Integrating Exponential Functions: A Guide to Solving e^(-(x^2)/2)

[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?

 

[dprimedx]

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.

 

[mathman]

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?

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).

 

[d_b]

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.

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

 

[d_b]

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).

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

 

[adriank]

The answer is

$$\int_{-\infty}^{\infty} e^{-x^2} \,dx = \sqrt{\pi}.$$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 [itex]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 $(r, \theta)$, 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.

 

[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 $$-x^2$$ ...

 

[adriank]

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.

 

[mathman]

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 $$-x^2$$ ...

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.

 

[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)

 

[adriank]

If you have the positive exponent e^x2, then the integral will diverge.

 

[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 $$e^x^2$$ ?

 

[d_b]

If you have the positive exponent e^x2, then the integral will diverge.

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??

 

[adriank]

No, it won't, but you won't be able to find it analytically. You'll have to resort to the (imaginary) error function.

 

[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

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

So your integral becomes

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

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

 

[d_b]

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

 

[adriank]

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?

 

[jacophile]

You already did that.
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.

 

[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.

 

[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.

 

[jacophile]

Hm, its amazing what a bit of sleep will do for you.

Ok, I see the problem now: the function is not well behaved at the lower limit of the integration so that is why the substitution is not viable.

 

[d_b]

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?

I posted a few of them. Which one are you talking about?