Integrate e^-(x^2) (1 Viewer)

[rama_v]

I have noticed in many textbooks people have written questions such as Integrate e^-x2..And then the answer is supposed to be -2xe^-x2..But is that even correct? The reason I ask is that when I do this on maple, I get things that involve the "Erfi" function, and I have read in books that there exists no simple primitive of e^-x2..

 

[darkliight]

If the book was right, we should be able to differentiate -2xe^-x2 and get back e^-x2. We don't even get close.

In fact, there is no elementary function that differentiates to e^-x2 (http://en.wikipedia.org/wiki/Error_function).

More reading if you're into it: http://en.wikipedia.org/wiki/Fresnel_integral

 

[Yip]

That is actually the differential of e^(-x^2). It is indeed impossible to get a closed form primitive for e^-x^2 unlike normal functions u see iin the hsc course, which is rather inconvenient since this function forms part of important functions such as probability distribution curves lol. I think u have to do weird things like express it in an infinite series before u integrate it

 

[Raginsheep]

Are you sure it wasn't differentiate and not integrate?

 

[Templar]

Are you sure it wasn't differentiate and not integrate?

Yeah, check it, most textbooks would not make such trivial mistake.

 

[rama_v]

Yeah, check it, most textbooks would not make such trivial mistake.

I have seen in an exam question, integrate xe^x2 ..I can't remember where I saw it though.

 

[dom001]

I have seen in an exam question, integrate xe^x2 ..I can't remember where I saw it though.

That one is possible though:

S(xe^(x^2)dx
=(1/2)S(2xe^(x^2)dx
=(1/2)(e^(x^2)) +C

Whereas the one posted first is the one posted first appears to have calculated the derivative instead of integral.

 

[rama_v]

That one is possible though:

S(xe^(x^2)dx
=(1/2)S(2xe^(x^2)dx
=(1/2)(e^(x^2)) +C

Whereas the one posted first is the one posted first appears to have calculated the derivative instead of integral.

hmm I seee..But the integral of e^x2 is not elementary is it? (And I have certainly seen this question before, somewhere..) Anyway I think my question has been answered. Thanks guys.

 

[icycloud]

hmm I seee..But the integral of e^x^2 is not elementary is it?

There's no closed form representation of the integral with elementary functions. However, by expansion using power series,

e^-x2 = 1 - x²/1! + x⁴/2! - ...
Thus, ∫e^-x2 dx = x - x³/(3 * 1!) + x⁵/(5 * 2!) - x⁷/(7 * 3!) + ... + C

 

[Riviet]

I have noticed in many textbooks people have written questions such as Integrate e^-x2..And then the answer is supposed to be -2xe^-x2..But is that even correct?

If you replaced "integrate" with "differentiate", then obviously it would be right, but that's a major error in a textbook, if that's what it had. I don't know how they check for errors, e.g getting someone to read through it, but this one should have been obvious enough.

 

[SeDaTeD]

If you could find a simple closed form integral then you'd be quite rich, well, maybe not, but famous.

 

[johnec10]

yeah guys best op is take the logarythm then using the Goldstone boson principle we integrate in terms of VALPHI using the equation:
{d\over dt} Q_A = {d\over dt} \int_x e^{-x^2\over 2A^2} J^0(x) = -\int_x e^{-x^2\over 2A^2} \nabla \cdot J = \int_x \nabla(e^{-x^2\over A^2}) \cdot J
peace out broskiez:jaw: :apig:

 

[Forbidden.]

omg why does this function come back once again

yeah guys best op is take the logarythm then using the Goldstone boson principle we integrate in terms of VALPHI using the equation:
{d\over dt} Q_A = {d\over dt} \int_x e^{-x^2\over 2A^2} J^0(x) = -\int_x e^{-x^2\over 2A^2} \nabla \cdot J = \int_x \nabla(e^{-x^2\over A^2}) \cdot J
peace out broskiez:jaw: :apig:

This stupidass forum doesn't support LaTeX syntax or whatever it is.

