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Integral of e^-x^2?

  1. Dec 6, 2009 #1

    Char. Limit

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    I've been told that there is no equation for the indefinite integral of e^-x^2. But how is that possible? Doesn't every equation have a indefinite integral?


    Also, what would happen if someone tried to solve it?
     
  2. jcsd
  3. Dec 6, 2009 #2
  4. Dec 6, 2009 #3

    djeitnstine

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    This is wrong. There are more functions that have no analytical anti derivative than there are that do.
     
  5. Dec 6, 2009 #4

    Char. Limit

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    Is there no antiderivative for e^u, though? Then say that u is -x^2 (integration by substitution). Where would that fail?
     
  6. Dec 6, 2009 #5

    Mute

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    You forgot the extra factor due to the change of variables.

    [tex]\int dx e^{-x^2} = \mp \int du \frac{e^u}{\sqrt{u}}[/tex]

    This integral is still not elementary.
     
  7. Dec 6, 2009 #6

    Redbelly98

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    u = -x^2
    du = -2x dx

    If you substitute into ∫eudu in this way, you get an extra factor of x due to the du=-2x·dx that is required when you make this substitution.

    So, this helps with integrating ∫e-x2x·dx, but not with ∫e-x2dx
     
  8. Dec 6, 2009 #7

    Char. Limit

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    I see now. Integration by substitution is still new to me...
     
  9. Dec 6, 2009 #8

    dextercioby

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    Every possible combination of <elementary> functions, seen as a function itself, leads, through differentiation, to another combination of <elementary> functions. However, the antidifferentiation rarely gives such a combination.

    This asymmetry is one of the oddities of mathematics. That's what i think...

    A simple example to the 2 assertions in the first paragraph :

    [tex] f(x) =\sqrt{\sin x} [/tex]
     
  10. Dec 6, 2009 #9

    Char. Limit

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    Strange indeed.

    I can't work that function's integral out...

    Let me see if Wolfram-α has an answer.

    EDIT: It doesn't.
     
  11. Dec 6, 2009 #10

    Gib Z

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    We'll it's not completely wrong. He didn't state every function must have an *elementary* indefinite integral. In the form he stated, it's actually a theorem! For every continuous function f(x), there exists a function [tex]\int^x_a f(t) dt[/tex] which is also continuous.

    To the OP: If someone tried to solve it, they would never get anywhere, or perhaps they may be able to prove there is no solution at all, such as via the Risch Algorithm. The important thing to understand is that the set of functions which we choose to call "elementary" and happen to come up in our study often is COMPLETELY arbitrary. That's how they were chosen : Nice properties, easy to manipulate, came up in our study often. Does that make any of the other "non-elementary" functions somehow less of a useful function?

    It turns out that many functions that have no "elementary" anti derivatives come up in mathematics and physics quite a lot, such as the Gamma Function, the Error function and the Exponential Integral. Nothing has stopped mathematicians and physicists from just *defining* these integrals as some new function with some name they gave it, and studying its properties from there.

    A simpler example would be to consider the scenario where you know nothing about the natural logarithm function, it's not elementary to you and you haven't come across it. Then you encounter the integral [tex]\int^x_1 1/t dt[/tex]. Should there be anything stopping you from investigating it, even though it has no "elementary" anti derivative? We can still work out this integrals properties. We can show that it satisfies F(xy) = F(x) + F(y), that F(x^n) = nF(x), etc etc.

    Sorry to ramble, but I'm just emphasizing, having no elementary anti derivative is really no limitation to studying a function. It's a pity many people seem to think it is.
     
    Last edited: Dec 6, 2009
  12. Dec 6, 2009 #11

    Char. Limit

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    Another question relating to the title topic:

    When looking at the definite integral of e^-x^2... where does the "2/sqrt(π)" come from? π shouldn't be involved at all, much less as an inverted square root, I would think...
     
