Fourier transform and integral

rayman · December 12th, 2011, 09:41 AM

I need to compute the integral
$\int_{-\infty}^{\infty} \frac{\cos(ax)}{x^2+1}dx$

by using the following Fourier transform

$\Bigl(e^{-|x|}\Bigr)^{\wedge}=\frac{2}{\xi^2+1}$

1)I start with the inversion

$e^{-|x|}=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2}{\xi^2+1}e^{i\xi x}d\xi=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2}{\xi^2+1}\cos(x\xi)d\xi +i\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2}{\xi^2+1}\sin(x\xi)d\xi$

$\int_{-\infty}^{\infty}\frac{\cos(ax)}{x^2+1}dx=\int_{-\infty}^{\infty}\frac{\cos(\xi a)}{\xi^2+1}d\xi$

and going back to my integral

$\int_{-\infty}^{\infty}\frac{\cos(ax)}{x^1+1}dx= \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2}{\xi^2+1}\cos(\xi a)d\xi + 0=\frac{1}{2}\int_{-\infty}^{\infty}\frac{2}{\xi^2+1}\cos(\xi a)d\xi+i\frac{1}{2}\int_{-\infty}^{\infty}\frac{2}{\xi^2+1}\sin(\xi a)d\xi= \frac{1}{2}\int_{-\infty}^{\infty}\frac{2}{\xi^2+1}(\cos(\xi a)+\sin(\xi a))d\xi=<br /> \pi\cdot \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2}{\xi^2+1}e^{i\xi a}d\xi=\pi\cdot f(a)=\pi\cdot e^{-|a|}$

and the result agrees with the answers

But I cannot find out how to compute this integral, could someone show me this?

$\int_{-\infty}^{\infty}\frac{\cos(ax)}{(x^2+1)^2}$ also using the same Fourier transform

thank you

wnvl · December 12th, 2011, 04:48 PM

Use

$\mathcal{F}_x^{-1}\left[\frac{1}{\left(x^2+1\right)^2}\right](t)=\frac{1}{4} e^{-|t|} (|t|+1)$

$\frac{1}{4} e^{-|t|} (|t|+1)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{\left(x^2+1\right)^2}e^{i tx}dx$
$\frac{\pi}{2} e^{-|t|} (|t|+1)=\int_{-\infty}^{\infty}\frac{cos(tx)}{\left(x^2+1\right)^ 2}dx$
$\frac{\pi}{2} e^{-|a|} (|a|+1)=\int_{-\infty}^{\infty}\frac{cos(ax)}{\left(x^2+1\right)^ 2}dx$

rayman · December 13th, 2011, 12:36 AM

thank you very much, I have been trying to solve it for hours maybe because I still do not know how did you compute this inversion of Fourier transform?

Quote:

Originally Posted by wnvl

$\mathcal{F}_x^{-1}\left[\frac{1}{\left(x^2+1\right)^2}\right](t)=\frac{1}{4} e^{-|t|} (|t|+1)$

wnvl · December 13th, 2011, 03:05 PM

Quote:

Originally Posted by rayman

thank you very much, I have been trying to solve it for hours maybe because I still do not know how did you compute this inversion of Fourier transform?

Quote:

Originally Posted by wnvl

$\mathcal{F}_x^{-1}\left[\frac{1}{\left(x^2+1\right)^2}\right](t)=\frac{1}{4} e^{-|t|} (|t|+1)$

The invers Fourier transform of f(x)^2 is equal to the invers Fourier transform of f(x) convoluted with itself.

rayman · January 6th, 2012, 07:38 AM

I actually did compute the same transform and got slightly different result

Our teacher uses this formula for convolution
$(f(x)g(x))^{\wedge}=\frac{1}{2\pi}f^{\wedge}(\xi)\ star g^{\wedge}(\xi)$
Do we always have this constant $\frac{1}{2\pi}$ ??

wnvl · January 6th, 2012, 07:42 AM

Quote:

Originally Posted by rayman

Do we always have this constant $\frac{1}{2\pi}$ ??

