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I was trying to write the function $\frac{1}{x\text{sin}(x)}$ in terms of partial fractions, so that I could integrate the function. If I had the form

$$\frac{1}{x\text{sin}(x)} = \frac{A}{x} + \frac{B}{\text{sin}(x)}$$

I then have

$$ 1 = A\text{sin}(x) + Bx $$

I think it is best to look at $0 < x < \pi$. The above equation should hold for all $0 < x < \pi$. If I choose $x = \frac{\pi}{2}$ and $x=\frac{\pi}{4}$ to find my coefficients, I find that

$$ A = 1+\sqrt{2}$$ $$ B = -\frac{2\sqrt{2}}{\pi}$$

However, if I choose $x=\frac{\pi}{2}$ and $x=\frac{\pi}{3}$ to find my coefficients, I find that

$$ A = \frac{6\sqrt{3}+8}{11}$$ $$ B = \frac{6-12\sqrt{3}}{11\pi}$$

I am wondering what is going wrong? The suspicion might be that it is wrong to decompose $\frac{1}{x\text{sin}(x)}$ in the manner as I have done, but it also means:

1) Can I NOT express it in terms of partial fractions?

2) How does one approach the integral then (I have tried partial fractions, substitution, and integration by parts)?

AstroInt
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    Only rational functions (and to a much lesser extent, certain algebraic functions) can be written as the usual sort of partial fractions. Weird things happen with trig functions: for example, $$ \frac{1}{\sin{x}\cos{x}} = \frac{\cos^2{x}+\sin^2{x}}{\sin{x}\cos{x}} = \frac{\cos{x}}{\sin{x}} + \frac{\sin{x}}{\cos{x}} $$ is one sort of "trigonometric partial fraction". – Chappers Apr 15 '17 at 03:19
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    (1) Partial fraction decomposition (PFD) relies on a very special property of polynomial. Since $\sin x$ is not a polynomial, PFD does not work in the way you guessed. On the other hand, $\sin x$ can be loosely thought as an infinite degree polynomial, and this suggest that the PDF would also involve in infinitely many terms. This is indeed rigorously proved in complex analysis.(2) I would be astounded if it has an elementary antiderivative. – Sangchul Lee Apr 15 '17 at 03:20
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    $$\frac{\pi^2}{\sin^2 (\pi z)} = \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}$$ This is an infinite partial fraction decomposition, which deserves to be called like that because it converges locally uniformly, so its poles are really or order $2$ at $z= n$. This is how Euler proved the Basel problem, after 2 years of work. – reuns Apr 15 '17 at 03:21

3 Answers3

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The technique of partial fractions is an extremely specific one. It applies in exactly one situation: when the function you are integrating is a fraction in which both the numerator and denominator are polynomials. What you're looking at is not of that form, so partial fractions can't apply. Here's a more familiar version of this phenomenon: to solve $x^2=2x$, you can use the "technique" of dividing by $x$ to obtain the result $x=2$ (and the "lost" solution $x=0$). But to solve $x^2=2x+1$, you can't use that technique - dividing by $x$ here doesn't get you anything useful. Likewise, the technique of partial fractions doesn't apply to every fraction, just a certain kind.

For this integral, it's unlikely that any technique will work; $\frac{1}{x\sin{x}}$ is not a well-behaved function. A sufficiently clever use of integration by parts might work, but I doubt it. Some integrals just can't be solved (technically, the phrase is "can't be solved with elementary functions", but that's a bit off-topic right now). If you have reason to believe you can solve this integral - for example, if you encountered this integral as part of an assignment - then you should look for a way to get around actually doing the integral. Maybe you made an error in determining which integral to do, or maybe there's a clever way to get the answer without integrating at all.

Reese Johnston
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    What do you mean by "it's not a well-behaved function"? I can see that it is hard because it is composed of an algebraic and a trig function, but how can you see weather it'll be hard to integrate from the behavior of the function itself? – Ovi Apr 15 '17 at 03:53
  • @Ovi It's just because it's "ugly". A trig function in a denominator, multiplied by an algebraic function also in the denominator, will resist all of the basic integration techniques. That's not a proof at all - but it's good reason to expect that the integral can't be done. – Reese Johnston Apr 15 '17 at 03:57
  • Ah I thought that when you mentioned the behavior of the function, you looked at the graph of $\dfrac {1}{x \sin x}$ and saw something strange. – Ovi Apr 15 '17 at 03:59
  • @reese How do you define a PFE? I provided one for $x\csc(x)$ hereon this page. In fact, any meromorphic function, $f(z)$, that approach $0$ as $z\to \infty$ can be expanded as a series of polynomials, $P_n\left( \frac1{z-z_n}\right)$, where $z_n$ is a pole of $f$. This is a PFE. – Mark Viola Apr 17 '17 at 22:05
  • @Dr.MV From context, it's reasonable to assume that OP is trying to use introductory calculus techniques. In particular, I'm taking PFEs to be finite. – Reese Johnston Apr 18 '17 at 01:10
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We can represent the function of interest in a partial fraction expansion, but the expansion is given in terms of an infinite series.

To show this we rely on THIS ANSWER, in which I showed, using real analysis only, that we could represent the cosecant function as the partial fraction expansion

$$\pi\csc(\pi y)=\sum_{n=-\infty}^\infty \frac{(-1)^n}{y-n} \tag 1$$

Now, letting $y=x/\pi $ in $(1)$ and multiplying the resulting equation by $1/\pi$ yields

$$\frac{1}{\sin(x)}=\sum_{n=-\infty}^\infty \frac{(-1)^n}{x-n\pi}\tag 2$$

Dividing both sides of $(2)$ by $x$ and using the expansion

$$\frac{1}{x(x-n\pi)}=\frac{1}{n\pi (x-n\pi)}-\frac{1}{n\pi x}$$

for $n\ne 0$, we obtain

$$\frac{1}{x\sin(x)}=\frac1{x^2}+\sum_{n=1}^\infty\left(\frac{(-1)^{n-1}}{n\pi(x+n\pi)}-\frac{(-1)^{n-1}}{n\pi(x-n\pi)}\right) \tag 3$$

Equation $(3)$ provides a partial fraction expansion representation of the function $\frac{1}{x\sin(x)}$ which converges for $x\ne k\pi$. We also see explicitly the second order pole at $x=0$ from the term $\frac1{x^2}$.

Mark Viola
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You might use the formula $$ \sin x = x\prod_{n\in\mathbb{N}}\left(1-\frac{x^2}{n^2\pi^2}\right)$$ so that $1/(x\sin x)$ is an infinite rational product, that might be writable as an infinite rational sum.

Empy2
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