  13. Dec 6, 2009 #12

    nrqed

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    This is such an important point that it should be emphasized in all introcutory calculus courses!! Unfortunately, it is usually not mentioned at all. Some students come out of the classes thinking that integrals that cannot be done in terms of elementary functions are pathological and of little use! Or even worse, some students don't even realize that some antiderivatives cannot be written in terms of elementary functions. It's not the students' fault, it's the teacher's.
     
  14. Dec 6, 2009 #13

    Char. Limit

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    Well, having an elementary integral does help to consider the equation...

    Also, technically speaking, I'm not done with calculus yet. I just learned integration by substitution and haven't even heard of, for example, Taylor polynomials. To me, elementary integrals are much better than non-elementary integrals where 2/sqrt(pi) pops up for no reason...
     
  15. Dec 6, 2009 #14

    jtbell

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    And in particular the (Gaussian) error function mentioned above is just the very integral we're discussing in this thread! :biggrin: (apart from a constant in front)

    Wikipedia has a graph, a table of values, and some of its mathematical properties.
     
  16. Dec 6, 2009 #15

    Char. Limit

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    Hmm... the graph of the integral suggests... that the integral from 0 to infinity of e^-x^2 is 1? but where does 2/sqrt(pi) come in? Seriously, pi has only a passing relation to e... because I can define pi in terms of a logarithm of base e. But why is it there, of all places?
     
  17. Dec 6, 2009 #16
    It's a shocking thing, really, but it's a matter of how you do it.

    [tex]
    I = \int e^{-x^2} dx
    [/tex]
    [tex]
    I^2 = ( \int e^{-x^2} dx)^2 =
    [/tex]

    Now, we'll do something crazy for no apparent reason. We'll rotate the function [tex] y = e^{-x^2} [/tex] around the y-axis and find the volume under that curve.

    By Calc I, you should have learned the Shell method, which would mean,

    [tex]

    2\pi \int xf(x) dx = \int x e^{-x^2} dx
    [/tex]

    Which you can evaluate easily. BUT! The volume can also be calculated by a double integral,

    i.e. [tex] \int \int e^{-x^2 -y^2} dx dy = \int e^{-x^2} dx \int e^{-y^2} dy = ( \int e^{-x^2} dx )^2 = I^2. [/tex]

    So...

    [tex]
    I = \sqrt{2 \pi \int xe^{-x^2} dx } [/tex]
     
  18. Dec 6, 2009 #17

    Char. Limit

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    Ok... so why do we use 2/sqrt(pi) and not sqrt(2*pi)?

    Shell method is next chapter... but looking at it, it looks understandable. But what does the volume tell us about the integral?
     
  19. Dec 7, 2009 #18
    We do. The function you are looking for is the erf(x), which, must be =1 when taking from 0 to infinity.

    Once you evaluate the integral, you'll end up with sqrt(pi)/2, which cancels the 2/sqrt(pi) and makes it 1.

    Don't forget that there is still that integral which must be evaluated, from -infinity to infinity (which yield 1/2) or from 0 to infinity, which yields half the answer, hence sqrt(pi)/2.
     
  20. Dec 7, 2009 #19

    Char. Limit

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    Right... the integral... that I didn't even consider.

    What does the volume around the y-axis show about the function, though? That part I still don't understand.
     
  21. Dec 7, 2009 #20

    D H

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    I concur with nrqed: This is a very important point. I don't know how many people I have had to convince that the orbits of the planets in our solar system *is* a solvable problem. Some teacher (or book) told them that the n-body problem is unsolvable in the elementary functions.

    That small print is very important. The n-body problem is solvable, as is [itex]\int_0^x \exp(-t^2)\,dt[/tex] (to get back to the original topic).

    Gib mentioned "special functions". The antiderivative of e-x2 is one of those special functions. It is called the error function.

    [tex]\mathrm{erf}(x) = \frac 2{\surd \pi} \int_0^x \exp(-t^2)\,dt[/tex]
     
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