# I need help on solving ∫e^(z^2)* fn(z) dz, where fn(z) = . . .

1. Jul 26, 2007

### T.Engineer

I need to work on solving ∫e^(z^2)* fn(z) dz.

where fn(z) = d^n/dz^n * e^(-z^2).

Thanks alot!

2. Jul 26, 2007

### cliowa

Try working out (inductively) the form of fn(z).

3. Jul 26, 2007

### T.Engineer

you mean to find the derivative for the
fn(z) = d^n/dz^n * e^(-z^2).

and then substitue it?

4. Jul 26, 2007

### cliowa

Indeed, yes. Try f1, then f2 and so on to see what's going on.

5. Jul 26, 2007

### T.Engineer

the derivative for e^-(Z^2) is
-2 Z e^-(z^2)

is not that true?

6. Jul 26, 2007

### cliowa

Yes it is. Do you notice anything special about the integral?

7. Jul 27, 2007

### T.Engineer

Now if I will substitute the derivative of fn(z)
Which is:
-2 z * e^(-z^2)
So, the integral equation will be like this:
-2 z * e^(z^2) * e^(-z^2) dz
here I get confused and I couldn’t find the integral?

8. Jul 27, 2007

### cliowa

What is $e^{a}\cdot e^{b}$ equal to?

9. Jul 27, 2007

### T.Engineer

you mean it will be somthing like this
e^[(z^2)*(-z^2)]

or it maybe e^(a+b)

Last edited: Jul 27, 2007
10. Jul 27, 2007

### cliowa

There's no guessing involved there! It's $e^a*e^b=e^{a+b}$.

11. Jul 27, 2007

### T.Engineer

So, the integral it will be like this:
-2 Z dz

is not that right?

12. Jul 27, 2007

### cliowa

That's correct.

13. Jul 27, 2007

### T.Engineer

Thanks alot!!