Results 1 to 5 of 5

Math Help - Integral Equation

  1. #1
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    469
    Thanks
    2
    Awards
    1

    Integral Equation

    Hi,
    I would like some help with this question :
     <br />
\int\limits_{-\infty}^{\infty} \frac{y(\xi)d\xi }{(t-\xi)^2+25}=\frac{1}{t^2+36}<br />
    Find  <br />
y(\xi)<br />
    This is a question from an old final exam that I'm struggling with. I think we haven't seen the necessary material this year. However, I want to be sure that I'm not missing some easy trick.
    Thanks
    Last edited by vincisonfire; December 10th 2009 at 10:01 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    469
    Thanks
    2
    Awards
    1
    Left side is a convolution product.
    Use a Fourier transform on both sides.
    Then isolate the Fourier transform of y(\xi) by dividing on both sides, and take the inverse Fourier transform of Y(\omega).
    I think you should get y(t)=\frac{1}{t^2+1}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Aug 2008
    Posts
    903
    That certainly makes sense to me now that you explained it. Thanks. However, when I back-substitute that solution into the integral equation and solve the integral numerically, I do not obtain values that are close enough to convince me it's right (I'd expect at least 2 or 3 digit agreement and preferably 6). And the numerical results reported by Mathematica do not change significantly when I up the precision of the numerical calculations. Here is an example for t=1:

    Code:
    In[230]:=
    y[t_] := 1/(t^2 + 1); 
    i1 = NIntegrate[y[s]/((t - s)^2 + 25) /. 
        t -> 1, {s, -Infinity, Infinity}]
    i2 = N[1/(t^2 + 36) /. t -> 1]
    
    Out[231]= 0.101889
    
    Out[232]= 0.027027
    Granted, there is the chance Mathematica is not correctly doing the numerical integration. However Mathematica is usually pretty good at that especially when it does not report any complications such as highly oscillatory or very slow rate of convergence. I've tried other values of t including complex ones and the expressions do not agree.
    Last edited by shawsend; December 19th 2009 at 07:36 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    469
    Thanks
    2
    Awards
    1
    You're right I typed :
    Integrate[1/((1 + x^2)*((x - a)^2 + 25)), {x, -Infinity, Infinity},
    Assumptions -> Im[a] == 0 ]
    in mathematica and got
    <br /> <br />
y(t)=\frac{6\pi}{5(t^2+36)}<br />
    The answer must thus be
    <br /> <br />
y(t)=\frac{5}{6\pi(t^2+1)}<br />
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Very good then and thanks for that Mathematica titbit. Didn't know how to do that.

    I've looked at it more carefully now and see where my problem was. This is actually a very interesting problem in contour integration as it involves the transform:

    F\left\{\frac{1}{x^2+25}\right\}=\int_{-\infty}^{\infty} \frac{e^{-isx}}{(x+5i)(x-5i)}dx

    and for now, lets not worry about the factor \frac{1}{\sqrt{2\pi}} which also enters the problem up there.

    Now normally, one uses a upper or lower half-disc contour and the residue at either 5i or -5i and for the integral to be zero on the half-circle part of the contour, we need to use the appropriate residue and that depends on the sign of s and that's how the expression:

    F\left\{\frac{1}{x^2+25}\right\}=\int_{-\infty}^{\infty} \frac{e^{-isx}}{(x+5i)(x-5i)}dx=\frac{\pi}{5}e^{-5|s|}

    is obtained I think.
    Last edited by shawsend; December 19th 2009 at 09:36 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. help integral equation
    Posted in the Calculus Forum
    Replies: 11
    Last Post: June 24th 2011, 01:57 AM
  2. Integral equation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 10th 2011, 12:50 PM
  3. integral equation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 4th 2009, 11:37 AM
  4. Replies: 2
    Last Post: April 28th 2009, 06:42 AM
  5. integral equation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 21st 2008, 03:43 AM

Search Tags


/mathhelpforum @mathhelpforum