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Finding the value of an integral

  1. Oct 11, 2007 #1

    quasar987

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    1. The problem statement, all variables and given/known data
    The question is worded in this simple form exactly: "Show that

    [tex]\frac{d}{dx}\int_0^{+\infty}e^{-xy}\frac{\sin y}{y}dy=-\frac{1}{1+x^2}[/tex]

    Deduce from this that

    [tex]\int_0^{+\infty}\frac{\sin y}{y}dy=\frac{\pi}{2}[/tex]"


    3. The attempt at a solution

    What I've accomplished by taking the derivative under the integral sign and then integrating by parts 2 times is to show the desired equality for x in ]0, +infty[. For x=0, integration by parts didn't work because of a limit of cos(y) as y-->+infty. For x<0, the integral obviously diverges.

    Setting

    [tex]I(x)=\int_0^{+\infty}e^{-xy}\frac{\sin y}{y}dy[/tex]

    I have shown that

    [tex]I'(x)=-\frac{1}{1+x^2}[/tex]

    and I want to show that I(0)=pi/2.

    Taking the antiderivative of I'(x) tells me that

    [tex]I(x)=-tan^{-1}(x)+cst[/tex]

    This is only on ]0, +infty[ however! I don't even know if I(0) exists!

    What do I do now?:confused:
     
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  3. Oct 11, 2007 #2

    morphism

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    Why is it only valid on (0, infty)? I believe what you did makes the formula for I(x) valid on [0, infty).

    And to finish things off, try considering the limit of I(x) as x goes infinity.
     
  4. Oct 11, 2007 #3

    quasar987

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    I don't understand what you mean.

    I know only that I'(x) = 1/1+x² on ]0, +infty[. Therfor, I can only conclude that I(x) = =-arctan(x) + cst on ]0, +infty[.

    And I've noted that my method only shows I'(x) = 1/1+x² on ]0, +infty[ because in the case x=0, integration by parts didn't work because of a limit of cos(y) as y-->+infty.
     
  5. Oct 11, 2007 #4

    morphism

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    Why? It's valid on [0, infty[.
     
  6. Oct 11, 2007 #5

    quasar987

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    Well, this is what I proceed to explain in the next paragraph:

    "And I've noted that my method only shows I'(x) = 1/1+x² on ]0, +infty[ because in the case x=0, integration by parts didn't work because of a limit of cos(y) as y-->+infty."
     
  7. Oct 11, 2007 #6

    morphism

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    Ah, I see it now. I guess what you can do is take the limit of I(x) as x approaches 0 from the right, i.e. I(0+).
     
  8. Oct 11, 2007 #7

    quasar987

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    I can, but for this to yield the desired result, I need the additional information that

    1) I(0) exists
    2) I is continuous at 0

    I can get 2) from 1) using Lebesgue's dominated convergence thm. Do you see how to get 1)?

    (By the way it has been noted by the professor earlier in the course that sin(y)/y is not Lebesgue-integrable, in the sense that it is not Riemann-absolutely integrable)
     
    Last edited: Oct 11, 2007
  9. Oct 11, 2007 #8

    morphism

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    What's T? And we don't need the fact that I(0) exists -- this is precisely why we're taking lim(x->0+) I(x).
     
  10. Oct 11, 2007 #9

    quasar987

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    T=I, sorry.

    A function can have a limit at a point and still not be defined there, can it not? So why is the hypothesis that I(0) exists unnecessary?
     
  11. Oct 11, 2007 #10

    morphism

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    We don't care about I(0). Try computing I(0+) in two different ways: use both your formulas.

    (From LDCT we know that I is continuous at 0 from the right.)
     
  12. Oct 11, 2007 #11

    quasar987

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    LDCT says,

    "Let fn:E-->R be a sequence of integrable functions converging to f. Suppose there exists an integrable function g:E-->R such that |fn|<=g on E. Then

    [tex]\lim_n\int_Ef_n = \int_Ef[/tex]"

    The idea is to let {x_n} be any sequence of positive numbers converging to 0 and let

    [tex]f_n(y)=e^{-x_ny}\frac{\sin y}{y}[/tex]

    but what do you take as g? You can't take sin(y)/y because we do not yet know that this function is integrable.
     
  13. Oct 11, 2007 #12

    Dick

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    I(0) does exist in the sense that if you change the upper limit to L and take the limit as L->infinity the limit exists. It's basically an alternating series argument. It's not, of course, lesbegue integrable or anything, any more than an alternating series is necessarily absolutely convergent. Is that enough? It's been a while since I took real analysis...
     
  14. Oct 11, 2007 #13

    morphism

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    Gah! I just spent the last 30mins LaTeXing my thoughts on the problems, and then I lost everything.

    I guess all I want to add is that if

    [tex]J(a,x) = - \int_0^a e^{-xy} \sin y \, dy = -\frac{1}{1+x^2} +\frac{e^{-ax}}{1+x^2}\left( x \sin a + \cos a)[/tex]

    then

    [tex]|J(a,x)| \leq \frac{1 + (1+x)e^{-x}}{1+x^2}.[/tex]

    So we can apply the LDCT.
     
  15. Oct 12, 2007 #14

    quasar987

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    On J(n,x)? How is this gonna help?:confused:
     
  16. Oct 13, 2007 #15

    Gib Z

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    I haven't taken a real analysis course, but I think that taking the limit of I(x) as x-->0+ would be sufficient as it represents the Cauchy Principle Value of the integral, not the actual integral though. As they expect you to deduce that the integral is equal to pi/2, even though the integral diverges on the lower bound implies that they want the Cauchy Principle Value!!

    Basically what I'm saying is, they used common notation rather than the most proper one:

    Deduce that [tex]\lim_{a\to 0^{+}, b\to \infty} \int^b_a \frac{\sin y}{y} dy= \frac{\pi}{2}[/tex].

    So really the question isn't really even asking for I(0), its asking for I(x) as x-->0+.
     
  17. Oct 14, 2007 #16

    Dick

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    I think Gib has it right. There is no absolute definition of the integral. It's divergent. You can only give it a meaning by specifying a regularization procedure. Multiplying by exp(-xy) and letting x->0 is a regularization procedure. In that sense the definition of the integral is limit I(x) as x->0.
     
  18. Oct 14, 2007 #17
    [tex]I(x)=\int_0^{+\infty}e^{-xy}\frac{\sin y}{y}dy[/tex]
    So [tex]I(\infty) = 0[/tex]
    now use this initial condition to find const C and u are done

    [tex]0 = -arctan(\infty)+C[/tex]
    [tex]C= \frac{\pi}{2}[/tex]
     
  19. Oct 14, 2007 #18

    morphism

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    Good job ignoring the entire thread.
     
  20. Oct 14, 2007 #19
    well the original poster wanted to show that it was equal to pi/2 maybe I misunderstood the question...
    EDIT: Also I(0) cant be used for finding C as u get the integral u want to evaluate.....
     
    Last edited: Oct 14, 2007
  21. Oct 16, 2007 #20

    quasar987

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    Why do you all say that the integral is divergent?

    Maple says it's worth pi/2. We don't have the same problem at x=0 as with the function 1/x because of the limit sinx/x-->1. I think this is an example of a function that is Riemann- integrable but not Riemann-absolutely integrable (or Lebesgue-integrable).
     
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