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I have to solve the equation

$\int_0^{\infty} f(x) \cos{(\alpha x)}\, dx=\frac{\sin{\alpha }}{\alpha}$

Using fourier transform. I know this is half of the usual fourier cosine transform, and so that I would get back $f(x)$ using $\frac{2}{\pi} \int_0^{\infty} \frac{\sin{\alpha} \cos{\alpha x}}{\alpha}d{\alpha}$

Is this correct? How do I do this integral?

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  • $\begingroup$ The inverse Fourier transform for the sinc function is the box function. $\endgroup$ – obataku Apr 29 '13 at 8:14
  • $\begingroup$ How can you show that? Hoe do you calculate $\int \frac{\sin{\alpha} \cos{\alpha x}}{\alpha}$? $\endgroup$ – user23238 Apr 29 '13 at 8:41
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$$ \begin{align} \frac{2}{\pi} \int_{0}^{\infty} \frac{\sin (\alpha) \cos (x \alpha)}{\alpha} \ d \alpha &= \frac{1}{\pi} \int_{0}^{\infty} \frac{\sin \big((1+x) \alpha \big)+\sin \big( (1-x) \alpha \big)}{\alpha} \ d \alpha \\ &= \frac{1}{\pi} \Big(\text{sgn}(1+x) \frac{\pi}{2}+\text{sgn}(1-x) \frac{\pi}{2} \Big) \\ &= \frac{1}{2} \Big(\text{sgn}(1+x) + \text{sgn}(1-x) \Big) \\ &= \begin{cases} \frac{1}{2}(-1+1) = 0 & \text{if} \ x <-1 \\ \frac{1}{2}(0+1) = \frac{1}{2} & \text{if} \ x = -1 \\ \frac{1}{2} (1+1) = 1 & \text{if} -1 < x <1 \\ \frac{1}{2}(1+0) = \frac{1}{2} & \text{if} \ x = 1 \\ \frac{1}{2} (1-1) = 0 & \text{if} \ x >1\end{cases} \end{align}$$

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  • $\begingroup$ Thanks! Please could you tell me how you got the sign function from the integral containing sin? $\endgroup$ – user23238 Apr 29 '13 at 16:09
  • $\begingroup$ $\int_{0}^{\infty} \frac{\sin at}{t} \ dt = \frac{\pi}{2} \text{sgn}(a)$ $\endgroup$ – Random Variable Apr 29 '13 at 16:12
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The RHS of your equation is real and even in $\alpha$. Therefore, its FT is real and even in $x$, its transform variable. Therefore, the equation is equivalent to

$$\int_{-\infty}^{\infty} dx \: f(x) e^{i \alpha x} = 2 \frac{\sin{\alpha}}{\alpha}$$

Inverting this transform, we get

$$f(x) = \begin{cases} \\ 1 & |x| < 1\\0 & |x| > 1\end{cases}$$

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  • $\begingroup$ I had done uptil this point, and was asking about the inverse fourier transform of my RHS, which I didn't know is a famous function called the sinc function, and who transform is the box function. BTW, should't the $\pi$ cancel out as $f(x)=\frac{1}{\pi}\int_0^{\infty}A(\xi)\cos{\xi x}$, where $A$ is the even part. The definitions vary, but I thought one of the integrals of the pair is defined with a $2 \pi$ $\endgroup$ – user23238 Apr 29 '13 at 15:59
  • $\begingroup$ @ramanujan_dirac: yes, you are correct: there is a factor of $1/(2 \pi)$ that is present in the inverse transformed that I did not apply. Good catch. $\endgroup$ – Ron Gordon Apr 29 '13 at 16:18
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$


