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Difficult improper integral!

  1. Apr 2, 2008 #1
    Could anyone please explain how to solve the improper integral below?
    I have no idea of how to do it.

    If f is a bounded non-negative function, then show that

    the integral from zero to infinity of f(x+1/x)*ln(x)/x dx=0.

    Thank you.
  2. jcsd
  3. Apr 2, 2008 #2
    i cant understand what the means
    f(x+1/x)*ln(x)/x dx=0

    what is f(x+1/x)??
    where is your whole function
    and the whole point of the integral is to find the area etc..
    here you have the answer
    unless you have some where a parameter i cant see what is the function

    any way
    in order to solve a omproper integral you need to put some variable instead of infinity
    and to solve a limit question for this whole integral where your variable goes to infinity
  4. Apr 2, 2008 #3
    Use a substitution and integration by parts:

    [tex]u=lnx , du=\frac{dx}{x}[/tex]

    applying integration by parts:

    [tex]\int_0^\infty f(x+\frac{1}{x})u du = \frac{u^2}{2}f(x+\frac{1}{x})\big]_0^\infty -\int_0^\infty \frac{u^2}{2} \frac{d}{dx}(f(x+\frac{1}{x}))dx[/tex]

    [tex]=\frac{u^2}{2}f(x+\frac{1}{x})\big]_0^\infty -\frac{u^2}{2}\int_0^\infty \frac{d}{dx}(f(x+\frac{1}{x}))dx[/tex]


    [tex]=\frac{u^2}{2}(0)\big]_0^\infty = 0[/tex]

    since f is bounded
  5. Apr 2, 2008 #4
    how can you ever integrate a function without knowing it form? what if f (x+1/x) = x; it becomes ln(x) dx which is divergent as x-> inf. unless you know the alzebraic form you can;t even tell if the function is even defined in the interval, let aside integrating it. and the by parts integration has a mistake dear. check out the formula. if your calculation is correct then you have proven that the inntegration is independent of f() :) which is not to be true
  6. Apr 3, 2008 #5


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    Welcome to PF!

    Hi davedave! Welcome to PF! :smile:

    Draw a diagram … it helps you see what's going on! :smile:

    You'll notice that f is negative for some time, then positive.

    So somehow, the positive has to balance out the negative, and this has to work for any positive bounded function f.

    Hint: what happens if you substitute 1/x for x? :smile:
    No … :frown: … it should be [tex]\frac{u^2}{2}\int_0^\infty \frac{d}{du}(f(x+\frac{1}{x}))dx\,.[/tex]
  7. Apr 3, 2008 #6
    Oh right, my bad, I was quite tired. I agree though, it's a strange question
  8. Apr 4, 2008 #7
    This problem doesn't seem well defined to me.
    Let for example f(t) be constant 1 for every t.
    Then just try to integrate from 0 to any epsilon (or from any A to infinity) using integration by part and it seems to me that you will end-up with an infinite result.
  9. Apr 5, 2008 #8


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    The function f isn't entirely arbitrary, though. You know that it is a nonnegative function that is bounded. This rules out any singularities that you can't integrate over. It doesn't rule out discontinuities, but that doesn't really present a problem.

    Using the following substitution.

    [tex]u=\ln x , du=\frac{dx}{x},[/tex]

    [tex]\int_0^\infty f\left(x+\frac{1}{x}\right)\frac{\ln x}{x} dx = \int_{-\infty}^\infty f\left(e^u+e^{-u} \right)u du = \int_{-\infty}^\infty f\left(2\cosh u \right)u du [/tex]

    The change of variables brought the integration region to be the entire real line. Now, cosh u is an even function: as u -> -u, it remains the same. u, on the other hand, is an odd function. So, for the original poster, what do you know about even and odd functions that lets you evaluate this integral?

    If f(t + 1/t) = 1, then the integrand is just ln x/x. u = ln x, du = dx/x, you get the integral from minus infinity to infity of u - does this really diverge? There might appear to be a problem if you integrate by parts, but what might happen doing so is you get an indeterminate form [itex]\infty - \infty[/itex], which doesn't mean the result is not actually a well defined number! If you were to integrate that by parts, you get

    [tex]\int_0^\infty dx \frac{\ln x}{x} = \left.(\ln x)^2\right|^\infty_0 - \int_0^\infty dx \frac{\ln x}{x}[/tex].


    [tex]\int_0^\infty dx \frac{\ln x}{x} = \frac{1}{2}\left.(\ln x)^2\right|^\infty_0 = \infty - \infty[/tex]. So, you need to figure out what that indetermine form is. By the solution of the OP's problem, it will be zero.
    Last edited: Apr 5, 2008
  10. Apr 5, 2008 #9


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    Welcome to PF!

