# $\int_{-\infty}^{+\infty}\phi\left(x\right)\Phi\left(\frac{a}{\mathrm{e}^x}\right)dx=\Phi\left(\frac{a}{\sqrt{2}}\right)$

I think I have found a solution to the integral below using similar logic I have found to an answer here https://mathoverflow.net/questions/101469/integration-of-the-product-of-pdf-cdf-of-normal-distribution. However, I am not certain the logic can also apply to log-normal as I have done below. Can you validate whether you think this result is correct?

$I_1=\int_{-\infty}^{+\infty}\phi\left(x\right)\Phi\left(\frac{a}{\mathrm{e}^x}\right)dx$

I observe that this is equivalent to:

$I_2=\int_0^{+\infty}f\left(y\right)F\left(\frac{\mathrm{e}^a}{y}\right)dy$

Where $f,F$ are the pdf and cdf of the log-normal distribution respectively.

This is the expectation $\mathbb{E}\left(F\left(\frac{\mathrm{e}^a}{y}\right)\right)=\mathbb{P}\left(Z\le\frac{\mathrm{e}^a}{Y}\right)$ where $Y,Z$ are independent log-normal.

Since $YZ$ is also log-normal with $\mu_{YZ}=\mu_Y+\mu_z=0,\sigma_{YZ}=\sqrt{\sigma_Y^2+\sigma_Z^2}=\sqrt{2}$, it follows that:

$\mathbb{P}\left(YZ\le \mathrm{e}^a\right)=\mathbb{P}\left(\mathrm{e}^{\sqrt{2}T}\le \mathrm{e}^a\right)=\mathbb{P}\left(T\le\frac{a}{\sqrt{2}}\right)=\Phi\left(\frac{a}{\sqrt{2}}\right)$
where $T$ is standard normal.

Therefore $I_1=\int_{-\infty}^{+\infty}\phi\left(x\right)\Phi\left(\frac{a}{\mathrm{e}^x}\right)dx=\Phi\left(\frac{a}{\sqrt{2}}\right)$

UPDATE: I have tried to generalise the result using the same logic applied to $\mathrm{e}^{\sigma x+\mu}$, which gives:

$\int_{-\infty}^{+\infty}\phi\left(x\right)\Phi\left(\frac{a}{\mathrm{e}^{\sigma x+\mu}}\right)dx=\Phi\left(\frac{a-\mu}{\sqrt{1+\sigma^2}}\right)$