# Tricky integral

1. Oct 11, 2005

### relatively_me

I'm having trouble finding an integral table w/ this integral, or understanding a method to solve it...any hints would really be appreciated (the book directs us to use integral tables)

Int[e^(x^2) dx] limits->{-inf, +inf}

2. Oct 11, 2005

### fahd

take x^2=t and proceed by parts...its fairly simple

3. Oct 11, 2005

### relatively_me

thanks, i'll try that now

4. Oct 11, 2005

### relatively_me

2x*e^(x^2) + x^2*e^(x^2)?

5. Oct 12, 2005

### Pietjuh

You don't even have to calculate it to see it diverges to infinity, since exp(x^2) is a strictly increasing function on (0,inf), and is also positive for negative x.

6. Oct 12, 2005

### amcavoy

Are you sure the problem wasn't with a negative x2?

$$\int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt{\pi}$$

Alex

7. Oct 12, 2005

### himanshu121

8. Oct 12, 2005

### HallsofIvy

I have no idea what you mean by that! What "parts" would you use? What would be u and what dv?

As others have pointed out, the infinite integral, as stated, does not exist. If that were $$e^{-x^2}$$ then it would be possible but certainly not "fairly simple"- there is no elementary anti-derivative.

9. Oct 27, 2005

### Taviii

e^x^2=e^x*e^x
so
e^x=t
e^xdx=dt......replace in your exercesise and it should be easy from now on!

10. Oct 27, 2005

### HallsofIvy

Yes, if I had to integrate
$$\int (e^x)^2dx$$
I guess I could do that- although I think I would be more likely to simply write
$$(e^x)^2= e^{2x}$$
$$\int e^{x^2}dx$$
and it is well known that that has no elementary anti-derivative.

11. Oct 27, 2005

### 1800bigk

if you are in analysis it can be solved using riemann-stieltjes or lebesgue theory, I saw this problem before and there was a theorem we used, it has slipped my memory.

it was something like if f(x) is differentiable on [a,b] and f prime is continuous on [a,b] then you can convert the integrand to somethin easier to integrate (i forget the rest) sorry

Last edited: Oct 27, 2005
12. Oct 27, 2005

### what

Yea you can put it into http://integrals.wolfram.com and it gives you an answer that has in it the "imaginary error function", i take this to mean that the described integral won't be solvable by any normal means, so if it is just about whether or not it diverges then this is pretty simple because it is obvious that the integral diverges.

13. Dec 1, 2005

### darkar

I got this....sqrt(pi)Erfi(x)/2

What is that Erfi?

14. Dec 3, 2005

### benorin

Erfi is the imaginary error function above quoted by what (cool psuedonym).

15. Dec 3, 2005

### BerkMath

My guess is that Tavi will think things through before typing incorrect answers, next time.

16. Dec 3, 2005

### HallsofIvy

All my errors are imaginary!

17. Dec 3, 2005

### darkar

Umm..........