# Solution to the integral,i.e, expected value of a function of normal variable

1. Jun 26, 2012

### ait.abd

I want to calculate $\int_a^b \frac{1}{\sqrt{2 \pi \sigma^2}} e^{(-(x-\mu)/\sigma^2)} log_2 (1 + e^{-x}) dx$

2. Jun 26, 2012

### D H

Staff Emeritus
You'll need to use numerical techniques. The above integral doesn't have a closed form solution in the elementary functions.

3. Jun 26, 2012

### ait.abd

4. Jun 26, 2012

### chiro

Hey ait.abd and welcome to the forums.

Numerical techniques are ones that give approximate answers in general. You can supply parameters to get a good enough approximation (like for example a number good enough to say 4 decimal places) for the better implementations.

If you are unsure, just use a common package for numerical calculation.

You should probably try searching online for a numeric integrator Java applet, or go to www.wolframalpha.com and enter in your expression to get an approximate answer.

5. Jun 26, 2012

### ait.abd

Thanks chiro. But, wolfram online integrator doesn't work for this expression as it tries to compute the exact expression. I can perform numerical integration but I want answer in terms of $a$ and $b$. Numerical integration will calculate the answer for a particular $a$ and $b$.

6. Jun 26, 2012

### Mute

If you need something analytical, you're going to have to develop some sort of approximation for the integral for certain parameter regimes. For example, for $\sigma \rightarrow 0$, the Gaussian essentially becomes a delta function and you would get

$$\log_2(1+e^{-\mu})\left[\Theta(b-\mu)+\Theta(\mu - a)\right]$$
as the result (the step functions $\Theta$ guarantee that mu is between a and b). So, for $\sigma$ small, a rather crude approximation might be

$$\log_2(1+e^{-\mu}) \int_a^b \frac{dx}{\sqrt{2\pi}\sigma} \exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right],$$
where the integral can be evaluated in terms of the error function. There is a more systematic way to generate this approximation called the method of steepest descent. (You'll have to look that up in a book; I'm afraid the wikipedia article isn't very helpful).

You might also be able to write down an infinite series for the integral. However, when I tried this by expanding the logarithm in powers of e^(-x), I got a sum which looks like it doesn't converge, indicating that either I made a mistake in my calculation or that switching the integral and sum isn't valid in this case.

7. Jun 26, 2012

### utkarsh1

Just looking at the form of equation. It seems that complex contour integral MAY work.

8. Jun 26, 2012

### HallsofIvy

Staff Emeritus
The integral in the original post does NOT involve $e^{-x^2}$. It is, rather, of the form $e^{-x}$.

9. Jun 26, 2012

### D H

Staff Emeritus
I suspect that that may have been a typo. Even if it is a typo, it doesn't help. Either way ( exp(-x2) vs exp(-x) ), this function is not integrable in the elementary functions. Since it's not an integral that is widely used, it's dubious that someone has come up with a nifty way to evaluate it.

To ait.abd: You need to learn how to do numerical integration sometime. If this is the right integral, that sometime is now.