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Are there tricks one can use while looking at integral expressions to determine if one is easily calculatable or not? For example: $$ A = \int \frac{dx}{\ln(1+x^2)} \quad B = \int \frac{dx}{x\ln(x^2)} $$ Wolfram finds an answer for $B$, but not one for $A$, even though $A$ looks easier. I would like to know why and also how/if one can look at an integral and see if it is a solvable one or not. With "solvable", I mean for a university student and not using non-real numbers or very advanced maths. :) I know most integrals can't be solved analytically but I'm thinking of those that can appear on exams.

Stand with Gaza
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  • see the Rish algorithm https://en.wikipedia.org/wiki/Risch_algorithm – Dr. Sonnhard Graubner Nov 11 '17 at 21:06
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    This question has been settled starting from the work of Liouville, in the XIXth Century (https://en.wikipedia.org/wiki/Liouville%27s_theorem_(differential_algebra)) and several others. The modern computer algebra systems integrate algorithms that answer the question in most of the cases. But the procedure is extremely complex, not doable by hand. As a rule of thumbs, all rational fractions (ratio of polynomials) integrate with a closed form, and integrands involving transcendental functions don't. –  Nov 11 '17 at 21:06
  • Related: https://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral – Hans Lundmark Nov 12 '17 at 08:59

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Integration is "difficult". There isn't one systematic approach that will work by hand for most integrals. When it comes to integrating in exams, your best bet is to just practice integrating lots of different things until you start to recognise patterns.

Many integrable functions are of the form:

$$\int f'(x)g'(f(x))\mathrm{d}x = g(f(x))+c$$

Your second example can be seen to be of this form:

$$\frac{1}{\ln(x^2)} = \frac{1}{2\ln(x)}$$

$$\frac{1}{2x\ln(x)} = \frac{1}{2}\frac{1}{x}\frac{1}{\ln(x)} = \frac{1}{2}\ln'(x)\ln'(\ln(x))$$

$$\int \frac{1}{x\ln(x^2)} = \int \frac{1}{2}\ln'(x)\ln'(\ln(x)) = \frac{1}{2}\ln(\ln(x))+c$$

The key is to be able to notice these patterns, and I think the only way you can do that is through practice.

As for integrability "in theory", this is entirely different to whether or not you can integrate by hand in an exam, and the comments below your question are quite good :)

Harambe
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  • Thank you. My second example I was also able to calculate. But I would also like to know how a mathematician can look at example A and see that it isn't "calculatable by hand." – Stand with Gaza Nov 11 '17 at 21:27
  • When integrating by hand, you have a "set of possible methods", and if none of them work you can't do it. It's a bit like a multichoice exam where you don't know what's correct so you go through and cross out wrong answers - but you have to keep going until you cross out all of them. As far as I know there isn't a simple test to check that you can't integrate something. What I usually do is what I wrote in the answer - if I can't reverse-chain-rule/integration-by-parts it, I usually just write a numerical integrator! – Harambe Nov 11 '17 at 21:37
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Risch’s algorithm will help as it turns a problem involving an integral into an algebraic problem.

It is based on the form of the function being integrated and on methods for integrating rational functions, radicals, trigonometrical, logarithmic, exponential and “special” functions, e.g. Riemann’s Zeta function.

Risch himself called it a decision procedure, because it is a method for deciding whether a function has an elementary function as an indefinite integral, and if it does, for determining that indefinite integral.

BE WARNED: The complete description of the Risch algorithm takes over 100 pages.

Integration is much more difficult than differentiation. Most fuctions, if you were to randomly combine them, cannot be integrated in terms of elementary functions e.g. f(x) = x^x is not integrable, however Risch’s algorithm gives its value at the point x=0, simplied(!!) by leaving out an infinite number of very small terms.

Most people will be surprised at it’s apparent complexity, but it’s value at x=0 is shown here:

https://i.stack.imgur.com/0DFNi.gif

Thank the good Lord for computer algorithm apps.

Kim Hutcheson
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