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Integration of ((e^x)/(x^2)) with respect to x

  1. Apr 2, 2008 #1

    Eus

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    1. The problem statement, all variables and given/known data

    [tex]
    \int{\frac{e^x}{x^2}dx}
    [/tex]

    2. Relevant equations

    • Integration by substitution
    • Integration by parts: [itex]\int{u\ dv}=uv\ -\ \int{v\ du}[/itex]

    3. The attempt at a solution

    Since it was clear that integration by substitution would not work, I tried integration by parts. Since the [itex]e^x[/itex] term would not be affected whatsoever with the application of differentiation or integration, I worked out the [itex]x^2[/itex] term instead. So, I took [itex]u=e^x[/itex] and [itex]dv=\frac{1}{x^2}\ dx[/itex]. It resulted in:

    [tex]
    \int{\frac{e^x}{x^2}dx}=-\frac{e^x}{x}+\int {\frac{e^x}{x}dx}
    [/tex]

    For the last term, I took [itex]u=e^x[/itex] and [itex]dv=\frac{1}{x}\ dx[/itex]. It resulted in:

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

    Beyond this point, if I took [itex]u=ln(x)[/itex] and [itex]dv=e^x\ dx[/itex], I would just undo the previous steps. If I took [itex]u=e^x[/itex] and [itex]dv=ln(x)\ dx[/itex], it resulted in:

    [tex]
    \int{\frac{e^x}{x^2}dx}=-\frac{e^x}{x}\ +\ e^x\ ln(x)\ -\ e^x\left ( x\ ln(x)\ -\ x \right )\ +\ e^x\ -\ x\ e^x\ +\ \int{x\ ln(x)\ e^x\ dx}}
    [/tex]

    The last term certainly shows that this technique won't solve the problem at hand because it will continue forever.

    How should I attack this problem?

    Thank you.

    Best regards,
    Eus
     
  2. jcsd
  3. Apr 2, 2008 #2

    Hootenanny

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    Although this may seem crazy, your right here with your substitution. You need to do integration by parts twice such that you arrive at something of the form,

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

    Then you may write,

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

    [tex]\int{\frac{e^x}{x^2}dx}=\frac{1}{2}f(x)[/tex]
     
    Last edited: Apr 2, 2008
  4. Apr 2, 2008 #3

    Eus

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    I have tried it and it really did not work; it really undid the previous steps resulting in [itex]0 = 0[/itex].

    Best regards,
    Eus
     
  5. Apr 2, 2008 #4

    Hootenanny

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    In which case, I believe that the integrand has no elementary anti-derivative. Do you have some reason do believe an elementary anti-derivative exists? One can of course find a series representation of the integral.
     
  6. Apr 2, 2008 #5
    You're right there is no elementary derivative. The answer is:

    [tex]\int \frac {e^x}{x^2}\;\rightarrow\; \frac {-e^x}{x}-Ei\;(1,-x)+c[/itex]

    Where [itex]Ei[/itex] is the exponential integral.
     
    Last edited: Apr 2, 2008
  7. Apr 2, 2008 #6

    Hootenanny

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    Cheers SD :smile:
     
  8. Apr 3, 2008 #7

    Eus

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    I have taken a look at the page on Wikipedia but it does not explain anything how a series of non-terminating integration by parts can be transformed to an exponential integral.

    Could you please tell me how to do the transformation? Or, is there any pointer to a webpage that shows how to do it?

    Thank you very much.

    Best regards,
    Eus
     
  9. Apr 3, 2008 #8

    Hootenanny

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    We simply define the exponential integral as

    [tex]Ei(x):=-\int^{\infty}_{-x}\frac{e^{-t}}{t}dt[/tex]

    There is no method to go from your integral to the exponential integral. This is what we mean when we say an integrand has no elementary anti-derivative, it means the integral cannot be evaluated using elementary functions, we cannot evaluate the integral and write it in terms of normal functions. Instead we simply say that if one integrates this, then this is what one will obtain.

    Does that make sense?
     
  10. Apr 4, 2008 #9

    Eus

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    Well, if I just tried to find out the answer, that fact would be enough for me. But, I want to know how in the past someone would be able to obtain that answer. Suppose a mathematician in the past encountered the same class of problem. Of course, the mathematician would have tried the integration by substitution and the integration by parts to no avail. So, the mathematician employed another method and found the answer to be something involving exponential integral. I want to know this "another method" and how it yields the exponential integral.

    Would you please tell me this "another method" and how it yields the exponential integral? Any pointer will also help.

    Thank you very much.

    Best regards,
    Eus
     
  11. Apr 4, 2008 #10

    Hootenanny

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    With all due respect Eus, did you read my previous post? There is no such method.
    The mathematician would have probably tried many different methods to evaluate the integral, but would have failed. The simple fact of the matter is that we do not know how to integrate such an integrand. So your mathematician would have said "I don't know how to integrate, instead I'll just say that if you do integrate it you get the exponential function".
     
  12. Apr 4, 2008 #11
    You just have to accept that some integrals are not solvable with any current method. A good example is

    [tex]x^{x^2}[/tex]

    Although we can see why this is generally insoluble as it becomes quickly very large, and does not converge [tex]\lim_{x\rightarrow\infty}[/tex].

    I suspect mathematicians would have found similar reasons why this integral is hard to pin down with elementary functions, and then given up and gone and had a cup of tea/coffee.

    [itex]
    \int \frac {e^x}{x^2}\;\rightarrow\; \frac {-e^x}{x}-Ei\;(1,-x)+c
    \equiv[/itex] I don't know.
     
    Last edited: Apr 4, 2008
  13. Apr 4, 2008 #12

    Gib Z

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    I'm afraid the reason is deeper than that we can not do it with any current method or that we just don't know; it's because its not possible! It can be proven, using something like the Risch algorithm, that certain integrals can NOT be expressed in terms of a finite combination of elementary functions.

    It's not that we can't find the integral of exp(-t) 1/t, but rather that no such function in terms of elementary functions exists! So instead we define it as a non-elementary function, which is perfectly fine.
     
  14. Apr 6, 2008 #13

    Eus

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    Yes, I read your previous post. I just want to ensure myself that there is really no "another method" out there. Thank you for telling me about what happened to the mathematician :smile:

    Thank you for all of your help.

    Best regards,
    Eus
     
  15. Apr 6, 2008 #14
    I don't believe it's impossible anyway, surely if we invent 8 extra dimensions it's soluble. :wink:

    For future reference though, no elementary solutions, means it's insoluble except in terms of a non explicit mathematical term which means, either "?" Or undefined or impossible. Except in the case of imaginary solutions, which I'm not sure whether they fall under elementary solutions or not, but are solutions.
     
  16. Apr 6, 2008 #15

    Gib Z

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    I'm not sure what this means, but have anyone even read my post? It can be proven that certain integrals can no have elementary anti derivatives. It's not a matter of it being really hard to find, its a matter if possibility. If it can be found but you just don't know how, that doesn't mean it has no elementary solutions, though it may not, it just means that you can't prove otherwise.
     
  17. Apr 6, 2008 #16
    That's exactly what I just said. Obviously I wasn't clear, and the 8 extra dimensions is a swipe at my favourite punching bag, string "theory". I'm in a silly mood, it's not your fault. :smile:

    Broadly speaking without redefining maths, some integrals are in fact impossible. But most I suppose are ? or undefined.
     
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