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Use the representation as Fourier integral to prove that $$\int_0^\infty \displaystyle\frac{cos(xw)}{1+x^2}dw=\frac{\pi}{2}e^{-x}$$

I am really confuse with this kind of problem, I do not know well how to use the Fourier Transform.

Somebody can give some advice to solve it?

EQJ
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1 Answers1

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the integral on left side does not coverge, i think you mistyped something there is something similar $\int_0^{\infty}\frac{cos(ωx)}{(1 + x²)} dx= \frac{1}{2}\int_{-\infty}^{\infty}\frac{cos(ωx)}{(1 + x²)} dx = \frac{1}{2}Re(\int_{-\infty}^{\infty}\frac{e^{i\omega x}}{(1 + x²)} dx ) $

use residue theory from Complex Analysis.

Consider closed curve C (counterclockwise ) [-R, R] U C', where C' is the top halfcircle of R > 1 from R to -R.

Then, $\int_C \frac{e^{i\omega z}}{1 + z^2} dz = \int_{C'} \frac{e^{i\omega z}}{1 + z^2} dz + \int_{-R}^{R} \frac{e^{i\omega z}}{1 + z^2} dz $

The singularity inside C iz z = i. $\int_C \frac{e^{i\omega z}}{1 + z^2} dz = 2\pi i\lim_{z\rightarrow i}\frac{(z-i)e^{i\omega z}}{1 + z^2} = \pi e^{-\omega}$ by residue theorem

$\int_{C'} \frac{e^{i\omega z}}{1 + z^2} dz = \int_{0}^\pi \frac{e^{i\omega Re^{it}}ie^{it}}{1 + R^2e^{2it}} dt $

lets show that this converge to 0 when $R \rightarrow \infty$

$|\int_{0}^\pi \frac{e^{i\omega Re^{it}}ie^{it}}{1 + R^2e^{2it}} dt |\le \int_{0}^\pi |\frac{e^{i\omega Re^{it}}iRe^{it}}{1 + R^2e^{2it}}| dt \le$

$\int_{0}^\pi \frac{|e^{i\omega R(cos t +i sin t)}|R}{|1 + R^2e^{2it}|} dt =\int_{0}^\pi \frac{|e^{-\omega R sin t)}|R}{|1 + R^2e^{2it}|} dt \le \int_{0}^\pi \frac{|e^{-\omega R sin t)}|R}{R^2-1} dt \le $

$\int_{C'} \frac{e^{i\omega z}}{1 + z^2} dz = \int_{0}^\pi \frac{e^{i\omega Re^{it}}ie^{it}}{1 + R^2e^{2it}} dt = |\int_{0}^\pi \frac{e^{i\omega Re^{it}}ie^{it}}{1 + R^2e^{2it}} dt |\le \int_{0}^\pi |\frac{e^{i\omega Re^{it}}iRe^{it}}{1 + R^2e^{2it}}| dt \int_{0}^\pi \frac{|e^{i\omega R(cos t +i sin t)}|R}{|1 + R^2e^{2it}|} dt =\int_{0}^\pi \frac{|e^{-\omega R sin t)}|R}{|1 + R^2e^{2it}|} dt \le \int_{0}^\pi \frac{|e^{-\omega R sin t)}|R}{R^2-1} dt \le \int_{0}^\pi \frac{R}{R^2-1} dt = \frac{\pi R}{R^2-1} \rightarrow0 $

when $ r\rightarrow \infty$ keep in mind sin is positive on $(0, 2\pi)$ and $R>1$ since $ R\rightarrow \infty$

so when $ R\rightarrow \infty$

$\pi e^{-\omega}=0 + \int_{-R}^{R} \frac{e^{i\omega z}}{1 + z^2} dz $

$\int_0^{\infty}\frac{cos(ωx)}{(1 + x²)} dx= \frac{1}{2}Re(\pi e^{-\omega} )=\frac{\pi e^{-\omega}}{2} $

jack
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  • Thanks. But I am searching for an answer using the Laplace transform. – EQJ Jun 12 '15 at 17:25
  • Fourier* I mean... – EQJ Jun 12 '15 at 19:03
  • hmm.. maybe do fourier transform of right side, you should get something like $A*1/(1+x^2)$ where A is some constant, then use first row i wrote to show that inverse fourieur transform that what you got is left side of equation. third integral in first row (the one inside real part) is inverse fouriur transform without $1/2\pi$ and if you swap x and $\omega $ its inverse fourieur transform like in books. Also check for $d\omega$ if its like that integral diverges for sure. good luck have fun – jack Jun 12 '15 at 22:46