Evaluating integrals or not?

1. Dec 6, 2007

pivoxa15

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??

Last edited: Dec 6, 2007
2. Dec 6, 2007

Hurkyl

Staff Emeritus
It's hard to answer your question without knowing just what you mean by "evaluated", "able", and "why". :tongue:

3. Dec 6, 2007

HallsofIvy

Staff Emeritus
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

4. Dec 6, 2007

pivoxa15

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. Dec 6, 2007

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?

6. Dec 6, 2007

morphism

$$\int_0^1 \sqrt{1 + \frac{1}{x^4}} \, dx = \int_0^1 \frac{\sqrt{x^4 + 1}}{x^2} \, dx \geq \int_0^1 \frac{1}{x^2} = \infty$$

7. Dec 6, 2007

HallsofIvy

Staff Emeritus
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.

8. Dec 6, 2007

Hurkyl

Staff Emeritus
What precisely do you mean by "compute"?

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 n-th digit in its decimal representation", then it is computable.

Last edited: Dec 6, 2007