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How to prove this integral is 0 ?

  1. Jun 27, 2015 #1

    qsk

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    For this function : Exp[-Cos[x]]*Cos[x/3]], integrated from 0 to 3*Pi. The numerical result suggests that this integral may gives 0. However, I couldn't find a way to analytically show it's 0. Could anyone tell me how to prove this ?
     
    Last edited by a moderator: Jun 27, 2015
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  3. Jun 27, 2015 #2

    mathman

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    If you look at the graph (analytically) you will see that one side the midpoint of the domain ([itex]\frac{3\pi}{2}[/itex]) is the same as the other with a sign switch. Therefore the integrals over each half cancel.
     
  4. Jun 28, 2015 #3

    HallsofIvy

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    If you make the change of variable [itex]u= x- 3\pi/2[/itex] (the midpoint of "0 to [itex]3pi[/itex]") then [itex]x= u+ 3\pi/2[/itex] so [itex]cos(x)= cos(u+ 3\pi/2)= sin(u)[/itex] and [itex]cos(x/3)= cos(u/3+ \pi/2)= -sin(u/3)[/itex]. Of course, dx= du so the integral becomes [itex]-\int_{-3\pi/2}^{3\pi/2} e^{sin(u)}sin(u) du[/itex]. Now the fact that sine is an "odd" function, sin(-x)= -sin(x) should make it clear that the integral is 0.
     
  5. Jun 28, 2015 #4

    qsk

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    I know that, but the shape of the two sides looks very different. How can I proof it analytically the integrals of the two sides cancels.
     
  6. Jun 28, 2015 #5
    This doesn't seem easy to prove. It seems that the definite integral of Exp[-Cos[x]]*Cos[x/n]] from 0 to n*Pi is zero for n integer>=2
     
  7. Jun 28, 2015 #6
    Equivalently the definite integral of Exp[-Cos[nx]]*Cos[x] from 0 to pi is zero
     
  8. Jun 28, 2015 #7
    Halls of Ivy, the presence of exp(sin u) makes the conclusion unclear to me. Also I think the integrand should have sin(u/3) not sin u
     
  9. Jun 28, 2015 #8

    mathman

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    I took another look and I believe I was wrong. The graphs on the two sides are the same without a sign switch.
     
  10. Jun 29, 2015 #9
    The only axis of symmetry of the graph is the line y = 3kπ with k integer.

    http://www.wolframalpha.com/input/?i=+Exp[-Cos[x]]*Cos[x/3]&x=0&y=0
    Somehow the part above the y-axis with 0 < x < 3π/2 has the same area as the part below the y-axis with 3π/2 < x < 3π, even tough the shapes are completely different.

    BTW, mathworld can't solve the integral symbolically and reports that it is equal to 0 with an error of 10-16
     
  11. Jun 29, 2015 #10

    verty

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    For ##t \ge 0##, let ##y(t) = \int_0^{3\pi} t^{- cos(x)} \; cos(x/3) \; dx##.

    Testing with Wolfram Alpha suggests ##y## is always zero. Needless to say, this is way beyond my knowledge.
     
  12. Jun 30, 2015 #11
    Sketch proof (apologies no latex):

    Integral [0,3pi] exp(-cos(x))*cos(x/3) dx. Put y=x/3
    = 3 Integral [0,pi] exp(-cos(3y))*cos(y) dy
    Expand exp(cos(3y))= Sum [n=0 to inf] (cos(3y))^n / n!)
    Now cos(3y)^n can be expressed as a sum of cosines of multiples of 3y
    So we have cos(y) [ a cos 3y + b cos 6y + etc]
    which equals a cos y cos 3y + b cosy cos 6y + ....
    and using the relationship cos (A-B) + cos(A+B) = 2 cosAcosB
    we will then have a simple sum of cosines of multiples of y
    Sum (n=1 to inf) of a(n) cos (ny)
    The integral of this is Sum (n=1 to inf) of a(n)/n sin (ny)
    which when evaluated between 0 and pi gives zero.

    This needs to be made rigorous, obviously.
     
  13. Jun 30, 2015 #12
    Very good. It's a professional answer.
     
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