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Bad integral

  1. Jan 10, 2006 #1
    Anyone has any idea how I could go about evaluating this one?

    [tex]\int_{-\infty}^{\infty} \frac{A^2}{(x^2+a^2)^2} dx[/tex]

    (I have the final answer, I need a method of solution though.)

    Thanks,
    Chen
     
  2. jcsd
  3. Jan 10, 2006 #2

    TD

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    It's possibe to do it by just finding the antiderivative (it'll clearly involve an arctan) but thanks to the interval you're integrating over, it's also possible to due it using complex analysis. Have you seen residue calculation?
     
  4. Jan 10, 2006 #3
    I don't believe so, no.

    I calculated the antiderivative in Mathematica and it's just ugly... but how can I go about finding it?

    Thanks. :)
     
  5. Jan 10, 2006 #4

    TD

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    Too bad, it makes it *a lot* easier :smile:

    Take a look at partial fractions. In the 4th post, there are 4 cases - yours is case 4. An example can be found at the bottom of that post.
     
    Last edited: Jan 10, 2006
  6. Jan 10, 2006 #5
    Ahh, great. Thanks! :)
     
  7. Jan 10, 2006 #6

    TD

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    No problem, give a shout if it doesn't work out.
     
  8. Jan 10, 2006 #7
    Well, I don't think that's the case. In that post 'x' is also in numerator which makes it quite easy to be solved with substitution, here in numerator there is just constant. I think the approach will have to be slightly different..
     
  9. Jan 10, 2006 #8

    TD

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    Case 4 is the general case where this falls under (due to the denominator), the coefficient of x in the numerator is just zero.
    But of course, that will influence the rest of the method a bit, it'll start like this (for example):

    [tex]
    \int {\frac{{A^2 }}
    {{\left( {x^2 + a^2 } \right)^2 }}dx} \mathop \to \limits_{dx = adt}^{x = at} \int {\frac{{aA^2 }}
    {{\left( {a^2 t^2 + a^2 } \right)^2 }}dt} = \int {\frac{{aA^2 }}
    {{\left( {a^2 \left( {t^2 + 1^2 } \right)} \right)^2 }}dt} = \frac{{A^2 }}
    {{a^3 }}\int {\frac{1}
    {{\left( {t^2 + 1} \right)^2 }}dt}
    [/tex]
     
  10. Jan 10, 2006 #9
    Actually, it worked out perfectly. And wasn't *too* complicated... ;-)

    Anyway, now a different problem - I need to find the inverse Fourier transform of the function:

    [tex]\frac{A}{x^2+a^2}[/tex]

    I tried integrating by parts and using the complex presentation of arctan to no avail. Any ideas?
     
  11. Jan 10, 2006 #10
    Thank you for the explanation.
     
  12. Jan 10, 2006 #11

    TD

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    You're welcome - the last integral can then be solved with integrating by parts.

    @Chen: what's your Fourier transform variable in your expression then? The [itex]i\omega [/itex]? It is your x?
     
    Last edited: Jan 10, 2006
  13. Jan 10, 2006 #12
    Doesn't matter, really... (does it?)

    I need to calculate:

    [tex]\frac{A}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} \frac{1}{x^2+a^2} e^{-\frac{i}{\hbar} p x}dx[/tex]

    But I want to know how to do this, because the final result I obviously can find myself. :)
     
  14. Jan 10, 2006 #13

    TD

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    Unfortunately, again, this becomes fairly easy using residues :blushing:

    It probably won't matter much, but I've seen the inverse Fouriertransform as

    [tex]\frac{1}
    {{\sqrt {2\pi } }}\int\limits_{ - \infty }^{ + \infty } {F\left( x \right)e^{ixt} dx} \to \frac{1}
    {{\sqrt {2\pi } }}\int\limits_{ - \infty }^{ + \infty } {\frac{A}
    {{x^2 + a^2 }}e^{ixt} dx} = \frac{A}
    {{\sqrt {2\pi } }}\int\limits_{ - \infty }^{ + \infty } {\frac{{e^{ixt} }}
    {{x^2 + a^2 }}dx} [/tex]
     
  15. Jan 10, 2006 #14
    Well, perhaps you could post the solution anyway? If it's not too long, of course. Maybe I did learn some of the methods you describe as "residues"...

    (Oh - different branches of science use different conventions for thigns such as the FT. In physics (quantum mechanics) this notation is often used. :smile:)
     
  16. Jan 10, 2006 #15
    By the way, it would be just as useful to me to find the IFT of arctan (in the same notation)... in case it's any easier.
     
