Let $X, Y$ be independent and identically distributed standard normal variables, with $aX+bY=u,$ what is $E\left[ cX+dY\frac{b}{\pi}\int_{\Bbb R}\frac {e^{itx}}{b^2 +(x-a)^2}\,\mathrm dxaX+bY=u\right]$? With $a,b,c,d,u$ fixed.
Currently what I have is:
\begin{align} & E[cX+dY\mid aX+bY=u] \\[10pt] = {} & cE\left[X\frac{b}{\pi} \int_{\Bbb R}\frac {e^{itx}}{b^2 +(x-a)^2}\,\mathrm dxaX+bY=u\right]+dE\left[Y\frac{b}{\pi}\int_{\Bbb R}\frac {e^{itx}}{b^2 +(x-a)^2}\,\mathrm dxaX+bY=u\right] \end{align}
and as $aX+bY=u$, then $X=\frac{u-bY}{a}$,$Y=\frac{u-aX}{b}$ So we have: $$cE\left[ X\frac{b}{\pi}\int_{\Bbb R}\frac {e^{itx}}{b^2 +(x-a)^2}\,\mathrm dxaX+bY=u \right] + dE \left[ Y\frac{b}{\pi}\int_{\Bbb R}\frac {e^{itx}}{b^2 +(x-a)^2}\,\mathrm dxaX+bY=u\right] = cE\left[\frac{u-bY}{a}\right] + dE\left[\frac{u-aX}{b}\right]$$ and as $E[X]=E[Y]=0$, we get $$u\left(\frac{c}{a}+\frac{d}{b}\right)$$ is that correct?
Observe that: $$(a^2+b^2)(cX+dY)=(ac+db)(aX+bY)+(bc-ad)(bX-aY)$$
So setting $U=aX+bY$ and $V=bX-aY$ we can write:$$cX+dY=\frac{(ac+db)U+(bc-ad)V}{a^2+b^2}$$ It is not difficult to find that: $$\mathsf{Cov}(U,V)=0$$ Because we are dealing with normal distribution this is sufficient for independence of $U$ and $V$.
Further we have $\mathbb EV=0$ and this together leads to:
$$\mathbb E(cX+dY\mid U=u)=\frac{(ac+db)u}{a^2+b^2}$$