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Integrate e^(x^2)

  • Thread starter sai2020
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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..
 

Answers and Replies

HallsofIvy
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There's a reason for that. That function, like most integrable functions, does not have an anti-derivative that can be written in terms of elementary functions.
 
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But someone told me that it's important in Quantum Mechanics. (I'm still first year)
 
But someone told me that it's important in Quantum Mechanics. (I'm still first year)
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.
 
tiny-tim
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Hi sai! :smile:

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. :smile:
 
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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.

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. :smile:
yeah i remember reading it somewhere. it's got something to do with the gaussian distribution if i'm not mistaken.
 
tiny-tim
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yeah i remember reading it somewhere. it's got something to do with the gaussian distribution if i'm not mistaken.
Yes, you're right … it is the gaussian distribution (aka the normal distribution)! :smile:
 
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Thanks :)
 
Hi sai! :smile:

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. :smile:
[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? I can't remember but I've seen that somewhere before? Reaching a bit now though, I wouldn't take my advice
 
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tiny-tim
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[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?
No … it's more like ∫e^x^2dx = √[∫e^x^2dx∫e^y^2dy], and then changing to dr and dθ. :smile:
 
HallsofIvy
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And, once again, it must be
[tex]e^{-x^2}[/tex]
in order to converge.
 
tiny-tim
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:redface: oops! :redface:

Thanks, HallsofIvy! :smile:
 
No … it's more like ∫e^x^2dx = √[∫e^x^2dx∫e^y^2dy], and then changing to dr and dθ. :smile:
Hehe in that case I don't know where that came from. :smile:

Humour me could you show me how you would go about doing that in more detail? I'm not sure quite if I get that?

Are you talking about spherical polars? If so how does that work? Or should I say why does that work?
 
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tiny-tim
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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

= … ? :smile:
 
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Hi sai! :smile:

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. :smile:
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:confused:
 
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

= … ? :smile:

I don't know [itex]e^{i\pi}+1[/itex] you're the teacher. :smile:

Please be gentle with me I haven't studied any of this in seven months, so I've forgotten most of it. :smile:
 
D H
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Humour me could you show me how you would go about doing that in more detail? I'm not sure quite if I get that?
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]
 
Ah right I get it thanks DH. That's what I was looking for. :smile:

[tex]\begin{aligned}\\[6pt]&= \int_0^{\infty}e^{-(x+y)^2}\,dxdy &\text{physicists have no problem w/ this step}\\[6pt]\end{aligned}[/tex]

Are the Secret Algebra Squad (SAS) onto them or something? :smile:

[tex]
\int_0^{\infty}e^{-x^2}dx=0.88622692545275801364908374167057
[/tex]

Nice and irrational. :smile:
 
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tiny-tim
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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:confused:
Hi Zachary! Welcome to PF! :smile:

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. :smile:
 
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Hi sai! :smile:

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. :smile:
Hi Zachary! Welcome to PF! :smile:

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. :smile:
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.
But D H has provided the answer, that is quite nice.
 
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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.
 
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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.
that's true. it is useful in the data analysis in the experiment..:smile:
 
HallsofIvy
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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]
Mathematicians have no problem with that step either- it follows from Fubini's theorem. It's taught in most Multi-variable Calculus classes.

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.
 
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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 didn't expect it to be easy. I just wanted to know :)

Btw, this is the friend
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:confused:
I was talking about :)
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.
 
HallsofIvy
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Strictly speaking, the "Gaussian function" on which the Normal distribution is based is
[tex]f(x)= e^{\frac{-x^2}{2}}[/tex]
rather than
[tex]e^{-x^2}[/tex]
The method of integration from [itex]-\infty[/itex] to [itex]\infty[/itex] is the same but the result is, of course, different.
 

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