Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have Calculus 2 background, so please try to keep your answers around that level. I inly want a brief explanation.

What is it about $\sin (\sin x)$ that makes it difficult to integrate? Also, what makes us think that we express $\int \sin (\sin x) dx$ in terms of already known functions? I imagine if we didn't think we can express it in known functions, then we would've already given it a name and defined it as a special function with its own name.

share|improve this question
9  
It provably cannot be integrated by elementary functions (look up differential galois theory), and it is not used often enough for it to get its own name. –  Steven Gubkin Mar 1 at 1:51
3  
Practically every elementary function has indefinite integral that is not elementary. So this is nothing "unuual" about the particular function $\sin(\sin x)$. –  GEdgar Mar 1 at 2:24

5 Answers 5

Well you can do

$$\int_0^{\pi}{\sin(\sin x)} = \pi H_0(1)\,,$$ where $H$ is a non-elementary function known as the Struve function. (For info on Struve functions, see http://mathworld.wolfram.com/StruveFunction.html.)

You might be wondering...what good does it do to express the integral in terms of a function that almost nobody has ever heard of?

Well, think back to a long time ago when the sine, cosine, and tangent functions themselves were as arcane as Struve functions are now. A math student might have asked a question analogous to your own, saying "Why hasn't anybody been able to integrate $\int{{dx\over 1+x^2}}$? The specialists at the time would have said, "Actually, we ${\it can}$ evaluate that integral:

$$\int{{dx\over 1+x^2}} = \tan^{-1} x\,.$$

If the student were to say, "What good does that do?" The expert would have said, well, this is helpful because see we have these books with values of the tangent function written down in them. So effectively we have "done the integral" because we know which book to look in.

Likewise today, it helps to express the integral $\int_0^{\pi}{\sin(\sin x)}$ as $\pi H_0(1)$ because now we know which book to look at. (Probably Abramowitz and Stegun's ${\it Handbook\ of\ Mathematical\ Functions}$, which I used to use a lot when I was a mathematical physicist.) More commonly, programs like Mathematica have all those values pre-loaded, along with approximation schemes for computing accurate values by interpolation.

And you know, that's all your calculator is doing when it computes $\sin x$ anyway. Once you're beyond the polynomial functions, you're pretty much in this territory.

share|improve this answer
1  
This is a great answer. –  LTS Mar 20 at 21:36
    
+1 It is a fine help. Thanks. –  Felix Marin Mar 20 at 21:59

The surprising thing (beyond Calc 2, unfortunately) are definite integrals with answers as Bessel functions... $$ \int_0^{\pi/2} \cos(\cos x)\;dx = \frac{\pi}{2}\;I_0(1) $$

share|improve this answer

The integration problems one gets in Calculus 2 have integrands specially tailored to admit sufficiently simple closed-form antiderivatives. In the platonic world of mathematics, integrands are more likely to have horrendous-looking antiderivatives than clean-looking, and much more likely still to not have any elementary closed-form. We have to be picky about defining special functions: the zoo is too large to name all non-elementary functions, so we need to pick the ones that (a) turn up the most, (b) are the most interesting, and (c) have the broadest utility.

In my experience, most special functions turn up in the contexts of differential equations, number theory, and combinatorics (with complex analysis as a given). The most general kind of special function that I am aware of is actually a family of them, multivariable hypergeometric functions, which are defined through infinite series expansions with ratios of rising factorials as coefficients.

The hypergeometrics are widely applicable because seemingly almost any elementary function and many non-elementary functions and their moms and dogs can all be expressed as one. For an easier example of how a single special function can "cover extra territory" than one might initially guess, consider the Lambert W function. It is the inverse of $xe^x$. So $W(xe^x)=x$, ignoring technical things like branch cuts in the complex plane. Initially the only kind of equation would seem to be useful for is $xe^x=a~\Rightarrow~ x=W(a)$. Somewhat surprisingly, some pretty simple manipulations allow things like $e^x+x=a$ to be solved (subtract $x$ from both sides, multiply both sides by $e^{a-x}$, ????, profit). We want our special functions to have the widest applicability, all else equal.