    Hi Gerard! Welcome to PF! :smile:

    Yes … and you don't even need integration by parts, since ∫ln(x)/x = ((ln(x))^2)/2, which as you say doesn't converge at 0 or ∞. :smile:

    Similarly, if f(x) = (ln(x))^a, with a ≥ -1, then as x -> ∞, ∫f(x+1/x)*ln(x)/x ~ ∫((ln(x))^(a + 1))/x = ((ln(x))^(a+2))/(a+2).

    I suspect the criterion is f(x)ln(x) -> 0 as x -> ∞.

    … anyone like to try proving or disproving that … ? :smile:

    Great start, Gerard! :biggrin:
  11. Apr 5, 2008 #10

    It seems perfectly well-defined to me. It's just wrong, as your counterexample f(t)=1 shows.:smile:
  12. Apr 5, 2008 #11


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    It's not wrong, and his counterexample isn't a counterexample. It yields an indeterminate form, as I demonstrated above in post #8. The general proof I gave in that post demands the result will be zero, since after the change of variables you are integrating an odd function over a symmetric interval. For the f(t+1/t) = 1 case, instead of the limits being zero and infinity, let them be 1/a and a, and take the limit as a -> infinity:

    [tex]\lim_{a\rightarrow\infty}\int_{1/a}^{a}dx~\frac{\ln x}{x} = \lim_{a\rightarrow\infty}\left.\frac{1}{2}(\ln x)^2\right|^{a}_{1/a}[/tex]

    [tex] = \lim_{a\rightarrow\infty} \frac{1}{2}\left[(\ln a)^2 - (\ln\frac{1}{a})^2\right] = \lim_{a\rightarrow\infty} \frac{1}{2}\left[(\ln a)^2 - (\ln a)^2\right] = 0[/tex]

    This is the Cauchy Principle Value of the integral.
    Last edited: Apr 5, 2008
  13. Apr 5, 2008 #12
    in the orginal statement of the problem there is no mention of Cauchy PV's.

    You integrate from 1/a to a which gives zero. If you had chosen 1/a and 2a instead, the result would have been different (in particular not zero). Do you have to adjust the integration bounds for each choice of f seperately or does 1/a and a always work?:smile:
    Last edited: Apr 5, 2008
  14. Apr 5, 2008 #13


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    … Cauchy Principal Value …

    Hi Mute! :smile:

    But is the Cauchy Principal Value consistent with this particular question?

    The question requires f to be bounded … is that necessary if Cauchy Principal Values are permitted?
  15. Apr 5, 2008 #14


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    I am admittedly not exceedingly familiar with CPVs, but I think the integration bounds have to reach their limiting value at the same rate in order for it to be the CPV. For my example 1/a reaches zero at the same rate a reaches infinity, but your 2a reaches infinity faster than 1/a reaches zero, so in your example you would not get the CPV.

    Perhaps I shouldn't have formulated the explicit integral for the case of f(t) = 1 as a CPV, but if you split the integral up from 0 to 1 and 1 to infinity (the first integral being -infinity, the second being +infinity) and you make the change of variables -u = lnx in the (0,1] integral and u = ln x in the [1,infinity] integral, then what you end up with is

    [tex]-\int_{0}^{\infty} du~u + \int_{o}^{\infty}du~u = 0[/tex].

    At least symbolically we would feel comfortable saying this is zero. Maybe there's some sneaky trick one might perform to get it to not be zero, but that would probably amount to doing a trick like letting the infinity in the first integral be 2a, but a in the second integral, and then taking the limit as a goes to infinity, would would not result in the CPV.

    At any rate, the integral I wrote down for the general case in post #8 is undoubtedly in the form of a CPV integral - you have the integrand evaluating to -infinity for (-infinity,0] and plus infinity for [0,infinity].

    Surely we all agree for an even function f(x)

    [tex]\int_{-\infty}^{\infty}dx~xf(x) = 0[/tex]

    if f(x) is bounded and nonnegative? f being bounded ensures it has no singularities where integration would fail, and being nonnegative ensures that it's not an oscillatory function like sine or cosine which would fail to converge to some definite value. Perhaps there is another condition on f someone thinks is necessary for this to be true?

    Once you make the change of variables [itex]u = \ln x[/itex] it sure seems to fall into the form of a CPV integral. Does someone see something wrong with this change of variables that I've overlooked?
    Last edited: Apr 5, 2008
  16. Apr 6, 2008 #15


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    … CPV can apply at any singularity …

    I don't agree …

    First, CPV is not the only possible definition.

    And second, f does not need to be bounded, provided any singularities are also "balanced" enough for a CPV to exist there also (CPV can apply at any singularity, not just ∞).
  17. Apr 6, 2008 #16


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    The fact that f need not be bounded for the problem isn't such a big issue - we're given that it is, so we wouldn't need to consider the possibility that if any singularities existed they would need to be balanced singularities such that a CPV existed for them too.

