Can anyone find a solution to the integral

ò₀^2p ln(cos²(x)+1) dx???
Thanks,

Brad

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:

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)=ò₀^2p 1/(cos²(x)+n) dx

Which you can solve using the 'e^{i x} definition' for cosine, then expanding with respect to e^{i x}. I then ended up with no terms being just a constant, and since e^{i x}×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

log(y+1)=y-1/2 x y² +1/3 x y³ -.......
where y is real and less than 1.....

love arun

or another idea....

find the indefinite integral....
òlog(cos² x+1)×1 dx using the product rule.....which gives...
log(cos² x+1)×x -something...

and then evaluated between limits 0 and 2p
love arun

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

lim h® 0

ln(h)+2×tanh-1 ((1+n)1/2)

, but this has managed to elude me thus far.
Brad

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

4p(Re(tanh^-1(2^1/2))-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

Thanks for your help,

Brad

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

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.

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

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

ò₀¹ 1/n+1/[n(n+1)^1/2]dn

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

2ò₀^21/2 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

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