  17. Jan 10, 2006 #16

    TD

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    Yes, which is why I thought it wouldn't matter :smile:

    Do you mean the method or just the solution? Using residues, I found the inverse transform to be

    [tex]f\left( t \right) = \frac{{Ae^{at} \sqrt {\frac{\pi }{2}} }}{a}[/tex]

    For some reason I haven't figured out yet, mathematica returns a minus in the exponent when I use the InverseFourierTransform.
     
  18. Jan 10, 2006 #17
    Mathematica also thinks there's a heaviside step function involved in there... the minus sign is probably because in the inverse transform, the exponent has a negative argument.

    And I mean the method, of course... :smile:
     
  19. Jan 10, 2006 #18

    TD

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    "My mathematica" doesn't give a Heaviside function...
    In the IFT fomula (the integral), the exponent doesn't have a negative sign (the FT does). At least not in my version, but Mathworld agrees on that.

    Well in order to understand the method, you'd need to know about residues of course. The idea is that we're going to evaluate the integral in the complex plane using a contour integral (closed path) which is a large semi-circle over the upper half plane, with say radius R. The idea is then to let R -> infinity so that you can split the integral in two parts: the large arc and the part over the x-axis (real axis).
    That last part is exactly what we're looking for (since we're going from -inf to +inf) and you can show that the integral over the arc will tend to zero (thanks to the second degree function in the denominator, this is generally not the case!).

    We can then evaluate the integral over the closed path by calculating all the residues of the poles which lie in our area of integration, which is in this case [itex]x=ia[/itex], the denominator becomes 0 there.
    Since that's the only pole we have to consider, that'll be the only residue as well. Then the integral is given by [itex]2\pi i Q[/itex] where Q means this residue (in general: the sum of all residues).

    Here, this residue is

    [tex] - \frac{{iAe^{at} }}
    {{2a}}[/tex]

    Try it for yourself in mathematica:
    Code (Text):
    Residue[A/(x^2 + a^2)*Exp[-\[ImaginaryI]*x*t],x,\[ImaginaryI]*a]
    Of course, these residues can be calculated by hand as well, using a limit:

    [tex]\mathop {\lim }\limits_{x \to ia} \left( {\frac{{Ae^{ixt} }}
    {{x^2 + a^2 }}\left( {x - ia} \right)} \right)[/tex]

    So

    [tex]2\pi i\left( { - \frac{{iAe^{at} }}
    {{2a}}} \right) = \frac{{\pi Ae^{at} }}
    {a}[/tex]

    So

    [tex]\frac{1}
    {{\sqrt {2\pi } }}\int\limits_{ - \infty }^{ + \infty } {\frac{A}
    {{x^2 + a^2 }}e^{ixt} dx} = \frac{1}
    {{\sqrt {2\pi } }}\frac{{\pi Ae^{at} }}
    {a} = \frac{{Ae^{at} \sqrt {\frac{\pi }
    {2}} }}{a}[/tex]

    It may still seem 'long' or 'complicated' but that's because it's new to you. When you're used to this (and you can skip the introduction), then this is a fast method for evaluating these kind of integrals without any need for 'real integration'.

    Edit: before some *real mathematicians* become angry with me, I'd like to point out that the above is not very rigourous and certainly not 100% 'mathematically correct'. There are theorems which give necessary and/or sufficient condition as to when you can do the method described above (e.g. conditions when that large arc tends to zero if we let R->infinity). It was just an attempt to briefly show how this could be solved using residues.
     
    Last edited: Jan 10, 2006
  20. Jan 11, 2006 #19

    benorin

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    It's not that bad, is it?

    Let a>0 and put [tex]I=\int_{-\infty}^{\infty} \frac{A^2}{(x^2+a^2)^2} dx[/tex]

    Make the substitution [itex]x=a\tan (y)\Rightarrow dx=a\sec^{2}ydy[/itex] to get

    [tex]I=\frac{A^2}{a^3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2}(y) dy = \frac{A^2}{2a^3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left( 1+ \cos (2y)\right) dy = \frac{A^2}{2a^3}\left[ y+ \frac{1}{2}\sin (2y)\right]_{y=-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{A^2\pi}{2a^3}[/tex]

    I must admit that I used Maple to find the substitution.
     
    Last edited: Jan 11, 2006
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