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ò0 2p ln(cos2(x)+1) dx
By Brad Rodgers on Friday, November 09, 2001 - 10:08 pm:

Can anyone find a solution to the integral

ò0 2p ln(cos2(x)+1) dx ???

Thanks,

Brad

By Dan Goodman on Saturday, November 10, 2001 - 02:00 am:

Hmm, I haven't done too well on this one. Mathematica gives an answer with a complex part, which is clearly rubbish. I think the problem is that the integral it finds involves multivalued complex functions such as log, and it chooses the wrong branches to make the answer real. I've included the Mathematica output below:

Mathematica output

The first line calculates the value numerically to 30 decimal places. The second line gives us the value of 4×Pi×(ln(1+sr(2))-ln(2)), which is the value that the inverse symbolic calculator gives for the first numerical integration. Unfortunately, I don't know what the Maple function sr(x) is. The third line gives the antiderivative, which is quite complicated and depends on many multivalued functions. The PolyLog[n,z] function is given by S¥ k=1zk/kn. Unfortunately, I don't know how the branches of PolyLog are related, otherwise it might have been possible to search for the presumably unique branch of the antiderivative which gives a real value for the integral. The last line is a plot of the function. Hope that helps a bit.

By Brad Rodgers on Saturday, November 10, 2001 - 02:47 am:

Thanks. I tried working with feeble mathCAD, alas to no avail as well. I tried my own approach to the problem, and it shows some promise, but I think I've gone wrong somewhere. I set

y(n)=ò0 2pln(cos2(x)+n) dx

then differentiated to get

y'(n)=ò0 2p1/(cos2(x)+n) dx

Which you can solve using the 'e^ix definition' for cosine, then expanding with respect to e^ix. I then ended up with no terms being just a constant, and since e^ix×n for some number n is in all the other terms, I evaluated the integral to be zero by Cauchy. Thus y'(n)=0, or y(n)=C. But this is surely wrong, as y(0)=-i4p2 (mathCAD evaluates this using dilogs).

Brad

By Arun Iyer on Saturday, November 10, 2001 - 06:14 am:

how about expanding log(cos2x+1) into a series and then integrating since we know that,
log(y+1)=y-1/2×y2+1/3×y3-....... ¥
where y is real and less than 1.....

love arun

By Arun Iyer on Saturday, November 10, 2001 - 06:20 am:

or another idea....

find the indefinite integral....
ò log(cos2x+1)×1 dx
using the product rule.....which gives...
log(cos2x+1)×x - something...

and then evaluate between limits 0 and 2p

love arun

By Brad Rodgers on Sunday, November 11, 2001 - 01:24 am:

I've been able to determine that (through differentiation under the integral)

ò0 2p ln(n×cos2(x)+1) dx = 2p(ln(n)+2tanh-1((n+1)1/2))+C

for C some unknown constant. The problem however, lies in finding C. It could be found if someone can find limh->0ln(h)+2×tanh-1((1+n)1/2), but this has managed to elude me thus far.

Brad

By Brad Rodgers on Sunday, November 11, 2001 - 01:33 am:

Using a very slightly different technique, I have determined that my original integral is equal to

4p(Re(tanh-1(2½))-ln(2))

(that technique still being differentiation under the integral, but afterwards simply integrating the equation you get from 0 to 1). Can this expression be simplified though?

Brad

By Brad Rodgers on Sunday, November 11, 2001 - 03:09 am:

Using the identity tanh-1(x)=½ln(1+x/1-x), we finally obtain a result for this integral as 4p(ln(1+2½)-ln(2)). Sorry for answering most of my own questions here, the answers didn't occur to me when I asked them.

Thanks for your help,

Brad

By Dan Goodman on Sunday, November 11, 2001 - 03:49 am:

Indeed. And that explains what the Maple function sr is, it's just square root. I should have guessed it :-).

By Dan Goodman on Sunday, November 11, 2001 - 03:56 am:

Pretty damn good work by the way. I can't do these integrals at this time of night. Do you have a neat (short) way of getting to the result of your 2nd last post (it's pretty easy from there)? By the way, what gave you the idea of throwing in the n term and differentiating with respect to n? Is this a standard trick I'm not aware of? It's pretty cunning.

By Sean Hartnoll on Sunday, November 11, 2001 - 11:56 am:

Brad, you might be interested to know that the idea of introducing constants and differentiating with respect to them under the integral before setting them equal to one again is something that was invented by Feynman in order to evaluate some difficult integrals that come up in quantum field theory. It is quite a nice trick (Feynman claimed that he could do any integral that other people did using contour integration with using contour integration). So if you have thought of this by yourself, then you're in very good company!! Also, even if you had seen the idea before, it is usually unobvious how to apply it in some given integral, so that's still pretty good.

Sean

By Brad Rodgers on Sunday, November 11, 2001 - 08:55 pm:

Dan, the method I used above was rather inelegant, and in fact, I had to resort to mathCAD quite often to check and perform operations. I realized this morning though, that there is a more elegant way to go about this integral. First, we know that we are trying to evaluate

ò0 1 1/n + 1/[n(n+1)1/2] dn

making the substitution n=z2-1, we can write the integral as

2ò0 2^½ z/(z2-1)-1/(z2-1) dz

Then just combine the two fractions inside, cancel terms, and the final result pops out quite nicely.

Sean, I'm afraid I can't claim credit for coming up with the differentiation under the integral on my own, I read Feynman's book "Surely you're joking Mr. Feynman", and for a while couldn't figure out what he was talking about when he said differentiation under the integral. It eventually hit me, and I was able to fairly easily come up with the proof behind the idea fairly easily, but still, I was influenced quite heavily.

Brad

By Michael Doré on Monday, November 12, 2001 - 12:03 pm:

Nice work, Brad! I think you need to be slightly careful in your last response - the first integral you've written down doesn't look like it's going to converge.

I don't think Feynman was the first to use this technique. Euler used it in the eighteenth century to evaluate:

ò0 ¥sin x / x

The way he did this was to write:

I(r) = ò0 ¥e-ax sin rx / x

where r is a constant, then calculated I'(r), hence I(r) and finally set r = 1 and took the limit as a->0.