# Homework Help: Integral of (1/(at+b))e^(-t^2)

1. Mar 27, 2013

### quantum_2000

1. The problem statement, all variables and given/known data

I need to calculate the integral of (1/(at+b))e^(-t^2), that is, a negative quadratic exponential divided by a linear function at+b. I need to integrate between some positive x and +infinity.

2. Relevant equations

3. The attempt at a solution

I could find the integral for b=0. In fact the integral of 1/(at) e^(-t^2) between x>0 and +infinity is equal to -1/(2a) Ei(-x^2) where Ei(x) is the exponential integral function defined as
-Ei(-x)=E1(x)=Integral of (1/t)e^-t between x and +infinity. Since Ei can be expressed through the Incomplete Gamma function and I have the latter in Excel, in this case my problem would be - at least computationally - solved.

However, when b is not zero I cannot find any solution. Any help guys?

2. Mar 27, 2013

### Staff: Mentor

Seems like a tough one. Do you have numerical values for a, b, and x? If yes, I would suggest a numerical integration.

3. Mar 28, 2013

### quantum_2000

The reason why I need an analytical solution is that I need to be able to compute the integral in a single Excel cell for any given set of parameter values. So, say that a=20 b=50 and x=80, I need to compute in a single cell (that means without being able to compute partial results in any other cell or set of cells) the integral of e^-t^2 / (20a+50) between 80 and + infinity.
If I understand correctly what you mean by numerical integration, it seems this would require to launch a calculation with a specific software for each instance, and this is not what I need :(

4. Mar 28, 2013

### Staff: Mentor

May I ask where the integral comes from?

5. Mar 28, 2013

### quantum_2000

It is the integral of the pdf of a normal distribution times 1/t. With substitution I get the integral in the title where b is the mean of the distribution. So, if anyone has a solution to compute the integral of (1/t)*PDF(NormalDistribution(mu,sigma),t) from x>0 to +infinity, this would make me very happy as well.

Now, I could also explain where this comes from, but it is not a short story...

6. Apr 2, 2013

### quantum_2000

So, in the end no one has a clue about it?