That depends on your definition of Fourier transform and inverse Fourier transform.
There are different definitions in use.

rayman · January 6th, 2012, 07:49 AM

ah okej

that is very confusing

December 12th, 2011, 09:41 AM	#1
rayman Senior Member Joined: Sep 2010 From: Germany Posts: 153 Thanks: 0	Fourier transform and integral I need to compute the integral $\int_{-\infty}^{\infty} \frac{\cos(ax)}{x^2+1}dx$ by using the following Fourier transform $\Bigl(e^{-\|x\|}\Bigr)^{\wedge}=\frac{2}{\xi^2+1}$ 1)I start with the inversion $e^{-\|x\|}=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2}{\xi^2+1}e^{i\xi x}d\xi=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2}{\xi^2+1}\cos(x\xi)d\xi +i\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2}{\xi^2+1}\sin(x\xi)d\xi$ $\int_{-\infty}^{\infty}\frac{\cos(ax)}{x^2+1}dx=\int_{-\infty}^{\infty}\frac{\cos(\xi a)}{\xi^2+1}d\xi$ and going back to my integral $\int_{-\infty}^{\infty}\frac{\cos(ax)}{x^1+1}dx= \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2}{\xi^2+1}\cos(\xi a)d\xi + 0=\frac{1}{2}\int_{-\infty}^{\infty}\frac{2}{\xi^2+1}\cos(\xi a)d\xi+i\frac{1}{2}\int_{-\infty}^{\infty}\frac{2}{\xi^2+1}\sin(\xi a)d\xi= \frac{1}{2}\int_{-\infty}^{\infty}\frac{2}{\xi^2+1}(\cos(\xi a)+\sin(\xi a))d\xi=<br /> \pi\cdot \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2}{\xi^2+1}e^{i\xi a}d\xi=\pi\cdot f(a)=\pi\cdot e^{-\|a\|}$ and the result agrees with the answers But I cannot find out how to compute this integral, could someone show me this? $\int_{-\infty}^{\infty}\frac{\cos(ax)}{(x^2+1)^2}$ also using the same Fourier transform thank you
$rayman is offline$

December 12th, 2011, 04:48 PM	#2
wnvl Senior Member Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0	Re: Fourier transform and integral Use $\mathcal{F}_x^{-1}\left[\frac{1}{\left(x^2+1\right)^2}\right](t)=\frac{1}{4} e^{-\|t\|} (\|t\|+1)$ $\frac{1}{4} e^{-\|t\|} (\|t\|+1)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{\left(x^2+1\right)^2}e^{i tx}dx$ $\frac{\pi}{2} e^{-\|t\|} (\|t\|+1)=\int_{-\infty}^{\infty}\frac{cos(tx)}{\left(x^2+1\right)^ 2}dx$ $\frac{\pi}{2} e^{-\|a\|} (\|a\|+1)=\int_{-\infty}^{\infty}\frac{cos(ax)}{\left(x^2+1\right)^ 2}dx$
$wnvl is offline$

January 6th, 2012, 07:38 AM	#5
rayman Senior Member Joined: Sep 2010 From: Germany Posts: 153 Thanks: 0	Re: Fourier transform and integral I actually did compute the same transform and got slightly different result Our teacher uses this formula for convolution $(f(x)g(x))^{\wedge}=\frac{1}{2\pi}f^{\wedge}(\xi)\ star g^{\wedge}(\xi)$ Do we always have this constant $\frac{1}{2\pi}$ ??
$rayman is offline$

January 6th, 2012, 07:49 AM	#7
rayman Senior Member Joined: Sep 2010 From: Germany Posts: 153 Thanks: 0	Re: Fourier transform and integral ah okej that is very confusing
$rayman is offline$

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