This might be a good example of a setting where using what the functions $\phi$ and $\Phi$ mean probabilistically, saves ink and tedious computations.

Recall that, for every suitable function $u$, $\int\limits_{-\infty}^\infty u(x)\phi(x)\mathrm dx=E(u(X))$ and that, for every real number $x$, $\Phi(x)=P(Y\leqslant x)$, where $X$ and $Y$ are standard normal random variables. Using this for $u:x\mapsto\Phi((x-b)/a)$ and assuming that $X$ and $Y$ are independent, one sees that the integral to be computed is $$ (\ast)=E(\Phi((X-a)/b))=P(Y\leqslant(X-b)/a)=P(Z\geqslant b), $$ where $Z=X-aY$ (this step uses the fact that $a\gt0$). Now, the random variable $Z$ is normal as a linear combination of independent gaussian random variables, with mean $0$ and variance $a^2\cdot1+1$, hence $Z=\sqrt{a^2+1}\cdot T$, where $T$ is standard normal. Thus, $$ (\ast)=P(T\geqslant b/\sqrt{a^2+1})=1-\Phi\left(b/\sqrt{a^2+1}\right). $$ Likewise, if $a\lt0$, then $(\ast)=\Phi\left(b/\sqrt{a^2+1}\right).$

In particular, if $b=0$ then, for every $a\ne0$, $(\ast)=\frac12$.

This might be a setting where relying on the probabilistic meaning of the functions $\phi$ and $\Phi$ saves ink and tedious computations.

Let $X$ and $Y$ denote standard normal random variables. Then $\int\limits_{-\infty}^\infty u(x)\phi(x)\mathrm dx=E(u(X))$ for every suitable function $u$ and $\Phi(x)=P(Y\leqslant x)$ for every real number $x$. Using this for the function $u:x\mapsto\Phi((x-b)/a)$ and assuming furthermore that $X$ and $Y$ are independent, one sees that the integral to be computed is $$ (\ast)=E(\Phi((X-a)/b))=P(Y\leqslant(X-b)/a). $$ Thus, $$ (\ast)=P(Z\geqslant b), $$ where $Z=X-aY$ (this step uses the fact that $a\gt0$). Now, the random variable $Z$ is normal as a linear combination of independent gaussian random variables, with mean $0$ and variance $1+a^2$, hence $Z=\sqrt{a^2+1}\cdot T$, where $T$ is standard normal. Thus, $$ (\ast)=P(T\geqslant b/\sqrt{a^2+1})=1-\Phi\left(b/\sqrt{a^2+1}\right). $$ Likewise, if $a\lt0$, then $(\ast)=\Phi\left(b/\sqrt{a^2+1}\right).$

In particular, if $b=0$ then, for every $a\ne0$, $(\ast)=\frac12$.

This might be a good example of a setting where using what the functions $\phi$ and $\Phi$ mean probabilistically, saves ink and tedious computations.

Recall that, for every suitable function $u$, $\int\limits_{-\infty}^\infty u(x)\phi(x)\mathrm dx=E(u(X))$ and that, for every real number $x$, $\Phi(x)=P(Y\leqslant x)$, where $X$ and $Y$ are standard normal random variables. Using this for $u:x\mapsto\Phi((x-b)/a)$ and assuming that $X$ and $Y$ are independent, one sees that the integral to be computed is $$ (\ast)=E(\Phi((X-a)/b))=P(Y\leqslant(X-b)/a)=P(Z\geqslant b), $$ where $Z=X-aY$ (this step uses the fact that $a\gt0$). Now, the random variable $Z$ is normal as a linear combination of independent gaussian random variables, with mean $0$ and variance $a^2\cdot1+1$, hence $Z=\sqrt{a^2+1}\cdot T$, where $T$ is standard normal. Thus, $$ (*)=P(T\geqslant b/\sqrt{a^2+1})=1-\Phi\left(b/\sqrt{a^2+1}\right). $$ Likewise, if $a\lt0$, then $(*)=\Phi\left(b/\sqrt{a^2+1}\right).$

This might be a good example of a setting where using what the functions $\phi$ and $\Phi$ mean probabilistically, saves ink and tedious computations.

Recall that, for every suitable function $u$, $\int\limits_{-\infty}^\infty u(x)\phi(x)\mathrm dx=E(u(X))$ and that, for every real number $x$, $\Phi(x)=P(Y\leqslant x)$, where $X$ and $Y$ are standard normal random variables. Using this for $u:x\mapsto\Phi((x-b)/a)$ and assuming that $X$ and $Y$ are independent, one sees that the integral to be computed is $$ (\ast)=E(\Phi((X-a)/b))=P(Y\leqslant(X-b)/a)=P(Z\geqslant b), $$ where $Z=X-aY$ (this step uses the fact that $a\gt0$). Now, the random variable $Z$ is normal as a linear combination of independent gaussian random variables, with mean $0$ and variance $a^2\cdot1+1$, hence $Z=\sqrt{a^2+1}\cdot T$, where $T$ is standard normal. Thus, $$ (\ast)=P(T\geqslant b/\sqrt{a^2+1})=1-\Phi\left(b/\sqrt{a^2+1}\right). $$ Likewise, if $a\lt0$, then $(\ast)=\Phi\left(b/\sqrt{a^2+1}\right).$

In particular, if $b=0$ then, for every $a\ne0$, $(\ast)=\frac12$.