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September 26th, 2018, 09:44 PM   #11
SDK
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That article is misunderstanding the difference between differentiating under the integral (DUIT) and Feynman's trick. The example in question is an example of DUIT but not an example of Feynman's trick.

The second example in that article illustrates both. DUIT is just a technique which allows (under some reasonable assumptions) to compute partial derivatives of an integral function. Specifically, given a function of the form
\[f(x,y) = \int_a^b g(x,y) \ dy \]
it is sometimes possible to interchange limit operations and write
\[\frac{\partial f}{\partial x} = \int_a^b \frac{\partial g}{\partial x} \ dy \]

Feynman's trick (also called parametric integration), while it does typically involve DUIT, is when you take an integrand in one variable, introduce a new variable which defines an integral function in that variable, and then differentiate it.

The magic of this technique is knowing when this will work, and where to insert the variable. Feynman had a near supernatural ability to just see these things immediately. For example, in the second example of the link you posted, the author makes the following replacement
\[ \int_0^1 \frac{\ln(x+1)}{x^2+1} \ dx \implies F(a) = \int_0^1 \frac{\ln(ax+1)}{x^2+1} \ dx \]

The question is, why define $F$ that way? Why did the $a$ get placed where it did? How do you know that after DUIT, $F'$ will be easy to compute? This was what Feynman was famous for and this is the hard part. Computing the derivative after adding the parameter is trivial.
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September 27th, 2018, 02:00 AM   #12
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Originally Posted by ProofOfALifetime View Post
. . . training myself for an integration problem
Are you able to divulge that problem?

The linked article, having evaluated that integral (referred to by SDK) in its second example, states "this way is by far the simplest and most elegant, not to mention the quickest", which is incorrect. It can be evaluated very simply and elegantly as shown below.

Substituting $\displaystyle x \to \frac{1 - x}{x + 1},\ I = \!\int_0^1\! \frac{\ln(x + 1)}{x^2 + 1}dx = \!\int_0^1 \!\frac{\ln(2/(x + 1))}{x^2 + 1}dx = \!\int_0^1 \!\frac{\ln 2}{x^2 + 1}dx - I$, so $\displaystyle I = \frac{\pi\ln 2}{8}$.
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September 27th, 2018, 10:02 AM   #13
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Originally Posted by SDK View Post
Feynman's trick (also called parametric integration), while it does typically involve DUIT, is when you take an integrand in one variable, introduce a new variable which defines an integral function in that variable, and then differentiate it.

The magic of this technique is knowing when this will work, and where to insert the variable.
This is a very good point, and you are right. Knowing DUIT is one thing, knowing how to use it is another.

Last year, I solved this one (also from the AMM), which is similar to the second example:

$\int_0^1 \frac{x\ln(x+1)}{x^2+1} \ dx \implies F(a) = \int_0^1 \frac{x\ln(ax+1)}{x^2+1} \ dx$. I put the parameter in the same spot. It was still tricky (for me because I'm far from a professional), but I was able to do it with this technique.

However, the one that's featured now in the (AMM), although it is similar, might turn out to be hard for me since it is not at all the exact same.

Quote:
Originally Posted by skipjack View Post
Are you able to divulge that problem?

The linked article, having evaluated that integral (referred to by SDK) in its second example, states "this way is by far the simplest and most elegant, not to mention the quickest", which is incorrect. It can be evaluated very simply and elegantly as shown below.
Wow, nice! Thanks!

I hope I can get through this problem I'm working on.

Thank you both, you are amazing, and I have learned a lot from you!

Last edited by skipjack; September 28th, 2018 at 01:14 AM.
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September 27th, 2018, 06:13 PM   #14
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.........

This is the integral: $\int_{0}^{\pi} \ln(2-2\cos{x}) dx$
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It's an improper integral. When x=0, ln0=-$\displaystyle \infty$

You have to integrate from b to $\displaystyle \pi$ and let b $\displaystyle \rightarrow 0$
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September 28th, 2018, 06:01 AM   #15
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It's an improper integral. When x=0, ln0=-$\displaystyle \infty$

You have to integrate from b to $\displaystyle \pi$ and let b $\displaystyle \rightarrow 0$
Of course it might not converge, in which case integral doesn't exist.

The way to test it, if someone has a math program to do integrals, is calculate it for b=.1, .01, .001 to see where it is going.
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