Expert: Paul Klarreich - 6/14/2009

Question
Hi Paul

I'm stuck on this integral:

(from -infinity to infinity) | e^(isx).ln[(e^2x + e^x + 1)/(e^2x - e^x + 1)] dx

(s is just a constant)

My first thought was to use e^(isx) = cos(sx) + isin(sx), and then spot that the sine integral was an odd function, which goes to zero. For the cosine integral (even fn), I did 2 lots of it, with the limits from -inifinity to 0. After much work, I got it down to:

(from -infinity to 0) | sin(sx).[(e^3x - e^x) / (e^4x + e^2x + 1)] dx

or equivalently

| sin(sx).[2sinhx / (2coshx + 1)] dx

(same limits)

Neither of which I'm having much luck integrating. Not sure if I'm overcomplicating things either. Any help much appreciated. Thank you.

Answer
Questioner:   Siva
Country:  United Kingdom
Category:  Calculus
Private:  No
 
Subject:  Integration_Calculus
Question:  Hi Paul

I'm stuck on this integral:

(from -infinity to infinity) | e^(isx).ln[(e^2x + e^x + 1)/(e^2x - e^x + 1)] dx

(s is just a constant)

My first thought was to use e^(isx) = cos(sx) + isin(sx), and then spot that the sine integral was an odd function, which goes to zero. For the cosine integral (even fn), I did 2 lots of it, with the limits from -inifinity to 0. After much work, I got it down to:

(from -infinity to 0) | sin(sx).[(e^3x - e^x) / (e^4x + e^2x + 1)] dx

or equivalently

| sin(sx).[2sinhx / (2coshx + 1)] dx

(same limits)

Neither of which I'm having much luck integrating. Not sure if I'm overcomplicating things either. Any help much appreciated. Thank you.
...................................................................
Hi, Siva,

I don't have a solution for you,but I think you are going in the wrong direction.  For one thing, the fact that  sin(sx) is odd does not make the integrand odd.

But here is a suggestion:

{+inf
|     e^(isx).ln[(e^2x + e^x + 1)/(e^2x - e^x + 1)] dx
}-inf

does kind of look like a Fourier transform of something.  A Fourier transform is:

{+inf
|     e^(2pi i s x) f(x) dx
}-inf

With a little massaging, you might be able to apply some standard transform formula to your integral.