Rock on.

 

[waxwing]

I notice nobody mentioned the incredibly beautiful solution to the problem of integrating this function between -infinity and plus infinity.
Since there is no latex here, I won't write it, but you can read a nice walkthrough here:
http://mathworld.wolfram.com/GaussianIntegral.html

It's called the Gaussian integral.
Now when you look at that technique (look at equations (1) to (8) ), you might be put off by the use of a double integral, and the change to polar coordinates. I guess if you've never seen either of those things it might be a bit too much to swallow; but if you at least have seen polar coordinates you should appreciate the genius of the idea.

Also of great interest is the fact that this is the basic functional form of the so-called "normal distribution" (or "Gaussian distribution") or bell curve which underlies all of modern statistics. If you do science at Uni you'll be swamped with these things.

 

[godzilla123]

I notice nobody mentioned the incredibly beautiful solution to the problem of integrating this function between -infinity and plus infinity.
Since there is no latex here, I won't write it, but you can read a nice walkthrough here:
Gaussian Integral -- from Wolfram MathWorld

It's called the Gaussian integral.
Now when you look at that technique (look at equations (1) to (8) ), you might be put off by the use of a double integral, and the change to polar coordinates. I guess if you've never seen either of those things it might be a bit too much to swallow; but if you at least have seen polar coordinates you should appreciate the genius of the idea.

Also of great interest is the fact that this is the basic functional form of the so-called "normal distribution" (or "Gaussian distribution") or bell curve which underlies all of modern statistics. If you do science at Uni you'll be swamped with these things.

I think there is an easier way to do this. We know that the probability density function of a standard normal distribution is:



We know that



Therefore,



Let


Therefore,


and



Then we have:


or

 

[waxwing]

Hi godzilla,
The problem is with your "We know that" step.
That integral

is, so to speak, "designed" to be one; its value was known to be the value described above and then it was divided by a constant to make it equal to one. This is called "normalizing" (not specifically because it's the "normal" distribution; physicists go around normalizing integrals all the time, for example).

 

[Dumbledore]

I have noticed in many textbooks people have written questions such as Integrate e^-x2..And then the answer is supposed to be -2xe^-x2..But is that even correct? The reason I ask is that when I do this on maple, I get things that involve the "Erfi" function, and I have read in books that there exists no simple primitive of e^-x2..

i'm pretty sure it said differentiate, -2xe^(-x^2) is the derivative of e^(-x^2)

 

[khorne]

What on "erf" are you talking about?

...I thought it was a good pun

 

[gurmies]

what on "erf" are you talking about?

...i thought it was a good pun

lol =]

 

[Uncle]

If you cannot evaluate integrals via integration there is a different method called numerical integration.
For those lost HSC kids who don't know what the fuck it is, Trapezoidal Rule and the Simpson's Rule is a form of numerical integration, the latter is quite accurate.
Of course the more terms of summation the better the accuracy.
Another reason why with advances in mathematics and computing we are developing more deadly machines like Metal Gear.


Also, there is a Taylor Polynomial approximation of order 20 and because of this high degree the approximation should be very good and at least CAN be integrated.

[maths]e^{-x^{2}} \approx 1-x^{2}+\frac{1}{2!}x^{4}-\frac{1}{3!}x^{6}+\frac{1}{4!}x^{8}-\frac{1}{5!}x^{10}+\frac{1}{6!}x^{12}-\frac{1}{7!}x^{14}+\frac{1}{8!}x^{16}-\frac{1}{9!}x^{18}+\frac{1}{10!}x^{20}[/maths]

Taylor polynomials were invented by the same guy who invented integration by parts, Brooke Taylor who actually stabbed Newton in the head and took the credit because he couldn't stand seeing Newton being in the spotlight of mathematics any longer.