\begin{align} &\bbox[5px,#ffd]{{2 \over \pi}\int_{0}^{\infty}{\sin\pars{\alpha} \cos\pars{\alpha x} \over \alpha}\,\dd\alpha} = {1 \over \pi}\int_{-\infty}^{\infty}{\sin\pars{\alpha} \cos\pars{\alpha x} \over \alpha}\,\dd\alpha \\[5mm] = &\ {1 \over \pi}\,\Re\int_{-\infty}^{\infty}\expo{-\ic \alpha x}\,\,{\sin\pars{\alpha} \over \alpha}\,\dd\alpha \\[5mm] = &\ {1 \over \pi}\,\Re\int_{-\infty}^{\infty}\expo{-\ic \alpha x} \pars{{1 \over 2}\int_{-1}^{1}\expo{\ic k\alpha}\dd k}\dd\alpha \\[5mm] = &\ \Re\int_{-1}^{1}\ \underbrace{ \int_{-\infty}^{\infty}\expo{\ic\pars{k - x}\alpha}\,\,\, {\dd\alpha \over 2\pi}}_{\ds{\delta\pars{k - x}}}\,\dd k = \bracks{\verts{x} < 1\vphantom{\Large A}} \end{align} The case $\ds{x = \pm 1}$ yields $\ds{\color{red}{1/2}}$ which can be directly derived from the proposed integral. Then, \begin{align} &\bbox[5px,#ffd]{{2 \over \pi}\int_{0}^{\infty}{\sin\pars{\alpha} \cos\pars{\alpha x} \over \alpha}\,\dd\alpha} = \left\{\begin{array}{lcl} \ds{1} & \mbox{if} & \ds{-1 < x < 1} \\[1mm] \ds{1 \over 2} & \mbox{if} & \ds{x = \pm 1} \\[1mm] \ds{0} & & \mbox{otherwise} \end{array}\right. \end{align}

enter image description here

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FORTRAN code for box function

c      www.uprh.edu/rbaretti
c integral for the box function
c https://math.stackexchange.com/questions/375960/solving-an-
c  integral-equation-using-the-fourier-transform

      implicit real*8(a-h,o-z)
      real*8 kernel1,kernel2, k
      data nx,nk/30,2000/
      kernel1(k,x)= (2.d0/pi)*(1.d0-k**2/6.d0)*dcos(k*x)
      kernel2(k,x)= (2.d0/pi)*dsin(k)*dcos(k*x)/k
      pi=2.d0*dasin(1.d0)
      xi= -7.d0/pi
      xf=-xi
      ki=0.d0
      kf=50.d0
      dx=(xf-xi)/dfloat(nx)
      dk=(kf-ki)/dfloat(nk)
      do 10 ix=0,nx
      x =xi+dx*dfloat(ix)
      sum=0.d0
      do 20 ik=1,nk
      k=ki+dk*dfloat(ik)
      if(ik.eq.1)sum=sum+(dk/2.d0)*(kernel1(k,x)  + kernel1(k-dk,x))
      if(ik.gt.1)sum=sum+(dk/2.d0)*(kernel2(k+dk,x)+ kernel2(k,x) )
20    continue
      print 100 ,x,sum
10    continue
100   format('x,f(x)=',2(3x,e10.3))
      stop
      end

x,f(x)=   -0.223E+01   -0.144E-01
x,f(x)=   -0.208E+01   -0.187E-01
x,f(x)=   -0.193E+01   -0.205E-01
x,f(x)=   -0.178E+01   -0.160E-01
x,f(x)=   -0.163E+01   -0.859E-02
x,f(x)=   -0.149E+01   -0.824E-02
x,f(x)=   -0.134E+01   -0.225E-01
x,f(x)=   -0.119E+01   -0.464E-01
x,f(x)=   -0.104E+01   -0.259E-01
x,f(x)=   -0.891E+00    0.951E+00
x,f(x)=   -0.743E+00    0.958E+00
x,f(x)=   -0.594E+00    0.983E+00
x,f(x)=   -0.446E+00    0.998E+00
x,f(x)=   -0.297E+00    0.994E+00
x,f(x)=   -0.149E+00    0.979E+00
x,f(x)=    0.833E-16    0.972E+00
x,f(x)=    0.149E+00    0.979E+00
x,f(x)=    0.297E+00    0.994E+00
x,f(x)=    0.446E+00    0.998E+00
x,f(x)=    0.594E+00    0.983E+00
x,f(x)=    0.743E+00    0.958E+00
x,f(x)=    0.891E+00    0.951E+00
x,f(x)=    0.104E+01   -0.259E-01
x,f(x)=    0.119E+01   -0.464E-01
x,f(x)=    0.134E+01   -0.225E-01
x,f(x)=    0.149E+01   -0.824E-02
x,f(x)=    0.163E+01   -0.859E-02
x,f(x)=    0.178E+01   -0.160E-01
x,f(x)=    0.193E+01   -0.205E-01
x,f(x)=    0.208E+01   -0.187E-01
x,f(x)=    0.223E+01   -0.144E-01
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