# How to compute $\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{e^{-{x^2}/{2}}}{1+e^{-(ax+b)}}\,\mathrm dx$ when $b \ne 0$

I'm trying to solve this expression when $b \ne 0$:

$$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{e^{-{x^2}/{2}}}{1+e^{-(ax+b)}}\,\mathrm dx$$

Using the answers provided to this question we can solve $$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{e^{-{x^2}/{2}}}{1+e^{-x}}\,\mathrm dx=\frac12$$

However, when we change the power of exponential term in the denominator to $-(ax+b)$ from $x$, the expression can not be solved using the same methods used in the post. I tried solving by substituting $z=ax+b$ and it also did not work.

• Approximation: if $\left| a\right| <<1$ then: $\frac{e^b}{e^b+1}$ – Mariusz Iwaniuk Aug 7 '18 at 12:46
• @MariuszIwaniuk can you please explain how that happens? – Nadheesh Aug 7 '18 at 12:54
• put a=0 and integrate. – Mariusz Iwaniuk Aug 7 '18 at 12:57
• Closed-form formula for: $a=1$ and $b=1$ is: $1-\frac{1}{2 \sqrt{e}}$. Another one: $a=1$ and $b=-1$ is: $\frac{1}{2 \sqrt{e}}$ – Mariusz Iwaniuk Aug 7 '18 at 16:31
• @MariuszIwaniuk thanks for the suggestion, but $a \ne 0$. $a$ and $b$ can be any value except for $0$. – Nadheesh Aug 7 '18 at 17:40