- #1
sai2020
- 26
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Homework Statement
Integrate e^(x^2)
Homework Equations
The Attempt at a Solution
I have no idea at all..
sai2020 said:But someone told me that it's important in Quantum Mechanics. (I'm still first year)
tiny-tim said: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 said:yeah i remember reading it somewhere. it's got something to do with the gaussian distribution if I'm not mistaken.
tiny-tim said: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.
Schrodinger's Dog said:[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?
tiny-tim said: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..tiny-tim said: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.
I don't know [itex]e^{i\pi}+1[/itex] you're the teacher.tiny-tim said: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 evaluateSchrodinger's Dog said: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?
Zachary said: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
tiny-tim said: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.
tiny-tim said: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.
Mindscrape said: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.D H said: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]
HallsofIvy said: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.
Zachary said: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
sai2020 said: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 said: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.
HallsofIvy said: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.
The purpose of integrating e^(x^2) is to find the area under the curve of the function e^(x^2). This can be useful in various applications in physics, engineering, and statistics.
To solve equations involving e^(x^2), you can use the technique of integration by parts. This involves rewriting the equation in a different form and using integration rules to solve for the variable.
Yes, most scientific calculators have a function for integrating e^(x^2). However, it is important to note that the accuracy of the result may vary depending on the complexity of the equation.
No, there is no specific formula for integrating e^(x^2). However, there are various techniques and integration rules that can be used to solve equations involving this function.
Yes, integrating e^(x^2) has many real-life applications. For example, it is used in probability and statistics to calculate the area under a normal distribution curve. It is also used in physics and engineering to find the energy and work done in systems with variable forces.