What is the Solution to the Advanced Integral?

In summary, lurflurf is suggesting that the integral be written as a product then integrating by parts. This may be helpful in solving the differential equation.
  • #1
JBrandonS
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Homework Statement



Find [itex]\int_0^\infty e^{-\beta x^2 - \alpha/x^2}dx[/itex]


Homework Equations



This is from a mathematical techniques of engineering and physics course by Dr. Feynman. Methods it uses are generalizing the function, differentiation under the integral sign and complexifying.

The Attempt at a Solution



I have tried everything I know over the course of a few days. Taylor series expansion does not work. differentiation wrt alpha or beta do not work. I am unable to find a general term I can introduce to allow me to change the formula into something that is workable. I am completely at a loss here.
 
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  • #2
To make your life easier you can expand the integral from -∞ to ∞ taking advantage of the fact that it is an even function.

Then I would just expect that you would want to think of the real line as being part of the complex plane and think of a half circle of radius r (r → ∞) centered around the origin.

This should make it easy as zero is a pole and you can use the Cauchy integral theorem.
 
  • #3
Jufro said:
To make your life easier you can expand the integral from -∞ to ∞ taking advantage of the fact that it is an even function.

Then I would just expect that you would want to think of the real line as being part of the complex plane and think of a half circle of radius r (r → ∞) centered around the origin.

This should make it easy as zero is a pole and you can use the Cauchy integral theorem.

Thanks, I'll look into doing it that way. I have never used the Cauchy integral theorem so its going to take some time. Oh well, time to learn something new. :)
 
  • #4
After differentiating you need to change variables to express the derivative in terms of the original integral. Also you will need to do the integral for one value of alpha.
 
  • #5
lurflurf said:
After differentiating you need to change variables to express the derivative in terms of the original integral. Also you will need to do the integral for one value of alpha.

Could you elaborate a bit more? I understand what you are say I am just not seeing how it can be applied to this problem.
 
  • #6
Jufro said:
To make your life easier you can expand the integral from -∞ to ∞ taking advantage of the fact that it is an even function.

Then I would just expect that you would want to think of the real line as being part of the complex plane and think of a half circle of radius r (r → ∞) centered around the origin.

This should make it easy as zero is a pole and you can use the Cauchy integral theorem.

Two problems with that. How does the integrand behave as r tends to infinity? Why is there a pole anywhere?
 
  • #7
JBrandonS said:
Could you elaborate a bit more? I understand what you are say I am just not seeing how it can be applied to this problem.
lurflurf is suggesting writing the integrand as a product then integrating by parts. See what you get.
Edit: But I'm not having much luck with that, so maybe I misinterpreted lurflurf's suggestion.
 
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  • #8
haruspex said:
lurflurf is suggesting writing the integrand as a product then integrating by parts. See what you get.
Edit: But I'm not having much luck with that, so maybe I misinterpreted lurflurf's suggestion.

I don't think so. Let me see if I can help without getting in Lurflur's way:

Let's work the problem backwards. Suppose we have the differential equation:

[tex]\frac{dI}{d\alpha}=kI,\quad I(\alpha_0)=g[/tex]

Then surely, [itex]I(\alpha)=e^{k\alpha}+c[/itex]

Now, let's just cheat a little bit in the interest of learning how to do this problem. Hope that's ok. When I blindly plug in that integral into Mathematica, I get for the solution:

[tex]I=\frac{e^{-2\sqrt{\alpha}\sqrt{-\beta}}\pi}{2\sqrt{-\beta}}[/tex]

Now compare that expression to the differential equation. That looks like we have to make some sort of substitution in the integral on the right of :

[tex]\frac{dI}{d\alpha}=\int_0^{\infty}\left(-\frac{1}{x^2}\right) e^{\beta x^2} e^{-\alpha/x^2} dx[/tex]

using [itex]u=f(\sqrt{-\beta},\sqrt{\alpha},x)[/itex]. Well there you go, that's what math is all about! You need to try things. How about:

[tex]u=\sqrt{-\beta}\sqrt{\alpha}x[/tex]

Huh? No? Alright, how about [itex]u=\frac{\sqrt{\alpha}}{\sqrt{-\beta}} x[/itex]? How about
[itex]u=\frac{\sqrt{\alpha}}{\sqrt{-\beta}} 1/x[/itex]

Now, if these don't work, just keep trying other substitutions to see if you can get it into the expression:

[tex]\frac{dI}{d\alpha}=kI[/tex]

Just worry about doing that much first. I don't have it yet either. I got it close though so I think we're on the right track and I just need to fiddle with it a little more, like you.

Edit: Ok, think I made a slight mistake. The DE we get I believe will be in the form:

[tex]\frac{dI}{d\alpha}=h(\alpha) I[/tex]

which is still first-order linear that we can easily solve. See what you get.
 
