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Evaluating integrals or not? 
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#1
Dec607, 03:40 AM

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1. The problem statement, all variables and given/known data
Why are some integrals not able to be evaluated? i.e. the integral of (1+1/x^4)^(1/2) is impossible to evaluate exactly from 0 to 1. Why?? 


#2
Dec607, 03:54 AM

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PF Gold
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It's hard to answer your question without knowing just what you mean by "evaluated", "able", and "why".



#3
Dec607, 04:07 AM

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PF Gold
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If you mean "find an expression for the antiderivative in an elementary form", that is true for almost all integrable functions. the problem is that we simply don't know enough functions. By "elementary functions" we typically mean rational functions, radicals, exponentials and logarithms, and trig functions. That is just a tiny part of all possible functions, even all possible analytic functions.
In a deeper sense, the problem is that while we have "formula" for the derivative, there is no "formula" for the antiderivative; it is simply defined as the "inverse" of the derivative. And "inverses" are typically very difficult. If we define [itex]y= x^5 3x^3+ 4x^3 5x^2+ x 7[/itex], the direct problem, to "evaluate" the function (Given x, what is y?) is relatively simple. The "inverse" problem, to "solve the equation" (Givey y, what is x) is much harder 


#4
Dec607, 06:06 AM

P: 2,268

Evaluating integrals or not?
evaluate by integrating from 0 to 1. which I have added in the OP. able as you computing it exactly. why as in why is it not able to be computed exactly. 


#5
Dec607, 06:08 AM

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#6
Dec607, 06:14 AM

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Well, your example diverges.
[tex]\int_0^1 \sqrt{1 + \frac{1}{x^4}} \, dx = \int_0^1 \frac{\sqrt{x^4 + 1}}{x^2} \, dx \geq \int_0^1 \frac{1}{x^2} = \infty[/tex] 


#7
Dec607, 07:58 AM

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PF Gold
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Of course, not all functions of interest are analytic i.e. have a Taylors series that converges TO the function on some interval around the central point. 


#8
Dec607, 03:45 PM

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PF Gold
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If you mean "write in decimal notation with finitely many digits", then isn't the answer obvious? If you mean "write down an algorithm that, given an integer n, outputs the nth digit in its decimal representation", then it is computable. 


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