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Question on changing interval of integration.

  1. Jul 17, 2013 #1
    Attached is a definite integral copy from page 20 of http://ia601507.us.archive.org/5/items/ATreatiseOnTheTheoryOfBesselFunctions/Watson-ATreatiseOnTheTheoryOfBesselFunctions.pdf

    It said

    I can understand ##\alpha=-\pi## and change the integral to

    But I can not agree with changing ##\theta## to ##-\theta## in half of the equation so

    You cannot just at will, substitute second part of the equation from ##\theta## to ##-\theta## and leave the first part ##\theta## still ##\theta##.

    Am I missing the point?


    Attached Files:

    Last edited: Jul 17, 2013
  2. jcsd
  3. Jul 17, 2013 #2
    Surely you can. It's a basic property of integration that if [itex]c \in [a,b][/itex],

    [tex]\int_a^b f(x) \mathrm{d}x =\int_a^c f(x) \mathrm{d}x + \int_c^b f(x) \mathrm{d}x[/tex]

    Now you have two separate integrals that can of course be treated completely independently, including changing variables. In particular, after bisecting the interval of your integral, the first part becomes

    [tex]\int_{-\pi}^{0} e^{i(n\theta-z\sin\theta)}\mathrm{d}\theta = - \int_{0}^{\pi} e^{i(n(-\theta)-z\sin(-\theta))}\mathrm{d}(-\theta) = \int_{0}^{\pi} e^{-i(n\theta-z\sin\theta)}\mathrm{d}\theta[/tex]

    so that

    [tex] \int_{-\pi}^{\pi} e^{i(n\theta-z\sin\theta)}\mathrm{d}\theta = \int_{-\pi}^{0} e^{i(n\theta-z\sin\theta)}\mathrm{d}\theta + \int_{0}^{\pi} e^{i(n\theta-z\sin\theta)}\mathrm{d}\theta[/tex]
    [tex]= \int_{0}^{\pi} e^{-i(n\theta-z\sin\theta)}\mathrm{d}\theta + \int_{0}^{\pi} e^{i(n\theta-z\sin\theta)}\mathrm{d}\theta = \int_0^{\pi} e^{-i(n\theta-z\sin\theta)} + e^{i(n\theta-z\sin\theta)}\mathrm{d}\theta[/tex]

    by the linearity of integration, resulting in the final expression given in your attachment.
    Last edited: Jul 17, 2013
  4. Jul 17, 2013 #3
    Yes, you beat me to it. I just figure it out and was going to ask to delete this whole post.

    Please take a look at this thread:https://www.physicsforums.com/showthread.php?t=701297, this is the very confusion on the interval of integration. I used series representation and proofed it's not equal. I have been stuck for days on this interval thing.

  5. Jul 18, 2013 #4
    I have another changing interval question. See attachment
    [tex] J_n(z)=\frac{1}{\pi}\int_0^{\pi}\cos(n\theta)\cos(z\sin\theta)d\theta\;+\;\frac{1}{\pi} \int_0^{\pi}\sin(n\theta)\sin(z\sin\theta)d\theta[/tex]
    The article claimed if we substitute [itex]\theta=\pi-\theta[/itex], if [itex]n[/itex] is odd, then:

    I just don't get this. I know
    [tex] J_n(z)=\frac{1}{\pi}\int_0^{\pi}\cos(n\theta)\cos(z\sin\theta)d\theta\;+\;\frac{1}{\pi} \int_0^{\pi}\sin(n\theta)\sin(z\sin\theta)d\theta[/tex]
    [tex] J_n(z)=\frac{1}{\pi}\int_0^{\pi}\cos(n\theta)\cos(z\sin\theta)d\theta\;+\;\frac{1}{\pi} \int_0^{\pi}\sin(n\theta)\sin(z\sin\theta)d\theta=\frac{1}{\pi}\int_{\pi}^{0}-\cos(n\theta)\cos(z\sin\theta)d\theta\;-\;\frac{1}{\pi} \int_{\pi}^{0}\sin(n\theta)\sin(z\sin\theta)d\theta[/tex]
    For [itex]n[/itex] is ODD.

