Is integration of $x\operatorname{cosec}(x)$ defined?

Question

Is integration of $x\operatorname{cosec}(x)$ possible? If yes, then what is its closed form; if not, then why is it non-integrable ?

Answer 1

We have $\sin x = x +O(x^3)$ in a neighbourhood of the origin, so $\frac{x}{\sin x}$ is integrable over any compact interval in $(-\pi,\pi)$. Its primitive, however, is not an elementary function:

$$ \int_{0}^{t}\frac{x}{\sin x}\,dx = t\log\tan\frac{t}{2}-\int_{0}^{t}\log\tan\frac{x}{2}\,dx$$ but has a nice Fourier series:

$$\forall t\in(0,\pi),\qquad \int_{0}^{t}\frac{x}{\sin x}\,dx = t\log\tan\frac{t}{2}+2\sum_{n\geq 0}\frac{\sin((2n+1)t)}{(2n+1)^2}.$$

This is strictly related with the inverse Gudermannian function.

Answer 2

Yes, $x\,\mathrm{cosec}\,x$ is integrable and has a closed form.

Hint:

Using the subsitution $u = \tan \left(\frac{x}{2}\right)$, we get:

$$\int \frac{1}{\sin x} \, \mathrm{d}x = \ln \left(\tan \left(\frac{x}{2}\right)\right) + \mathrm{C}$$

Combine this result with the use of integration by parts, specifically, let $u= x \implies \mathrm{d}u = \mathrm{d}x$ and $\mathrm{d}v = \frac{1}{\sin x}$.

Using integration by parts with the above , we get that $$\int \frac{x}{\sin x} \, \mathrm{d}x = x \ln \left(\tan \left(\frac{x}{2}\right)\right) + \int \ln \left(\tan \left(\frac{x}{2}\right)\right) \, \mathrm{d}x$$

The above evaluates to

$$i\left(\mathrm{Li}_2(e^{-ix}) - \mathrm{Li}_2 (e^{ix})\right) + x \left(\ln(1-e^{ix}) - \ln(1+e^{ix})\right) + \mathrm{C}.$$

Where $\mathrm{Li}_2$ is the Polylogarithm function.

Answer 3

Why is it non-integrable ?

The answer to this question is given by Liouville's theorem and the Risch algorithm. However, understanding them requires advanced knowledge of abstract algebra, such as $($ differential $)$ Galois theory, for instance.

Answer 4

Using $u=\tan(x/2)$ so that $\sin(x)=\frac{2u}{1+u^2}$, $\mathrm{d}x=\frac{2\,\mathrm{d}u}{1+u^2}$ $$ \begin{align} &\int\frac{x}{\sin(x)}\,\mathrm{d}x\\ &=2\int\frac{\arctan(u)}u\,\mathrm{d}u\\ &=2\sum_{k=0}^\infty(-1)^k\int\frac{u^{2k}}{2k+1}\,\mathrm{d}u\\ &=2\sum_{k=0}^\infty(-1)^k\frac{u^{2k+1}}{(2k+1)^2}+C\\ &=\frac{\mathrm{Li}_2(iu)-\mathrm{Li}_2(-iu)}i+C\\[6pt] &=\frac{\mathrm{Li}_2(i\tan(x/2))-\mathrm{Li}_2(-i\tan(x/2))}i+C \end{align} $$ where $\mathrm{Li}_2(x)=\sum\limits_{k=1}^\infty\frac{x^k}{k^2}$ is the Dilogarithm Function.