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Math Help - (Integral of f(x) dx) * (Integral of 1/f(x) dx) = -1

  1. #1
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    (Integral of f(x) dx) * (Integral of 1/f(x) dx) = -1

    (Integral of f(x) dx) * (Integral of 1/f(x) dx) = -1
    How can I find all equations f(x) that satisfy this equation? Its obvious the f(x)=e**x works, but how can I prove this? I tried differentiating but that doen't seem to be helping. Thanks for help
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  2. #2
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    Re: (Integral of f(x) dx) * (Integral of 1/f(x) dx) = -1

    The wording of the problem is ambiguous because we don't know if the integrals are definite or undefinite integrals.
    If the integrals were definite, the bounds should be specified and the problem may have a solution, or no solution, depending on the given bounds.
    If the integral were undefinite, there is no solution. For example f(x)=exp(x) doesn't works :
    Integral of exp(x)*dx = exp(x)+C1
    Integral of (1/exp(x))dx = -exp(-x)+C2
    the product = (exp(x)+C1)*(-exp(-x)+C2) is not equal to -1.
    Can you clarify the wording of the problem about the bounds of the integrals ?
    Thanks from topsquark
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    Re: (Integral of f(x) dx) * (Integral of 1/f(x) dx) = -1

    I gave all the information I am given. That part was kinda confusing me too. So I guess I would take the limits of integration to be the same for each integral.
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    Re: (Integral of f(x) dx) * (Integral of 1/f(x) dx) = -1

    Also if they are indefinite integrals, the constants should be the same since they are from the same function so they would cancel out for e^x and it would work. So I'm fairly certain now the integrals are indefinite.
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    Re: (Integral of f(x) dx) * (Integral of 1/f(x) dx) = -1

    Quote Originally Posted by Shanter View Post
    Also if they are indefinite integrals, the constants should be the same since they are from the same function so they would cancel out for e^x and it would work. So I'm fairly certain now the integrals are indefinite.
    I don't agree.
    For example, with f(x)=exp(x) :
    Integral of exp(x)*dx = exp(x)+C1
    Integral of (1/exp(x))dx = -exp(-x)+C2
    Suppose C1=C2=C
    the product = (exp(x)+C)*(-exp(-x)+C) is not equal to -1 except if C=0.
    But if C=0, the integrals are no longer undefinite integrals because you have to define some particular values of the bounds so that C could be computed and equal to 0. In that case, the integrals are definite integrals with one bound =x and with the other bound = a particular value to be defined (not necessarily the same for both integrals).
    If you raise differently the problem (in taking account of the preceeding remark), I could show you how to solve it.
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    Re: (Integral of f(x) dx) * (Integral of 1/f(x) dx) = -1

    And how would you solve it, making your necessary assumptions to the problem?
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  7. #7
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    Re: (Integral of f(x) dx) * (Integral of 1/f(x) dx) = -1

    See the attachments.
    The integrals are definite integrals.
    The bounds must not be the same : The only case where they are solutions is if one of the bounds of an integral is -infinity and one of the bounds of the other integral is +infinity.
    On these conditions, the solutions are y(x)=c*exp(x) or y(x)=c*exp(-x) with c=constant different from 0.
    It is proved that no other solution exists.
    Attached Thumbnails Attached Thumbnails (Integral of f(x) dx) * (Integral of 1/f(x) dx) = -1-first-part.jpg   (Integral of f(x) dx) * (Integral of 1/f(x) dx) = -1-second-part.jpg  
    Thanks from hollywood and topsquark
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  8. #8
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    Re: (Integral of f(x) dx) * (Integral of 1/f(x) dx) = -1

    Looks good. Very interesting problem and solution.

    Thanks,
    Hollywood
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  9. #9
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    Re: (Integral of f(x) dx) * (Integral of 1/f(x) dx) = -1

    Can you solve this with differential equations???
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  10. #10
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    Re: (Integral of f(x) dx) * (Integral of 1/f(x) dx) = -1

    Of course, one can solve this with a differential equation. The differential equation given at end of the attachment to post #7 is very easy to solve. The solutions are explicitly written in post #7.
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