How to integrate this function How to integrate the following double integral:
$$\int_0^{+ \infty}\int_{- \infty}^{+ \infty}e^{(ax+by)} e^{-x} e^{-0.5(y-x)^2} \; dy \; dx$$
 A: $$\begin{align*}
\int_0^\infty\int_{-\infty}^\infty e^{ax+by}e^{-x}e^{-0.5(y-x)^2}\ dy\ dx &= \int_0^\infty\int_{-\infty}^\infty e^{(a-1)x}e^{by}e^{-\frac12(y^2-2xy-x^2)}\ dy\ dx \\
 &= \int_0^\infty e^{(a-1)x+\frac12x^2}\int_{-\infty}^\infty e^{-\frac12y^2+(b+x)y}\ dy\ dx
\end{align*}
$$
Now, evaluate the inner integral first, treating $x$ as a constant.
A: Hint
Expand the $(y-x)^2 = y^2 + x^2 - 2xy$ then separate the 3 integrals, factor out $x,y$ integration into separate problems and use integration by parts.
If the intent was to have $e^{0.5 \cdot (y-x)^2}$ instead, expand the exponent, separate the $x,y$ integration and compute.
A: Formally, at least, (i.e., interchanging the order of integration and making substitutions at will), we have
$$\begin{align}
\int_0^\infty\int_{-\infty}^\infty e^{ax+by}e^{-x}e^{-{1\over2}(y-x)^2}\,dy\,dx
&=\int_0^\infty\int_{-\infty}^\infty e^{ax+bx+bu}e^{-x}e^{-{1\over2}u^2}\,du\,dx\\
&=\int_0^\infty e^{ax+bx-x}\,dx \int_{-\infty}^\infty e^{-{1\over2}(u-b)^2+{1\over2}b^2}\,du\\
&=e^{{1\over2}b^2}\int_0^\infty e^{(a+b-1)x}\,dx\int_{-\infty}^\infty e^{-{1\over2}v^2}\,dv
\end{align}$$
The integral over $x$ is easy to evaluate, but it diverges if $a+b\ge1$.  The integral over $v$ is standard; it integrates to $\sqrt{2\pi}$, I think.
