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

ait.abd

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

D H

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

ait.abd

can any approximation be made?

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.

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 $b$.

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 [itex]\sigma \rightarrow 0[/itex], the Gaussian essentially becomes a delta function and you would get

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

[tex]\log_2(1+e^{-\mu}) \int_a^b \frac{dx}{\sqrt{2\pi}\sigma} \exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right],[/tex]
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.

utkarsh1

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

HallsofIvy

The integral in the original post does NOT involve [itex]e^{-x^2}[/itex]. It is, rather, of the form [itex]e^{-x}[/itex].

D H

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.