    I can't get (6) in the attached copy. Please help.


    Attached Files:

  6. Jul 18, 2013 #5
    Let's see. The result (6) only applies for odd n, so keeping that in mind we can show some symmetry properties:

    [tex]J_n(z)= \frac{1}{\pi} \int_0^\pi \cos n\theta\cos(z\sin\theta)\mathrm{d}\theta + \frac{1}{\pi} \int_0^\pi \sin n\theta\sin(z\sin\theta)\mathrm{d}\theta[/tex]
    [tex]=\frac{1}{\pi} \int_\pi^0 \cos n(\pi-\theta)\cos(z\sin(\pi-\theta))\mathrm{d}(-\theta) + \frac{1}{\pi} \int_\pi^0 \sin n(\pi-\theta)\sin(z\sin(\pi-\theta))\mathrm{d}(-\theta)[/tex]
    [tex]=\frac{1}{\pi} \int_0^\pi \cos n\pi \cos n\theta\cos(z\sin\theta)\mathrm{d}\theta + \frac{1}{\pi} \int_0^\pi \sin n\pi\sin n\theta\sin(z\sin\theta)\mathrm{d}\theta[/tex]

    For odd n,

    [tex]=-\frac{1}{\pi} \int_0^\pi\cos n\theta\cos(z\sin\theta)\mathrm{d}\theta + \frac{1}{\pi} \int_0^\pi \sin n\theta\sin(z\sin\theta)\mathrm{d}\theta[/tex]

    So, let's define the initial integrals as [itex]I_1, I_2[/itex]. The last line gives us

    [tex]J_n(z)=I_1+I_2=-I_1+I_2 \implies I_1=0[/tex].

    If this is what you didn't get, it's because the two integrals are independent, the general result is only achieved if [itex]I_1=-I_1=0[/itex], independent of z.

    In addition, the very last equality in the attachment holds as what the change of variables "did" was essentially to mirror the interval from "0 to π" to "π to 0". What's written above shows that it does not effect the value of the integrand in any point of the interval, and so the integrand must be symmetric over the interval, so that

    [tex]J_n(z)=I_2=\frac{1}{\pi} \int_0^\pi \sin n\theta\sin(z\sin\theta)\mathrm{d}\theta=\frac{2}{\pi} \int_0^\frac{\pi}{2} \sin n\theta\sin(z\sin\theta)\mathrm{d}\theta[/tex]

    I hope I didn't make any careless mistakes here, this was a bit lengthier than the previous one to write out.

    What was your problem in the previous thread, I couldn't quite figure it out? I only saw was the definition for [itex]J_0[/itex] you were concerned about, but that is, like stated in the thread, incorrect (or, at least it's very non-standard...).

    EDIT: Fixed the bit about the second integral.
    Last edited: Jul 18, 2013
  7. Jul 18, 2013 #6
    Thanks for the detail answer. As for the last step, do you imply

    [tex]\frac{1}{\pi}\int_0^{\pi}sin\theta d\theta=\frac{2}{\pi}\int_0^{\frac{\pi}{2}}sin\theta d\theta[/tex]
    Because integral of sine from 0 to π equal to twice of 0 to π/2 because the area of sine function from π/2 to π is the mirror image of area of sine function from 0 to π/2?

    Is what you can't figure out is in this thread:https://www.physicsforums.com/showthread.php?t=701297 ?

    If so, I used series expansion of ##e^{jx\sin\theta}## to show that for m=odd, the function is complex and cannot get the result like here.
  8. Jul 18, 2013 #7
    Basically yes, but it's obviously not just the sine function but the sine times the nested sine we're looking at. But yes, the area under the curve for 0 to π/2 is the same as for π/2 to π.
    Cf. n=3, z=π: Plot to see what I mean.

