# Integrate e^(x^2)

1. Apr 9, 2008

### sai2020

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

Integrate e^(x^2)

2. Relevant equations

3. The attempt at a solution

I have no idea at all..

2. Apr 9, 2008

### HallsofIvy

Staff Emeritus
There's a reason for that. That function, like most integrable functions, does not have an anti-derivative that can be written in terms of elementary functions.

3. Apr 9, 2008

### sai2020

But someone told me that it's important in Quantum Mechanics. (I'm still first year)

4. Apr 9, 2008

### Schrodinger's Dog

You can solve it using a lambert function, but that is no elementary solution. I only know that as by a fluke someone asked me the same question yesterday. I already knew it was insoluble though with elementary functions.

5. Apr 9, 2008

### tiny-tim

Hi sai!

e^(-x^2) is important in Quantum Mechanics.

But it's usually integrated from 0 to ∞, for which there's a trick (from memory, the result is something like √π), and so an anti-derivative isn't needed.

6. Apr 9, 2008

### sai2020

I see. one of my friends is very good at maths and he's trying to solve this one for days. I'll tell him.

yeah i remember reading it somewhere. it's got something to do with the gaussian distribution if i'm not mistaken.

7. Apr 9, 2008

### tiny-tim

Yes, you're right … it is the gaussian distribution (aka the normal distribution)!

8. Apr 9, 2008

Thanks :)

9. Apr 9, 2008

### Schrodinger's Dog

$\int_0^{\infty} e^{x^2}\;dx$ where $e^{x^2}={\frac{2}{x}}e^{x^2}$

Is it something like this? I can't remember but I've seen that somewhere before? Reaching a bit now though, I wouldn't take my advice

Last edited: Apr 9, 2008
10. Apr 9, 2008

### tiny-tim

No … it's more like ∫e^x^2dx = √[∫e^x^2dx∫e^y^2dy], and then changing to dr and dθ.

11. Apr 9, 2008

### HallsofIvy

Staff Emeritus
And, once again, it must be
$$e^{-x^2}$$
in order to converge.

12. Apr 9, 2008

### tiny-tim

oops!

Thanks, HallsofIvy!

13. Apr 9, 2008

### Schrodinger's Dog

Hehe in that case I don't know where that came from.

Humour me could you show me how you would go about doing that in more detail? I'm not sure quite if I get that?

Are you talking about spherical polars? If so how does that work? Or should I say why does that work?

Last edited: Apr 9, 2008
14. Apr 9, 2008

### tiny-tim

Well, ∫e^-x^2dx and ∫e^-y^2dy are two different ways of writing the same thing, aren't they?

So each is the √ of their product, ie:

√∫∫e^-(x^2 + y^2)dxdy

= √∫∫e^-r^2dxdy

= √∫∫r.e^-r^2drdθ

= … ?

15. Apr 9, 2008

### Zachary

You are right! It is integrated from negative infinity to positive infinity. I get one method to do that, but have another problem..
e^(x^2)=e^(-i)*e^(ix^2)=e^(-i)*(cos(x^2)+isin(x^2))
I try to integrate sin(x^2) or cos(x^2) instead, however, i havent get it yet

16. Apr 9, 2008

### Schrodinger's Dog

I don't know $e^{i\pi}+1$ you're the teacher.

Please be gentle with me I haven't studied any of this in seven months, so I've forgotten most of it.

17. Apr 9, 2008

### D H

Staff Emeritus
The goal is to evaluate

$$I=\int_0^{\infty}e^{-x^2}dx$$

This is obviously a positive value (if the integral converges), so $I=\sqrt{I^2}$, or

\begin{aligned} I^2 &= \left(\int_0^{\infty}e^{-x^2}dx\right)\left(\int_0^{\infty}e^{-x^2}dx\right) \\[6pt] &= \int_0^{\infty}e^{-(x+y)^2}\,dxdy &\text{physicists have no problem w/ this step}\\[6pt] &= \int_0^{\pi/2}\int_0^{\infty}e^{-r^2}r\,dr\,d\theta & \text{conversion to polar}\\[6pt] &= \frac{\pi}4 \int_0^{\infty} e^{-u} du & \text{with substitution  u=r^2 } \\[6pt] &= \frac{\pi} 4 & \text{and thus}\\[6pt] I &= \frac{\sqrt{\pi}}2 \end{aligned}

18. Apr 9, 2008

### Schrodinger's Dog

Ah right I get it thanks DH. That's what I was looking for.

\begin{aligned}\\[6pt]&= \int_0^{\infty}e^{-(x+y)^2}\,dxdy &\text{physicists have no problem w/ this step}\\[6pt]\end{aligned}

Are the Secret Algebra Squad (SAS) onto them or something?

$$\int_0^{\infty}e^{-x^2}dx=0.88622692545275801364908374167057$$

Nice and irrational.

Last edited: Apr 9, 2008
19. Apr 9, 2008

### tiny-tim

Welcome to PF!

Hi Zachary! Welcome to PF!

No … e^(x^2) doesn't equal e^(-i)*e^(ix^2).

And ∫e^(ix^2)dx can be dealt with in the same way as ∫e^(-x^2)dx, except that you end up with √∫e^(iu)du, which doesn't converge.

20. Apr 9, 2008

### Zachary

sorry.. I meant x^2=-(-1)*x^2=-(i^2)*(x^2)=-i*(i*x^2) so that i can use e^ix=cosx+isinx.
But D H has provided the answer, that is quite nice.