Differentiating Under Integral Sign

In summary, the differential equation for I(α) is -\frac{1}{\sqrt \alpha}I(\alpha) and it can be solved for I(α) by taking the derivative with respect to α and integrating over all values of α.
  • #1
iggyonphysics
6
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I(α) = 0∫e-(x2+α/x2) dx

Differentiating under the integral sign leads to:

I(α) = 0∫-e-(x2+α/x2)/x2 dx

Here I am supposed to let u = sqrt(a)/x, but the -x2 doesn't cancel out,

Wolfram-Alpha tells me the answer is: e(-2 sqrt(α) sqrt(π))/(2 sqrt(α)). I understand where the sqrt(π))/(2)sqrt(α) comes from, but not the 2 sqrt(α)) in the numerator.

Thanks!
 
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  • #2
Differentiating what with respect to what, and why?
Your integral is missing something ("dx"?).
 
  • #3
First, please, please, please, include a "dx" in the integral!
Yes, the derivative, with respect to "a", of [itex]\int_0^\infty e^{-x^2- a/x^2}dx[/itex] is [itex]-\int_0^\infty \frac{e^{-x^2- a/x^3}}{x^2}dx[/itex].

Is your question about actually doing that integration? Are you required to?
 
  • #4
Yes, with dx. Sorry! (Also, how do you format equations?)

For this problem, I am supposed to find I(1) using u = sqrt(x)/a and the answer is e-2sqrt(π)/2 .
 
  • #5
You have ##\displaystyle l(a)=\int_0^\infty e^{-x^2- a/x^2}dx##
Then ##\displaystyle l'(a)=-\int_0^\infty \frac{e^{-x^2- a/x^2}}{x^2}dx##
Apply the substitution ##u=\frac{\sqrt a}{x}## to this last integral, and compare what you get with the expression for ##l(a)##.
This will give you an easy differential equation for ##l##.

You will have to use the value of the Gaussian integral at some point:
##\displaystyle l(0)=\int_0^\infty e^{-x^2}dx=\frac{\sqrt \pi}{2}##.

iggyonphysics said:
Also, how do you format equations?
See the LaTeX link at the bottom left of the edit box.
 
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  • #6
Samy_A said:
You have ##\displaystyle l(a)=\int_0^\infty e^{-x^2- a/x^2}dx##
Then ##\displaystyle l'(a)=-\int_0^\infty \frac{e^{-x^2- a/x^2}}{x^2}dx##
Apply the substitution ##u=\frac{\sqrt a}{x}## to this last integral, and compare what you get with the expression for ##l(a)##.
This will give you an easy differential equation for ##l##.

You will have to use the value of the Gaussian integral at some point:
##\displaystyle l(0)=\int_0^\infty e^{-x^2}dx=\frac{\sqrt \pi}{2}##.

See the LaTeX link at the bottom left of the edit box.
Right, so I have [itex] I'(α) = 1/\sqrt{α}\int_0^∞ e^{-x^2 - u^2} du = 1/\sqrt{α}\int_0^∞ e^{-a/u^2 - u^2} du [/itex], which looks just like the [itex]l(a)=\int_0^\infty e^{-x^2- a/x^2}dx[/itex]

I don't understand what to do next.
 
  • #7
iggyonphysics said:
Right, so I have [itex] I'(α) = 1/\sqrt{α}\int_0^∞ e^{-x^2 - u^2} du = 1/\sqrt{α}\int_0^∞ e^{-a/u^2 - u^2} du [/itex], which looks just like the [itex]l(a)=\int_0^\infty e^{-x^2- a/x^2}dx[/itex]

I don't understand what to do next.
I also have a minus sign, giving:

##\displaystyle I'(\alpha)=-\frac{1}{\sqrt \alpha}\int_0^\infty e^{-x^2- \alpha /x^2}dx=-\frac{1}{\sqrt \alpha}I(\alpha)##

Now solve the differential equation ##I'(\alpha)=-\frac{1}{\sqrt \alpha}I(\alpha)##.
 
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  • #8
Got it, thanks so much!
 

FAQ: Differentiating Under Integral Sign

1. What is the concept of Differentiating Under Integral Sign?

The concept of Differentiating Under Integral Sign, also known as Leibniz's rule, is a method for finding the derivative of a function that is defined by an integral. It allows us to differentiate a function with respect to one of its parameters while keeping the other parameters constant.

2. When is it appropriate to use Differentiating Under Integral Sign?

Differentiating Under Integral Sign is appropriate to use when the function being integrated depends on a variable that is also in the limits of integration. This method can also be used when the function being integrated is difficult or impossible to integrate directly.

3. What is the formula for Differentiating Under Integral Sign?

The formula for Differentiating Under Integral Sign is:
d/dx [∫ab f(x,t) dt] = ∫ab ∂f(x,t)/∂x dt + f(x,b)∂b/∂x - f(x,a)∂a/∂x

4. Can Differentiating Under Integral Sign be used for indefinite integrals?

No, Differentiating Under Integral Sign can only be used for definite integrals. This is because the limits of integration are crucial in the differentiation process and cannot be ignored.

5. What are the benefits of using Differentiating Under Integral Sign?

Using Differentiating Under Integral Sign can greatly simplify the process of finding the derivative of a function, especially when the function is difficult to integrate directly. It also allows us to differentiate a function with respect to a parameter without having to solve the integral first.

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