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1. Homework Statement
Integrate e^(x^2)
2. Homework Equations
3. The Attempt at a Solution
I have no idea at all..
Integrate e^(x^2)
2. Homework Equations
3. The Attempt at a Solution
I have no idea at all..
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.But someone told me that it's important in Quantum Mechanics. (I'm still first year)
yeah i remember reading it somewhere. it's got something to do with the gaussian distribution if i'm not mistaken.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.
Yes, you're right … it is the gaussian distribution (aka the normal distribution)!yeah i remember reading it somewhere. it's got something to do with the gaussian distribution if i'm not mistaken.
[itex]\int_0^{\infty} e^{x^2}\;dx [/itex] where [itex]e^{x^2}={\frac{2}{x}}e^{x^2}[/itex]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.
No … it's more like ∫e^x^2dx = √[∫e^x^2dx∫e^y^2dy], and then changing to dr and dθ.[itex]\int_0^{\infty} e^{x^2}\;dx [/itex] where [itex]e^{x^2}={\frac{2}{x}}e^{x^2}[/itex]
Is it something like this?
Hehe in that case I don't know where that came from.No … it's more like ∫e^x^2dx = √[∫e^x^2dx∫e^y^2dy], and then changing to dr and dθ.
You are right! It is integrated from negative infinity to positive infinity. I get one method to do that, but have another problem..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.
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
= … ?
The goal is to evaluateHumour me could you show me how you would go about doing that in more detail? I'm not sure quite if I get that?
Hi Zachary! Welcome to PF!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
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.
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.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.
that's true. it is useful in the data analysis in the experiment..I'd like to point out that Gaussian's are useful for more than just quantum mechanics. They are good for error analysis and optics just to name a couple.
Mathematicians have no problem with that step either- it follows from Fubini's theorem. It's taught in most Multi-variable Calculus classes.The goal is to evaluate
[tex]I=\int_0^{\infty}e^{-x^2}dx[/tex]
This is obviously a positive value (if the integral converges), so [itex]I=\sqrt{I^2}[/itex], or
[tex]
\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}
[/tex]
I didn't expect it to be easy. I just wanted to know :)Sai2020, the fact that something is "important in Quantum Mechanics" (or, more generally probability and statistics) doesn't mean it is going to be easy to calculate. For x other than infinity, you look up values in a table generated by numerical integration.
I was talking about :)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
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.