    About the other thread, yeah, similar arguments don't immediately apply there, you can't halve the interval and expect to get the same result. You have to integrate to 2π to get the Bessel function, the integral to π is indeed complex in general.
  9. Jul 18, 2013 #8
    The problem is most books just make
    That cannot be right!!!

    Again, thanks for all your help.
  10. Jul 18, 2013 #9
    Hm. I don't know where you've seen that, I've never seen such a relation, and it just seems to be incorrect (even assuming you mean to write [itex]J_0(x)[/itex] there). If you'd have [itex]\Re[/itex] in front of them, it'd be correct, as the real part of the integrals do indeed give the Bessel functions.

    Based on your usage of j, I assume that the material you're reading is engineering-oriented. Is it possible that all these functions are meant to represent a physical quantity and there's a general convention that the real part of a function is always taken to represent the physical function or something like that, and the imaginary parts are just 'ignored'? It's often done with waves and such, and I guess one could do that in general as well, although it does strike me as an odd thing to do.
    Last edited: Jul 18, 2013
  11. Jul 18, 2013 #10

    Attached is p57 of Introduction to Bessel Functions by Frank Bowman. Also it is in page 7 of this paper http://www.math.kau.se/mirevine/mf2bess.pdf

    This is where I got stuck for days!!! Thanks for your help.

    Attached Files:

  12. Jul 18, 2013 #11
    The paper seems to have a typo in it (the upper limit should be [itex]2\pi[/itex]), and the attachment has a cosine, not sine, in the exponent of eq. (4.3): It is indeed true that [itex]J_0(z)=\frac{1}{\pi}\int_0^\pi e^{iz\cos\theta} \mathrm{d}\theta[/itex], but earlier you wrote that the books said the zeroth Bessel function is [itex]\frac{1}{\pi}\int_0^\pi e^{iz\sin\theta} \mathrm{d}\theta[/itex] – just some carelessness, it seems. Eq. (4.4) also seems to be correct as it's written.
  13. Jul 18, 2013 #12
    I am confused, you said [itex]J_0(z)=\frac{1}{\pi}\int_0^\pi e^{iz\cos\theta} \mathrm{d}\theta[/itex] is true? I wrote my reasoning in the other post that this cannot be true for interval [0,π]. Please explain.

  14. Jul 18, 2013 #13
    Care to point me a post number? I can only see you consider [itex]\frac{1}{\pi}\int_0^\pi e^{iz\sin\theta} \mathrm{d}\theta[/itex] instead of [itex]\frac{1}{\pi}\int_0^\pi e^{iz\cos\theta} \mathrm{d}\theta[/itex] in the first thread.
  15. Jul 18, 2013 #14

    Sorry, I mis read your post, I was looking at the interval of integration. Now I saw [itex]\frac{1}{\pi}\int_0^\pi e^{iz\cos\theta} \mathrm{d}\theta[/itex]....which is ##\cos(z\sin\theta)##. I never commented on this at all. Still stuck at the interval, never get to (4.3) and (4.4) yet.

    So bottom line ONLY [itex]J_n(z)=\frac {1}{2\pi}\int_0^{2\pi}e^{jz\cos\theta}d\theta[/itex] is true, not [itex]J_n(z)=\frac {1}{\pi}\int_0^{\pi}e^{jz\cos\theta}d\theta[/itex]. This is all I want to know at this time.

    BTW, I am studying Bessel Function ONLY because in my antenna theory book, it said:
    [tex]\pi j^n J_n(z)=\int_0^{\pi} \cos(n\phi)e^{+jz\cos\phi}d\phi[/tex]
    Now I got stuck in this ##J_0(z)## thing for a few days and I really need to move on. Can you help me on this as I really run out of time?

    I understand:
    [tex]J_n(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\phi-n\theta)} d\theta[/tex]
    Can you show me how do I get to
    [tex]\pi j^n J_n(z)=\int_0^{\pi} \cos(n\theta)e^{+jz\cos\theta}d\theta[/tex]
    Last edited: Jul 18, 2013
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