$$
Thus $P(X \leq x)=1-e^{-x}$ only holds for $x \geq 0$. It follows that $P(X \leq y+1) = 1-e^{-(y+1)}$ only holds for $y \geq -1$; for $y<-1$ we have instead $P(X \leq y+1)=0$. Therefore the next step should restrict the integral bounds to go from $y=-1$ to $y=\infty$:
$$P(X <Y+1) = \int_{-1}^\infty (1-e^{-(y+1)}) \frac{1}{\sqrt{2\pi}} e^{-\frac{y^2}{2}} \ dy.$$
This should give you the correct answer.
