# Integrating exponential

1. Feb 7, 2009

### d_b

Hi!!

I want to know how to intergrat e^(-(x^2)/2) . Someone told me to use the intergration of e^x= 1+ x + (x^2) /2!+ ....(X^n)/n! ..... is that sounds right?? and if it is i got pi as an answer for e^(-(x^2)/2) , is that the right answer???

2. Feb 7, 2009

### dprimedx

Re: intergrating exponential

We know that $$\int{e^x}dx = e^x$$ so try to get it into that form. Substitute u = (-(x^2)/2) and get $$\int{e^u}du = e^u * du/dx$$. Solving du/dx gives you -x. You should get an answer of -xe^((-x^2)/2) - I'm pretty sure, anyway.

3. Feb 7, 2009

### mathman

e-x2 cannot be integrated (indefinite) analytically. If you expand it into a power series, you can get the power series of the integral. There is no closed form solution.
The integral (-oo,x) (divided by sqrt(2pi)) is called erf(x).

Last edited: Feb 7, 2009
4. Feb 7, 2009

### d_b

Re: intergrating exponential

I don't think you can intergrate it like that because you have x^2 and so when trying to use another variable du it will gives dx equals du*fraction of x and you can't solve for x then

5. Feb 7, 2009

### d_b

I actually meant to intergrate from negative infinity to positive infinity.

6. Feb 8, 2009

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

Here's a proof: Let

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

Clearly $I$ is positive, so if we can show that $I^2 = \pi[/tex] we are done: $$I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2} \,dx \right) \left(\int_{-\infty}^{\infty} e^{-y^2} \,dy \right) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)} \,dx \,dy.$$ Switching to polar coordinates [itex](r, \theta)$, we have $x^2 + y^2 = r^2$ and $dx\,dy = r\,dr\,d\theta$. Thus,

$$I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} r e^{-r^2} \,dr \,d\theta.$$

Substituting $u = -r^2$ so that $r \,dr = -du/2$ will give the result.

7. Feb 8, 2009

### d_b

Thank you for clearing up the problem, I haven't learn how to do double intergral yet so is there any other way of doing it without using double intergral??

also $$\int_{-\infty}^{\infty} e^{-x^2/2} \,dx = \sqrt{\pi/2}.$$ ....is this correct??? what i did i just use $$e^{-x^2}$$ and substitute $$-x^2/2$$ with $$-x^2$$ ....

Last edited: Feb 8, 2009
8. Feb 8, 2009

No, you should have
$$\int_{-\infty}^{\infty} e^{-x^2/2} \,dx = \sqrt{2\pi}.$$

Unfortunately I don't know of any way to prove the Gaussian integral without using double integration.

9. Feb 8, 2009

### mathman

If you look at adriank's method (which is the standard for this integral), you will see that it is simply a product of two integrals, both before and after the change of variables. The only tricky part is the change from rectangular to polar coordinates.

10. Feb 8, 2009

### d_b

i found http://en.wikipedia.org/wiki/Gaussian_integral this online which helps....but I was just thinking if i was go have$$e^{x^2}$$ can i still use the double intergral?? or is there other way of doing it? (ie. using the theory property of exponential)

11. Feb 8, 2009

If you have the positive exponent ex2, then the integral will diverge.

12. Feb 8, 2009

### d_b

$$(\int_{-\infty}^{\infty}e^{-x^2}dx)^2=(\int_{-\infty}^{\infty}e^{-x^2}dx)\cdot (\int_{-\infty}^{\infty}e^{-y^2}dy)= (\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy)= \int_0^{2\pi}dw\int_0^\infty dr r e^{-r^2}=-\pi e^{-r^2}\vert^\infty_0=\pi$$

THis is what i got....Also would $$e^x^2$$ be different if it is not from negative infinity to positive infinity. Could it intergral exists if i was to intergrat each term of $$e^x^2$$ ????

Last edited: Feb 8, 2009
13. Feb 8, 2009

### d_b

it would be diverge if i was to take from negative infinity or zero to positive infinity. What if i was to take the intergral for a small section. let say from 1 to 5. Would it still be diverge??

14. Feb 9, 2009

15. Feb 9, 2009

### jacophile

The other way to look at this is to use the definition of the error function:

Then split your integral into 2 sections: $$-\infty \rightarrow0 \mbox{ and }0 \rightarrow\infty\\$$

$$\int_{-\infty}^{\infty} e^{-x^2} \,dx=\int_{0}^{\infty} e^{-x^2} \,dx-\int_{0}^{-\infty} e^{-x^2} \,dx\\$$

since

$$\int_{a}^{b} f(x) \,dx=-\int_{b}^{a} f(x) \,dx\\$$.

$$\lim_{x\to\infty}\, (erf(x)-erf(-x))\frac{\sqrt{\pi}}{2}$$

which is $$\sqrt{\pi}$$ because $$\lim_{x\to\infty}\, erf(x) = 1$$ and $$\lim_{x\to-\infty}\, erf(x) = -1$$

Last edited: Feb 9, 2009
16. Feb 9, 2009

### d_b

is this the only way of proving $$\int_{0}^{a} e^{x^2} \,dx$$

17. Feb 9, 2009

jacophile: But first you need to prove that $$\lim_{x\to\infty} \operatorname{erf}(x) = 1$$!

d_b: What about that integral are you trying to find?

18. Feb 9, 2009

### jacophile

My aim was to solve

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

using the erf identity, not to prove it. Like I said in my post, just another way of looking at it.

Last edited: Feb 10, 2009
19. Feb 10, 2009

### jacophile

Actually, can someone help me to understand why using the following substitution is wrong?
At least I am pretty sure it is...

$$\int_{0}^{x}e^{-z^2}\,dz$$

let $$y=z^{2}\;\rightarrow\; dy=2 z dz$$

$$\rightarrow\;z=\sqrt{y}\;\mbox{ }(\forall\,y\geq\;0)\mbox{ }\;\rightarrow\,dz= \frac{1}{2} y^{-\frac{1}{2}}\;dy$$

So that the integral becomes

$$\frac{1}{2} \;\int_{0}^{x^{2}}\;e^{-y}\;y^{-\frac{1}{2}}\,dy$$

If it was ok to do this one could then integrate by parts using

$$u=y^{-\frac{1}{2}}\;\rightarrow\;du=-\frac{1}{2} \;y^{-\frac{3}{2}}\;dy$$

$$dv=e^{-y}dy\;\rightarrow\;v=-e^{-y}$$

and then build a power series (with increasingly -ve powers of y), by successive integration by parts, of the form

$$e^{-x^{2}}\sum_{i=1}^{\infty}f(x,i)\;\;+\;\lim_{i\to\infty}I_{i}$$

Assuming you could show that the second term was zero...

I'm just unsure about the first substitution.

Last edited: Feb 10, 2009
20. Feb 10, 2009

### NoMoreExams

I didn't really check your work but how's that different than just rewriting

$$e^{x^2} = \sum_{n=0}^{\infty} \frac{x^{2n}}{n!}$$

And then integrating it out.