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  • #9
jackmell's suggestion is good, but I think it is hard to guess the form without knowing the answer in advance.
The backwards method would be to change variable u=sqrt(a/b)/x then recognize the integral and its derivative are related by multiplication by a constant so e^(k x) is the integral with k=sqrt(a/b)

so first find
$$\left. \int_0^\infty \! e^{-\beta x^2-\alpha /x^2} \,\mathrm{dx} \right] _{\alpha=0}= \int_0^\infty \! e^{-\beta x^2-0 /x^2} \,\mathrm{dx}$$
then show
$$\dfrac{d}{d \alpha}\int_0^\infty \! e^{-\beta x^2-\alpha /x^2} \,\mathrm{dx}=\int_0^\infty \! (-1/x^2) e^{-\beta x^2-\alpha /x^2} \,\mathrm{dx}$$
note it is not always possible to move the derivative inside the integral, but it is in this case as the convergence is rapid.
change variables to see that
$$\dfrac{d}{d \alpha}\int_0^\infty \! e^{-\beta x^2-\alpha /x^2} \,\mathrm{dx}=\int_0^\infty \! (-1/x^2) e^{-\beta x^2-\alpha /x^2} \,\mathrm{dx}=k \int_0^\infty \! e^{-\beta x^2-\alpha /x^2} \,\mathrm{dx} $$
for some k
finally deduce the integral
 
  • #10
The idea was already good. Let's treat the problem as differential equation initial-value problem as a function of [itex]\alpha[/itex]. We have from #1 the function
[tex]I(\alpha)=\int_0^{\infty} \mathrm{d} x \exp(-\beta x^2-\alpha/x^2).[/tex]
Now take the derivative
[tex]I'(\alpha)=-\int_0^{\infty} \mathrm{d} x \frac{1}{x^2} \exp(-\beta x^2-\alpha/x^2).[/tex]
In this integral substitute
[tex]x=\sqrt{\frac{\alpha}{\beta}}1/u,[/tex]
which leads you to the equation
[tex]I'(\alpha)=-\sqrt{\frac{\beta}{\alpha}}I(\alpha).[/tex]
This you can easily solve by separation. Together with the initial condition (Gaussian integral!)
[tex]I(\alpha=0)=\frac{\sqrt{\pi}}{2 \sqrt{\beta}}[/tex]
you find the unique result
[tex]I(\alpha)=\frac{\sqrt{\pi} \exp(-2\sqrt{\alpha \beta})}{2 \sqrt{\beta}}.[/tex]
 
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  • #11
vanhees71 said:
The idea was already good. Let's treat the problem as differential equation initial-value problem as a function of [itex]\alpha[/itex]. We have from #1 the function
[tex]I(\alpha)=\int_0^{\infty} \mathrm{d} x \exp(-\beta x^2-\alpha/x^2).[/tex]
Now take the derivative
[tex]I'(\alpha)=-\int_0^{\infty} \mathrm{d} x \frac{1}{x^2} \exp(-\beta x^2-\alpha/x^2).[/tex]
In this integral substitute
[tex]x=\sqrt{\frac{\alpha}{\beta}}1/u,[/tex]
which leads you to the equation
[tex]I'(\alpha)=-\sqrt{\frac{\beta}{\alpha}}I(\alpha).[/tex]
This you can easily solve by separation. Together with the initial condition (Gaussian integral!)
[tex]I(\alpha=0)=\frac{\sqrt{\pi}}{2 \sqrt{\beta}}[/tex]
you find the unique result
[tex]I(\alpha)=\frac{\sqrt{\pi} \exp(-2\sqrt{\alpha \beta})}{2 \sqrt{\beta}}.[/tex]

You rock, Never thought of that substitution.

Thanks for all the help guys.
 

1. What is an advanced integral question?

An advanced integral question is a mathematical problem that involves finding the integral, or the area under a curve, of a more complex function. It often requires advanced techniques and concepts from calculus, such as integration by parts or substitution.

2. How is an advanced integral question different from a regular integral question?

An advanced integral question typically involves more complex functions and may require the use of advanced integration techniques. It may also involve multiple variables or require the use of methods such as partial fractions or trigonometric substitutions.

3. What are some common strategies for solving advanced integral questions?

Some common strategies for solving advanced integral questions include using integration by parts, substitution, trigonometric identities, and partial fractions. It is also important to have a solid understanding of the fundamental properties and rules of integration.

4. Are there any tips for approaching advanced integral questions?

One tip for approaching advanced integral questions is to carefully analyze the given function and determine which integration technique would be most effective. It can also be helpful to break down the problem into smaller, more manageable parts and then integrate each part separately.

5. How can mastering advanced integral questions benefit a scientist?

Mastering advanced integral questions can benefit a scientist in several ways. It can improve their understanding of complex mathematical concepts, which can be useful in many scientific fields. It can also help them solve more complex problems and make more accurate predictions and calculations in their research.

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