"Integrate e^(x^2): Solving the Equations

[sai2020]

Homework Statement

Integrate e^(x^2)

Homework Equations

The Attempt at a Solution

I have no idea at all..

 

[HallsofIvy]

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.

 

[sai2020]

But someone told me that it's important in Quantum Mechanics. (I'm still first year)

 

[Schrodinger's Dog]

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]

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.

 

[sai2020]

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.

yeah i remember reading it somewhere. it's got something to do with the gaussian distribution if I'm not mistaken.

 

[tiny-tim]

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)!

 

[sai2020]

Thanks :)

 

[Schrodinger's Dog]

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.

$\int_0^{\infty} e^{x^2}\;dx$ where $e^{x^2}={\frac{2}{x}}e^{x^2}$

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

 

[tiny-tim]

$\int_0^{\infty} e^{x^2}\;dx$ where $e^{x^2}={\frac{2}{x}}e^{x^2}$

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.

 

[HallsofIvy]

And, once again, it must be
$$e^{-x^2}$$
in order to converge.

 

[tiny-tim]

oops! ​

Thanks, HallsofIvy!

 

[Schrodinger's Dog]

No … it's more like ∫e^x^2dx = √[∫e^x^2dx∫e^y^2dy], and then changing to dr and dθ.

Hehe in that case I don't know where that came from.

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?

 

[tiny-tim]

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

= … ?

 

[Zachary]

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.

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 haven't get it yet

 

[Schrodinger's Dog]

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

= … ?

I don't know $e^{i\pi}+1$ you're the teacher.

Please be gentle with me I haven't studied any of this in seven months, so I've forgotten most of it.

 

[D H]

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

$$I=\int_0^{\infty}e^{-x^2}dx$$

This is obviously a positive value (if the integral converges), so $I=\sqrt{I^2}$, or

$$\begin{aligned} I^2 &= \left(\int_0^{\infty}e^{-x^2}dx\right)\left(\int_0^{\infty}e^{-x^2}dx\right) \\[6pt] &= \int_0^{\infty}e^{-(x+y)^2}\,dxdy &\text{physicists have no problem w/ this step}\\[6pt] &= \int_0^{\pi/2}\int_0^{\infty}e^{-r^2}r\,dr\,d\theta & \text{conversion to polar}\\[6pt] &= \frac{\pi}4 \int_0^{\infty} e^{-u} du & \text{with substitution u=r^2 } \\[6pt] &= \frac{\pi} 4 & \text{and thus}\\[6pt] I &= \frac{\sqrt{\pi}}2 \end{aligned}$$

 

[Schrodinger's Dog]

Ah right I get it thanks DH. That's what I was looking for.

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

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

$$\int_0^{\infty}e^{-x^2}dx=0.88622692545275801364908374167057$$

Nice and irrational.

 

[tiny-tim]

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 haven't get it yet

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.

 

[Zachary]

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.

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.

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.

 

[Mindscrape]

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.

 

[Zachary]

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..

 

[HallsofIvy]

The goal is to evaluate

$$I=\int_0^{\infty}e^{-x^2}dx$$

This is obviously a positive value (if the integral converges), so $I=\sqrt{I^2}$, or

$$\begin{aligned} I^2 &= \left(\int_0^{\infty}e^{-x^2}dx\right)\left(\int_0^{\infty}e^{-x^2}dx\right) \\[6pt] &= \int_0^{\infty}e^{-(x+y)^2}\,dxdy &\text{physicists have no problem w/ this step}\\[6pt] &= \int_0^{\pi/2}\int_0^{\infty}e^{-r^2}r\,dr\,d\theta & \text{conversion to polar}\\[6pt] &= \frac{\pi}4 \int_0^{\infty} e^{-u} du & \text{with substitution u=r^2 } \\[6pt] &= \frac{\pi} 4 & \text{and thus}\\[6pt] I &= \frac{\sqrt{\pi}}2 \end{aligned}$$

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.

 

[sai2020]

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 haven't get it yet

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]

Strictly speaking, the "Gaussian function" on which the Normal distribution is based is
$$f(x)= e^{\frac{-x^2}{2}}$$
rather than
$$e^{-x^2}$$
The method of integration from $-\infty$ to $\infty$ is the same but the result is, of course, different.

 

[Schrodinger's Dog]

Strictly speaking, the "Gaussian function" on which the Normal distribution is based is
$$f(x)= e^{\frac{-x^2}{2}}$$
rather than
$$e^{-x^2}$$
The method of integration from $-\infty$ to $\infty$ is the same but the result is, of course, different.

Ah that's what I was thinking about earlier, I knew I'd seen it before earlier.

 

[D H]

Strictly speaking, the "Gaussian function" on which the Normal distribution is based is
$$f(x)= e^{\frac{-x^2}{2}}$$
rather than
$$e^{-x^2}$$
The method of integration from $-\infty$ to $\infty$ is the same but the result is, of course, different.

The error function, on the other hand, is based on $f(x)=e^{-x^2}$. In particular,

$$\text{erf}(x) = \frac 2{\sqrt{\pi}}\int_0^x e^{-t^2}\,dt$$