Many elementary functions (expressions obtained through the elementary operations applied to compositions of exponentials, logarithms, trigonemtric and inverse trigonometric functions - if we work with complex variables, the latter two are special cases of the former two) that we find do not have an elementary antiderivative. And we've seen we want there to be some strategy in which of these we pick to define as special functions. It is not just that we haven't yet found antiderivatives and suspect there aren't any, we can prove there aren't any.

Some examples are $e^{e^x}$ and $e^{-x^2}$. I think $\sin\sin x$ is another example, but I can't locate a source and I am not personally familiar with the machinery to determine if it is. A heuristic or rule of thumb is that composing exponential or trigonometric functions is walking a tightrope when it comes to having elementary antiderivatives. Perhaps the reason $\int \sin\sin xdx$ doesn't have a name is that it's not something that really comes up in practice. However, it turns out there are some instances of trig functions composed with trig functions occurring "in nature." The special functions needed for this situation are the Bessel functions (which appear in another answer here).

There are historically some other impossibility issues that have turned up. The ancients could bisect an angle but could not trisect one. At the turn of the $18$th century, we could solve cubics and quartics but could not solve a general quintic. The Abel-Ruffini theorem proved the impossibility of solving generic fifth degree or higher polynomials using radicals. Galois developed his theory of symmetry of equations which explained why this was true. It turns out that the impossibility of trisecting angles is a corollary to Galois theory. Very fittingly the current subject which deals with the impossibility of integrating some elementary functions is an outgrowth called differential Galois theory. There are various theorems and algorithms of significance here; see this thread.

(With sufficiently general hypergeometric functions one can exactly solve arbitrary polynomials.)

The real question here, then, is why is it so easy to differentiate and so hard to integrate? There are various rules to differentiate things like products and compositions. The closest product rule for integrals is by-parts and the closest chain rule for integrals is substitution. Why the asymmetry?

This very question was posed here. My favorite answer is that local operations are easier than global operations (differentiation is local and integration is global).

share|improve this answer
    
This is a deep answer and deserves more upvotes. –  AndrewG Aug 18 at 19:38

If you view the integral by series approach, you will find that the integral can solve quite easily:

For the maclaurin series of $\sin x$ , $\sin x=\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}}{(2n+1)!}$

$\therefore\int\sin(\sin x)~dx=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+1}x}{(2n+1)!}dx$

Now for $\int\sin^{2n+1}x~dx$ , where $n$ is any non-negative integer,

$\int\sin^{2n+1}x~dx$

$=-\int\sin^{2n}x~d(\cos x)$

$=-\int(1-\cos^2x)^n~d(\cos x)$

$=-\int\sum\limits_{k=0}^nC_k^n(-1)^k\cos^{2k}x~d(\cos x)$

$=\sum\limits_{k=0}^n\dfrac{(-1)^{k+1}n!\cos^{2k+1}x}{k!(n-k)!(2k+1)}+C$

$\therefore\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+1}x}{(2n+1)!}dx=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k+1}x}{k!(n-k)!(2n+1)!(2k+1)}+C$

If you don't like the summation symbol, you may express the above result in term of the Kampé de Fériet function.

share|improve this answer
    
Lovely elegant formula +1 –  LTS Mar 20 at 21:40

We can express the function in terms of known functions, if you use power series (which is typically taught at the end of Calculus 2). I do not know of a particular name for the function represented by this expansion (or use for it) but if you find a regular application somewhere, feel free to call it something catchy. The following approximation for the power series of the antiderivative of $\sin(\sin(x))$is: $$\int \sin(\sin(x)) dx = C+x^2/2-x^4/12+x^6/60-x^8/315+ \ldots$$

This results from integrating term-for-term the power series expansion $$\sin(\sin(x))=x-x^3/3+x^5/10-(8 x^7)/315+\ldots$$

share|improve this answer
    
Can this be expressed in $\Sigma$-notation? –  Sanath Mar 1 at 3:31
    
Interesting question. I'm not sure, off hand, what the answer to that would be. –  Darrin Mar 1 at 3:39
1  
@Sanath Devalapurkar, see my answer. –  doraemonpaul Mar 2 at 1:39
    
@Sanath Devalapurkar, or you may use the Kampé de Fériet function. –  doraemonpaul Mar 20 at 21:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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