0
$\begingroup$

I know about elementary functions and their integrals, also some elementary function does not have elementary anti-derivative. For example $\frac{\sin(x)}{x}$ , $x^x$ does not have elementary anti-derivatives in spite of being elementary functions.

The given integral $\int \frac{1}{(x\sin(x))^2}\,dx$ , if I start solving it by putting,

$$\sin^2(x) = (1 - \cos(2x))/2$$

$$\int \frac{1}{x^2(1 - \cos(2x))/2}\,dx$$

$$\int \frac{2}{x^2(1 - \cos(2x))}dx$$

dividing and multiplying by $(1 - \cos(2x))$ , we get

$$2\int \frac{1-\cos(2x)}{x^2(1 - \cos(2x))^2}\,dx$$

replacing numerator $(1 - \cos(2x))$ with $2 \sin^2(x)$

$$2\int \frac{2 \sin^2(x)}{x^2(1 - \cos(2x))^2}\,dx$$

after rearranging the above equation we get:

$$4\int \frac{ \sin^2(x)}{x^2}\frac{1}{(1 - \cos(2x))^2}\,dx$$

$$4\int \left(\frac{ \sin(x)}{x}\right)^{\!2}\frac{1}{(1 - \cos(2x))^2}\,dx$$

Now as the given integral contains $\frac{\sin(x)}{x}$ which has no elementary anti-derivative, is this the sufficient condition to say that the given function cannot be integrated?

If this is so, then can we say that any function containing $\frac{\sin(x)}{x}$ , $x^x$ or any such functions cannot have elementary anti-derivative?

EDIT : Please also provide a solution for this problem as the result is calculated by @Dr.SonnhardGraubner in comments.

$\endgroup$
  • $\begingroup$ not integrable in which sense ? That you can't compute by hand ? That is not Riemann integrable over $\mathbb R$ ? Not Lebesgue integrable over $\mathbb R$ ? $\endgroup$ – Surb Feb 28 '18 at 16:49
  • $\begingroup$ No it's not enough, $$ \int \frac{\sin x}{x}x\mathrm dx $$ also contains $\sin x/x$, it can be integrated in terms of elementary functions. $\endgroup$ – qbert Feb 28 '18 at 16:50
  • $\begingroup$ @Surb because the anti-derivative cannot be calculated in terms of elementary fucntions $\endgroup$ – sharq Feb 28 '18 at 16:52
  • $\begingroup$ you will get this here $${\frac {-2\,i}{ \left( \left( {{\rm e}^{ix}} \right) ^{2}-1 \right) { x}^{2}}}-4\,\int \!{\frac {i}{{x}^{3} \left( \left( {{\rm e}^{ix}} \right) ^{2}-1 \right) }}{dx} $$ $\endgroup$ – Dr. Sonnhard Graubner Feb 28 '18 at 16:53
  • $\begingroup$ @qbert I got your point but how to prove that this function (given in question) has not elementary anti-derivative ? $\endgroup$ – sharq Feb 28 '18 at 16:54
3
$\begingroup$

You are correct that $\int (x\sin(x))^{-2} \mathrm{d}x$ does not have an elementary antiderivative.

However, your reasoning is incorrect. It is entirely possible for $f(x)$ to not have an antiderivative but $f(x)g(x)$ to have one. @qbert gave an excellent example with $f(x) = \sin(x)/x$ and $g(x) = x$. But a somewhat less trivial example would be what occurs when both $f(x)$ and $g(x)$ are $\sqrt{1 - k^2 \sin^2(x)}$, which is the form of the incomplete elliptic integral of the second kind.

Typically, the method computers use to find the antiderivative or prove the non-existence of an antiderivative is the Risch algorithm.

For us humans, there are other options for certain functions such as the theorem of Liouville. Which states that if you have rational functions $f$ and $g$, $g$ nonconstant, the antiderivative

$$\int [f(x)\exp(g(x))] \, \mathrm dx$$

can be expressed in terms of elementary functions if and only if there exists some rational function $h$ such that it is a solution to the differential equation:

$$f = h' + hg$$

This is from this answer, which seems to address your question in somewhat more broad terms.

Somewhat off-topic. I believe you may have made an error in computation. I don't see how you suddenly got $(1 - \cos(2x))^4$, shouldn't this be $(1- \cos(2x))^2$?

$\endgroup$
  • $\begingroup$ Thanks , I have made the corrections. $\endgroup$ – sharq Feb 28 '18 at 17:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.