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How do I solve this integration problem?

  1. Dec 17, 2016 #1
    • Member advised that the homework template must be used
    ln(e^{Φ^2}+1)dΦ

    I am a high school math student, so my calculus knowledge is that of high school. I tried to solve this problem, but nothing I have learnt seemed to work so far, substitution didn't work, integration by parts didn't work. I presume this problem is beyond high school level calculus. I will be satisfied if anyone can solve this problem. And I apologise for the way I presented the problem, I don't know latex.
     
  2. jcsd
  3. Dec 17, 2016 #2
    Well, even wolfram fails to solve it analytically ( I suppose you already have tried it there). However a good approximation (if ##\phi## takes large enough values) is ##\phi^3/3## because for large ##\phi## we have ##ln(e^{\phi^2}+1)\approx \phi^2##.
     
  4. Dec 17, 2016 #3

    Jonathan Scott

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    I typed "integral of ln(e^(phi^2)+1) d phi" into Wolfram Alpha and it appears to interpret it as expected but it says "No result found in terms of standard mathematical functions". So not even Wolfram Alpha can solve that one.
     
  5. Dec 17, 2016 #4

    Ray Vickson

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    Maple cannot find a formula for the integral, either. Because Maple and Mathematica (Wolfram Alpha) both fail, I suspect strongly that this integral is one of the "non-elementary" integrals that cannot be expressed as a finite number of terms involving all the standard functions.

    There are numerous examples that are provably non-elementary; this means that people have proved that they are impossible in terms of standard functions. It is not just a matter of nobody so far having been smart enough to get the answer; rather, it is the fact that you can prove that a finite answer is absolutely impossible in terms of standard functions.

    Your function is not among the standard non-elementary examples, but the fact that powerful computer packages fail on it suggests very strongly that it is not do-able. These packages use powerful advanced methods based on some deep theorems due to Risch and others about possibility/impossibility. They go way beyond anything you will find in introductory calculus textbooks.

    Note, however, that various approximation methods exist, including series expansions plus many numerical methods, so getting a numerical answer for ##F(x) = \int_0^x \ln(1 + e^{\phi^2})\,d \phi## for numbers ##x > 0## is not difficult using modern methods.
     
    Last edited: Dec 17, 2016
  6. Dec 18, 2016 #5
    Depends on what you mean by "solve". Just because an integral cannot be expressed in terms of simple functions like polynomials or radicals or trig functions doesn't mean it can't be "solved". There are many special functions that represent such integrals. For example:
    ##\int \frac{\sin(x)}{x}dx=\text{Si}(x)##

    And the sine function Si(x) is just a function like ##\sin(x)##. Why then cannot we just say:

    ##\int \log(e^{x^2}-1)dx=\text{myfunction}(x)##

    Also, how about a (convergent) power series solution? Is that not a legitimate solution just like sine or cosine? For example:

    ##\log(e^{x^2}+1)=\log(2)+\frac{x^2}{2}+\frac{x^4}{8}+\frac{x^8}{192}+\frac{x^{12}}{2280}+\cdots##

    then can't we write:

    ## \int \log(e^{x^2}+1)dx=\int \big[ \log)(2)+\frac{x^2}{2}+\frac{x^4}{8}+\frac{x^8}{192}+\frac{x^{12}}{2280}+\cdots \big]dx##

    Can you integrate that power series solution and come out with a closed-form solution like:

    ##\int \log(e^{x^2}+1)dx=\log(2)x+\sum_{n=1}^{\infty} (\text{some expression here})##?
     
  7. Dec 18, 2016 #6

    Ray Vickson

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    Let
    $$F(x) = \int_0^x \ln \left( 1 + e^{\phi^2} \right) \, d\phi .$$
    The series solution proposed above will converge only for small values of ##|x|##. A much more effective series solution (involving non-elementary but well-studied and readily-available functions) is obtained by writing
    $$ \ln\left(1+e^{\phi^2}\right) = \ln\left( e^{\phi^2} \right) + \ln \left( 1 + e^{-\phi^2} \right) = \phi^2 + \ln \left( 1 + e^{-\phi^2} \right).$$
    Thus
    $$F(x) = \frac{x^3}{3} + \int_0^x \ln \left( 1 + e^{-\phi^2} \right) \, d\phi.$$
    Now expand the second logarithm, using ##\ln(1+u) = u - u^2/2 + u^3/3 - \cdots## with ##u = e^{-\phi^2} < 1## (so: convergent). Note that ##u^n = e^{-n \phi^2}##, so we can express the ##n##th term of the series for ##F(x)## as
    $$t_n(x) = \frac{(-1)^{n-1}}{n} \int_0^x e^{-n \phi^2} \, d\phi = \frac{(-1)^{n-1} \sqrt{\pi}}{n^{3/2}} \text{erf}( x \sqrt{n}) .$$
    Here, "erf" is the non-elementary function
    $$\text{erf}(w) = \frac{2}{\sqrt{\pi}} \int_0^w e^{-t^2} \, dt$$.
    The infinite series above converges for all ##x > 0.##
    In sophisticated computer packages the function "erf" is built-in, so numerical evaluation of the terms of ##F(x)## is not very difficult. (However, to be fair, I must say that for sizable values of ##x>0## we may need to include a large number of terms in order to obtain good accuracy.)
     
    Last edited: Dec 19, 2016
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