    CPV may not be the only possible definition, but which other definition would you prefer to use? Is it your expectation that the proof is to be done such that any method of the evaluation of the original integral is to result in zero? Do we at least agree that the CPV value of the original integral is zero?
  18. Apr 6, 2008 #17
    If we integrate from any a != 0 to 1/a we get result zero (we may want a to me small).
    But, if we integrate from the same a to 2/a for example we don't get result zero.
    My guess is that the original problem was an integral from a to 1/a and someone
    changed it by another problem that doesn't make any sense.

    To check that integral from a to 1/a is zero, do the following:
    1) Break the integral into 2 terms, respectively integral from a to 1 and from 1 to 1/a
    2) In (only) one of the 2 integrals subsitute x=1/u and simplify it (don't forget to also
    change the bounds of the integral).

    You will end up with 2 bounded terms (given the hypothesis on f) that compensate exactly, thus result zero.

    Sorry for not using Latex yet. I still need some training prior to being confortable with
    its syntax.

    (BTW, if the above is not enough for somebody to post the solution from a to 1/a in Latex, I will try to do it)
  19. Apr 6, 2008 #18
    Mute has been promoting this solution in posts #8, #11 and #14. You might want to read those.
    Last edited: Apr 6, 2008
  20. Apr 6, 2008 #19


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    I've been convinced that I should scale back some of the claims I've made and concede that the solution I proposed is only the Cauchy Principle Value of the integral, and if the statement of the OP's problem is to be proved generally, for all possible interpretations of the integral, then I've not done that. Originally, though not stated explicitly, I had also made the assumption that an odd function, at least one that is of only one sign on either side of it's point of symmetry, is always zero when integrated about any symmetric interval, even an infinite one, but of course there's no reason why that should be true for an infinite integration region, as the discussion of CPVs has revealed.

    So, I think that as far as the original question goes, if we consider the CPV as the value of the integral to be obtained, then what I did was a solution showing that value was zero. If the lack of mention of CPV in the original problem was not due to it being an implicit assumption of the problem but that the value of the integral should be zero any way you take it, then I didn't show that and I am also not convinced that is true, which seems to be the stance others have taken.
  21. Apr 6, 2008 #20
    I agree.
    Maybe we should wait for the OP to come back, or otherwise leave it at that.:smile:
  22. Apr 7, 2008 #21


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    … stealing LaTeX …

    Hi Gerard! :smile:

    Useful tip: click the "QUOTE" button, and steal the original code for someone else's LaTeX!

    For example, click on post #11 (or on this post … I've changed it slightly), which includes:
    and you see:

    [noparse][tex]\lim_{a\rightarrow\infty}\int_{1/a}^{a}dx~\frac{\ln x}{x} = \lim_{a\rightarrow\infty}\left[\frac{1}{2}(\ln x)^2\right]^{a}_{1/a}[/tex][/noparse] :smile:

    Also bookmark
    http://www.physics.udel.edu/~dubois/lshort2e/node61.html#SECTION008100000000000000000 [Broken]
    and maybe
    http://www.physics.udel.edu/~dubois/lshort2e/node54.html#SECTION00830000000000000000 [Broken]

    erm … anyone want to check whether the criterion is f(x)ln(x) -> 0 as x -> ∞? :biggrin:
    Last edited by a moderator: May 3, 2017
  23. Apr 8, 2008 #22

    Hi everyone,

    Thanks for all your great ideas for analyzing the problem I posted. Let me tell you where I got it from. If you search under Google for "challenging integrals", and click on INTEGRATION BEE, CHALLENGING INTEGRALS: VISHAL LAMA'S BLOG, you will see a list of integrals. The 3rd one which is this problem is the only one I cannot do. That is why I need you guys' help.

    I really like the idea of the Cauchy Principle Value for proving this integral by Mute.

    I just thought of a way of doing this integral using the quote above.
    But, I am not sure if this way would make any sense.

    From the integral in the quote above, I let u=tan(z) du=(sec(z))^2 dz

    I=integral from -infinity to +infinity of f(2*cosh(u))*u du

    =integral from -pi/2 to +pi/2 of f(2*cosh(tan(z)))*sin(z)/(cos(z))^3 dz

    Since f is bounded, let K be a constant > 0 such that
    the absolute value of f is < or = K for all z in the domain of f

    I< or = K*integral from -pi/2 to +pi/2 of sin(z)/(cos(z))^3 dz

    Now, I put (cot(z))^4 in the integrand to cancel (cos(z))^3
    With the extra term in the integrand, the integral will be bigger
    than or equal to the one without it.

    < or = K*integral from -pi/2 to +pi/2 of cos(z)/(sin(z))^3 dz

    = -K/2 * (1/(sin(z))^2]from -pi/2 to pi/2 = 0.

    Please comment on my approach.

    Once again, the Cauchy Principle Value is a really neat way to solve the problem.
    I will keep that solution. Thanks to Mute and others.
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