by JBrandonS
 P: 20 1. The problem statement, all variables and given/known data Find $\int_0^\infty e^{-\beta x^2 - \alpha/x^2}dx$ 2. Relevant 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. 3. 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.
 P: 80 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.
P: 20
 Quote by Jufro 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. :)

HW Helper
P: 2,149

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.
P: 20
 Quote by lurflurf 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.
Homework
HW Helper
Thanks ∞
P: 9,154
 Quote by Jufro 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?
Homework
HW Helper
Thanks ∞
P: 9,154
 Quote by JBrandonS 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.
P: 1,666
 Quote by haruspex 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:

$$\frac{dI}{d\alpha}=kI,\quad I(\alpha_0)=g$$

Then surely, $I(\alpha)=e^{k\alpha}+c$

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:

$$I=\frac{e^{-2\sqrt{\alpha}\sqrt{-\beta}}\pi}{2\sqrt{-\beta}}$$

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 :

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

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

$$u=\sqrt{-\beta}\sqrt{\alpha}x$$

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

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

$$\frac{dI}{d\alpha}=kI$$

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:

$$\frac{dI}{d\alpha}=h(\alpha) I$$

which is still first-order linear that we can easily solve. See what you get.
 HW Helper P: 2,149 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
 Sci Advisor Thanks P: 2,132 The idea was already good. Let's treat the problem as differential equation initial-value problem as a function of $\alpha$. We have from #1 the function $$I(\alpha)=\int_0^{\infty} \mathrm{d} x \exp(-\beta x^2-\alpha/x^2).$$ Now take the derivative $$I'(\alpha)=-\int_0^{\infty} \mathrm{d} x \frac{1}{x^2} \exp(-\beta x^2-\alpha/x^2).$$ In this integral substitute $$x=\sqrt{\frac{\alpha}{\beta}}1/u,$$ which leads you to the equation $$I'(\alpha)=-\sqrt{\frac{\beta}{\alpha}}I(\alpha).$$ This you can easily solve by separation. Together with the initial condition (Gaussian integral!) $$I(\alpha=0)=\frac{\sqrt{\pi}}{2 \sqrt{\beta}}$$ you find the unique result $$I(\alpha)=\frac{\sqrt{\pi} \exp(-2\sqrt{\alpha \beta})}{2 \sqrt{\beta}}.$$
P: 20
 Quote by vanhees71 The idea was already good. Let's treat the problem as differential equation initial-value problem as a function of $\alpha$. We have from #1 the function $$I(\alpha)=\int_0^{\infty} \mathrm{d} x \exp(-\beta x^2-\alpha/x^2).$$ Now take the derivative $$I'(\alpha)=-\int_0^{\infty} \mathrm{d} x \frac{1}{x^2} \exp(-\beta x^2-\alpha/x^2).$$ In this integral substitute $$x=\sqrt{\frac{\alpha}{\beta}}1/u,$$ which leads you to the equation $$I'(\alpha)=-\sqrt{\frac{\beta}{\alpha}}I(\alpha).$$ This you can easily solve by separation. Together with the initial condition (Gaussian integral!) $$I(\alpha=0)=\frac{\sqrt{\pi}}{2 \sqrt{\beta}}$$ you find the unique result $$I(\alpha)=\frac{\sqrt{\pi} \exp(-2\sqrt{\alpha \beta})}{2 \sqrt{\beta}}.$$
You rock, Never thought of that substitution.

Thanks for all the help guys.

 Related Discussions Calculus & Beyond Homework 5 Advanced Physics Homework 3 Introductory Physics Homework 3 Advanced Physics Homework 4