TITLE: $\int^\infty_{-\infty} x \exp\{ -\frac{1}{2(1-\rho^2)} (x-y\rho)^2 \} \, dx$
QUESTION [0 upvotes]: How do I integrate the inner integral on 2nd line? 


$$\int^\infty_{-\infty} x \exp\{ -\frac{1}{2(1-\rho^2)} (x-y\rho)^2 \} \, dx$$
I know I can use integration by substitution, let $u = \frac{x-y\rho}{\sqrt{1-\rho^2}}$ resulting in
$$\sqrt{1-\rho^2}\int^{\infty}_{-\infty} [u\sqrt{1-\rho^2} + y\rho] e^{-u^2/2} \; du$$
Thats the 3rd line in the image, but how do I proceed?

REPLY [1 votes]: \begin{align}
&\phantom{=\,\,}\int^\infty_{-\infty}x
\exp\left(%
-\,\frac{\left[x - y\rho\right]^{2}}{2\left[1 - \rho^{2}\right]}
\right)\,dx
\tag{1}
\\[3mm]&=
\int^\infty_{-\infty}\left(x + y\rho\right)
\exp\left(-\,\frac{x^{2}}{2\left[1 - \rho^{2}\right]}\right) \, dx
\tag{2}
\\[3mm]&=
y\rho\,\sqrt{2\pi\,}\,\sqrt{1 - \rho^{2}\,}\
\underbrace{\quad\left[%
{1 \over \sqrt{2\pi\,}\,\sqrt{1 - \rho^{2}\,}}\int^\infty_{-\infty}
\exp\left(-\,\frac{x^{2}}{2(1-\rho^2)}\right) \, dx
\right]\quad}_{{\LARGE =\ 1}}
\tag{3}
\\[3mm]&=
\color{#ff0000}{\large y\rho\,\,\sqrt{\,2\pi\left(1 - \rho^{2}\right)\,\,\,}}
\end{align}
{"set_name": "stack_exchange", "score": 0, "question_id": 526820}