# Evaluating integrals or not?

by pivoxa15
Tags: evaluating, integrals
 P: 2,267 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??
 Emeritus Sci Advisor PF Gold P: 16,091 It's hard to answer your question without knowing just what you mean by "evaluated", "able", and "why".
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,683 If you mean "find an expression for the anti-derivative 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 anti-derivative; it is simply defined as the "inverse" of the derivative. And "inverses" are typically very difficult. If we define $y= x^5- 3x^3+ 4x^3- 5x^2+ x- 7$, 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
P: 2,267
Evaluating integrals or not?

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

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.
P: 2,267
 Quote by HallsofIvy If you mean "find an expression for the anti-derivative 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 anti-derivative; it is simply defined as the "inverse" of the derivative. And "inverses" are typically very difficult. If we define $y= x^5- 3x^3+ 4x^3- 5x^2+ x- 7$, 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
So we will have an infinite series as an antiderivative for the function? If so then we can compute the series? It might diverge?
 Sci Advisor HW Helper P: 2,020 Well, your example diverges. $$\int_0^1 \sqrt{1 + \frac{1}{x^4}} \, dx = \int_0^1 \frac{\sqrt{x^4 + 1}}{x^2} \, dx \geq \int_0^1 \frac{1}{x^2} = \infty$$
Math
Emeritus
Thanks
PF Gold
P: 39,683
 Quote by pivoxa15 So we will have an infinite series as an antiderivative for the function? If so then we can compute the series? It might diverge?
Given an analytic function to be integrated, we certainly can calculate its Taylor's series and integrate that term by term to get a power series for the anti-derivative. If I remember correctly, it should converge on the same radius of convergence as the original function.

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.
Emeritus