ò0¥ 1/(1+xn)dx=p/(n sin p/n)
By Andrew Hodges on Monday, June 17, 2002 - 02:06 pm:

How do you integrate 1/(1+xn ) in general, where n is a positive integer?

By Julian Pulman on Monday, June 17, 2002 - 02:59 pm:

How about this line of reasoning:

if we can expand (1 + xn ) into the product of linear factors (x - r1 )(x - r2 )...(x - rn ), where ri are the roots of (1 + xn ). Then we can merely expand this fraction into a sum of partial fractions, hence:


ò1/(1+xn) dx=òk1/(x-r1)+k2/(x-r2)+...+kn/(x-rn) dx

Which has the obvious solution of a sum of logarithms.

Now, the roots will obviously be of form ra=eip(2a-1)/n, the coefficients ka will be more complicated though...see if you can work it out.
Julian

By David Loeffler on Monday, June 17, 2002 - 03:55 pm:
There is a general result:

ò0¥ 1/(1+xn) dx=p/(n sin p/n)

I don't know of any proofs of this by truly elementary methods, not offhand. It's very easy using complex analysis, or the properties of various special functions (b and G functions).

On the other hand, I think one can continue Julian's line of working. For each of the roots of -1, ri, consider

lim
x® r 
 i(x-ri)/(1+xn)

. This will be precisely ki; and it can be worked out by various methods - indeed, it is just 1 divided by the derivative of f(x)=1+xn at ri, so 1/(n rin-1), or -ri/n. This should allow you to work out the indefinite integral, in principle at least.

These quantities ki are called the residues of the function at the singular points ri and are crucial to the complex-analysis approach.

David

By David Loeffler on Wednesday, June 19, 2002 - 08:14 pm:

There's a fair bit about complex analysis in the archive here .

David

By David Loeffler on Wednesday, June 19, 2002 - 08:21 pm:

Actually on rereading that thread in full, it doesn't define a residue; and there is a link to an external site (the second link supplied by Arun, which Brad refers to in the last post on the thread) which does give a definition of a residue, but it's hopelessly wrong, and the site is incredibly badly written.

I will have a trawl of the web to see if I can find any better sites on basic complex analysis. I should perhaps warn you that it'll be hard going - officially this is a second year course in the Cambridge maths degree.

David

By David Loeffler on Wednesday, June 19, 2002 - 08:33 pm:

There's a set of lecture notes from Heriot-Watt University here - it's correct, but it's rather sketchy and a lot of the important proofs are missing.

This one , from the University of Chicago, seems a little better. It's more clearly explained. There's a lot there, but that's par for the course - learning enough complex analysis to do this problem in general took us about eight hours of lecturing. I never said this would be easy - look upon it as a little light reading for your summer holidays!

David

By Andrew Hodges on Friday, June 21, 2002 - 07:17 pm:

Thanks very much for your time! Ill enjoy reading up on this topic after my step exams are out of the way!


Thanks

Andrew :0)