Integration of Parts and the Wallis formula for pi

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cruche

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Ed Sandifer demonstrates how Euler derived Wallis formula for pi by a repeated integration by parts on an integral that yields arsin. This is a definite integral from 0 to 1:
∫ dx / (1-x2 ) 1/2 = 2/1 ∫x2 dx / (1-x2 ) 1/2

The uv portion of the integration of parts cancels out due to the end points of the definite integral.

He applies the general form that he uses to create the integration by parts.

∫ xm-1dx / (1-x2 ) 1/2 = (m+1)/m ∫xm+1 dx / (1-x2 ) 1/2

I am trying to calculate his general formula, but I do not seem to have the right dv and u.
dv = xm-1dx


u = 1/ (1-x2 ) 1/2


From that I calculated:

du = x dx / (1-x2 ) 3/2

v = 1/ m xm
 
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Good link but

Glad that you found the link. But that is where I found the problem in the first place. Ed Sandifer's explanation didn't make sense to me. I was having trouble figuring out which u and v he picked. I emailed him twice to no avail. If anyone can help decipher the Sandifer's application of Euler's method, that would be great. Thanks.
 
Sorry about that. I did not notice. I do not know what he chose either.

But, there are other ways of going about showing Wallis.
 
More on Wallis.

I've seen Newton's approach. And I've read a host of articles on other ways of constructing Wallis's formula. Euler's approach had special appeal to me, because of its apparent simplicity. It also has an interesting tie-in to some of his other work on series.

Sandifer sheds more light on the problem in another article.

http://www.southernct.edu/~sandifer/Ed/History/Preprints/Talks/NYU Basel Problem Transparencies.PDF

There it is clear that he uses the (1-x^2) ^1/2 as the dv. But I still can't fit everything together. I feel that I understand things a little better. But I want to work out all the details. Am I missing something very basic?

I wish that Ed Sandifer would answer an email.
 
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