How do I change integration order?

Question

I need to bound (or calculate in a closed form) this integral: $$\int_{R}^{\infty}dz\int_{a-bz}^{\infty}dy\cdot \frac{1}{\sqrt{2\pi}}e^{\frac{-y^2}{2}} e^{-z}$$ as a function of $R,a,b$. The result can be expressed in terms of: $$ Q(x) = \int_x^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}dt$$

Answer 1

To change the integration order, it is useful to draw the integration domain. On the first figure, the domain is bounded by $y=R$ and $y=a-bz$. On the second, these are the same except that we have exchanged the axes.

If $b>0$ we therefore get $$I=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{a-bR}\mathrm e^{-y^2/2}\mathrm dy\int_{\frac{a-y}b}^\infty \mathrm e^{-z}\mathrm dz +\frac{1}{\sqrt{2\pi}}\int_{a-bR}^\infty \mathrm e^{-y^2/2}\mathrm dy\int_R^\infty\mathrm e^{-z}\mathrm dz.$$ This easily becomes $$I=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{a-bR}\mathrm e^{-y^2/2} \mathrm e^{(y-a)/b}\mathrm dy+Q(a-bR)\mathrm e^{-R}$$ and finally $$I=Q\left(bR-a+\frac1b\right)\mathrm e^{\frac{1}{2b^2}-\frac ab}+Q(a-bR)\mathrm e^{-R}.$$

Answer 2

Remember that the Gauss' integral is $$G:=\int_{\mathbb{R}}\mathrm{e}^{-y^2/2}\mathrm{d}y=\sqrt{2\pi}$$ (see https://en.wikipedia.org/wiki/Gaussian_integral). So we have $Q\leq G/\sqrt{2\pi}=1$ (because we increase the domain of integration and the integrand is non-negative) and then \begin{align} \int_{R}^{+\infty}\left(\int_{a-bz}^{+\infty}\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-y^2/2}\mathrm{d}y\right)\mathrm{e}^{-z}\mathrm{d}z & = \int_{R}^{+\infty}Q(ax-z)\mathrm{e}^{-z}\mathrm{d}z \\ & \leq \int_{R}^{+\infty}\mathrm{e}^{-z}\mathrm{d}z \\ & = \mathrm{e}^{-R}. \end{align}