Integral of cos(tx)/(x^2 - 2x + 2) from 0 to infinity?

...for t real. Feel free to do the case when t=1.

It's easy to do the integral from 1 to infinity with residue calculus, but I have no idea how to do it from 0 to infinity. Mathematica can only integrate it in terms of special functions (SinIntegral and CosIntegral). There's a chance the question just has a typo, but in case I'm missing something, I thought I'd ask.

2013-07-17T22:27:09Z

@Shivaji: Certainly the integral exists: the denominator is not zero for real x, and it grows as 1/x^2, which is integrable. Mathematica certainly tried integration by parts and all the other basic techniques and didn't get anything nice.

2013-07-18T19:58:08Z

Correction: it's not easy to do the integral from 1 to infinity. It's easy to do "half" of it: substitute y = x-1 and replace cos(tx) with Re(e^(itx)). One can find the real part of the integral of e^(itx)/(x^2+2) dx from 0 to infinity easily (use a half-circle contour -R to R to i; assume t>0), but the imaginary part seems out of reach because it's odd.

Eugene2013-07-18T21:24:04Z

Favorite Answer

By residue calculus, ∫[1, ∞) cos(tx)/(x^2 - 2x + 2) dx = (π/2)e^(-|t|)cos(t)-sin(t)f(t), where f(t) = ∫[0, ∞) sin(tx)/(x^2 + 1) dx. Now ∫[0,1] cos(tx)/(x^2 - 2x + 1) dx equals

cos(t) ∫[-1,0] cos(tx)/(x^2 + 1) dx - sin(t) ∫[-1,0] sin(tx)/(x^2 + 1) dx

= cos(t) A(t) + sin(t) B(t),

where A(t) = ∫[0,1] cos(tx)/(x^2 + 1) dx and B(t) = ∫[0,1] sin(tx)/(x^2 + 1) dx. I don't think f(t), A(t) or B(t) can be simplified any further by elementary means.

11

?2013-07-18T05:23:37Z

I honestly don't feel like doing the integration, but it seems plausible. I mean, the function converges to zero for all values of t. So I don't see anyway that the integral doesn't converge. Try integration by parts or partial fractions.

02

?2013-07-18T18:30:03Z

Waste of time.

03

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