How does this integral based on the $\Phi$ function equal $x$?

Question

So a stats problem involving normal random variables has a solution involving a step where the following simplification occurs (based on inspection):

$$\int_{-\infty}^{ \infty } y\frac{e^{-(y-x)^2/2}}{\sqrt{2\pi}} dy = x$$

But I don't understand how they arrived at that.

I know that:

$$\Phi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{-\frac{t^2}{2}}dt$$

and:

$$\int_{-\infty}^{\infty}e^{-x^2}dx = \sqrt{\pi}$$

Context: problem solution that involves this simplification

Answer 1

Applying the substitution $u=(y-x)$ yields;

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(u+x)e^{-\frac{u^2}{2}}du$$

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}ue^{-\frac{u^2}{2}}du+\frac{x}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{u^2}{2}}du$$

By symmetry, the first integral is equal to $0$. Which leaves us with...

$$\frac{x}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{u^2}{2}}du=\frac{x}{\sqrt{2\pi}}\cdot\sqrt{2\pi}=x$$

$$\therefore\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}ye^{-\frac{(y-x)^2}{2}}dy=x$$

as expected.

Answer 2

Making the substitution $y\rightarrow y+x$ gives $$ \int_{-\infty}^{\infty}y\frac{e^{-(y-x)^2/2}}{\sqrt{2\pi}}dy=\int_{-\infty}^{\infty}(y+x)\frac{e^{-y^2/2}}{\sqrt{2\pi}}dy = x \int_{-\infty}^{\infty}\frac{e^{-y^2/2}}{\sqrt{2\pi}}dy + \int_{-\infty}^{\infty}y\frac{e^{-y^2/2}}{\sqrt{2\pi}}dy=x+0=x. $$ (The second integral is zero by symmetry: the integrand is an odd function.)