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Integral of sin(e^x)

  1. Jun 12, 2008 #1


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    The function is continuous so the integral exists, but how do you find it :)?
  2. jcsd
  3. Jun 12, 2008 #2
    It is highly unlikely that a closed form expression in terms of elementary functions exists, however, if you only wish to evaluate it as a definite integral then there are numerous methods of doing that.
  4. Jun 12, 2008 #3


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    [tex]\int sin(e^t) dt[/tex]

    Let [itex]x=e^t \Rightarrow \frac{dx}{dt}=e^t=x[/itex]

    [tex]\int sin(e^t) dt \equiv \int \frac{sinx}{x}dx[/tex]

    and that doesn't exist in terms of elementary functions
  5. Jun 12, 2008 #4

    Why can't you use the method of intergation by Parts for [tex]\int sin(e^x) dt[/tex]?

    We CAN think that this integral is in the form [tex]\int f(x) g(x) dx[/tex], right?

    I think it can be done using the formula of integration by parts, [tex]\int uv' = uv - \int v u'[/tex]. This might be done that way imo.
    Last edited: Jun 12, 2008
  6. Jun 12, 2008 #5

    D H

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    Go ahead and try. Be prepared to be frustrated, however because ...

    This latter integral is encountered in math and science quite frequently, so frequently that it has been given a name -- the sine integral. For more info, see

    http://planetmath.org/encyclopedia/SinusIntegralis.html" [Broken]
    http://mathworld.wolfram.com/SineIntegral.html" [Broken]
    http://en.wikipedia.org/wiki/Sine_integral" [Broken]
    Last edited by a moderator: May 3, 2017
  7. Jun 13, 2008 #6


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    Why would you think so? If it were exsin(x), f and g would be obvious but here the only functions "multiplied" together are 1 and sin(ex). If you take u= 1 and dv= sin(ex)dx, you are back to the original problem. If you take u= sin(ex) and dv= dx you get du= cos(ex)exdx and v= x so you have gone from [itex]\int udv= \int sin(e^x)dx[/itex] to [itex]uv- \int v du= xsin(e^x)- \int x cos(e^x)e^x dx[/itex] which doesn't look any easier to me.
  8. Jun 13, 2008 #7


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    Well if you keep on doing integration by parts you get an infinitely long result, don't think that's of any use :eek:
    It's just weird.. because my math teacher says that if the funciton is continuous the integral exists, but that must only be with a definite integral then.
  9. Jun 13, 2008 #8


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    I think we all agree the integral exists. Nobody here has said it doesn't.

    What was said was:
    The integral exists, but we can't express it in terms of elementary functions.
  10. Jun 13, 2008 #9

    D H

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    Just because some function cannot be expressed as a sum of a finite number of combinations of elementary functions does not mean the function does not exist. It just means that the function in question is not an elementary function, and that is all it means. The function can still be expressed as an infinite series, for example.
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