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

Zain Patel
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Dhiraj Barnwal
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4 Answers4

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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.

Jack D'Aurizio
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  • just curious ... How does $\log \tan(x/2)$, or its integral, have a Fourier Series for $t<0$? Do we not need to restrict the domain to $t>0$ since the tangent is an odd function and for real-values, the logarithm is not defined for negative argument. – Mark Viola Jul 03 '15 at 22:26
  • @Dr.MV: sure, my bad. Now fixed. – Jack D'Aurizio Jul 03 '15 at 22:53
  • It is interesting and kind of cool that $\int_0^{\pi}\csc x,\sin (nx),dx=-\pi$ for odd $n$ and $0$ otherwise. And it is very straightforward to show. Of course the final result can also be represented in terms of the dilogarithm with comparable convergence speed I believe. I gave this a +1 even before the typo fix. – Mark Viola Jul 03 '15 at 23:00
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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.

robjohn
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  • Since $\mathrm{Li}_2(i)=-\frac{\pi^2}{48}+iG$ and $\mathrm{Li}_2(-i)=-\frac{\pi^2}{48}-iG$, we get that $\int_0^{\pi/2}\frac{x}{\sin(x)},\mathrm{d}x=2G$ where $G$ is Catalan's Constant. – robjohn Aug 30 '20 at 23:08
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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.

Zain Patel
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  • I am quite new with Integration, can you please post the full solution, I will try to figure that out. – Dhiraj Barnwal Jul 03 '15 at 16:30
  • How do you find the integral on the right side, though? – user84413 Jul 03 '15 at 16:34
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    Are you comfortable with the polylogarithm function? – Zain Patel Jul 03 '15 at 16:34
  • No, heard it for the first time. Any reference link? – Dhiraj Barnwal Jul 03 '15 at 16:48
  • https://en.wikipedia.org/wiki/Polylogarithm – Zain Patel Jul 03 '15 at 16:49
  • What you found is a «closed form» only because you defined it to be. Te polylogarithm function is defined (amng other ways) as an integral. You could decide that the integral the OP wants defines the Patel function and call that a closed form, too... – Mariano Suárez-Álvarez Jul 03 '15 at 20:57
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    @MarianoSuárez The dilogaritm function is well-know, has a "nice" power series, several useful properties such as duplication and inversion, has several know special values, etc. If one wishes to study the antiderivative of $x\csc x$ and reports any notable properties, then it indeed becomes a well-known special function. Until then, the relationship it has to the dilogarithm is as acceptable as if it had a representation in terms of any other well-known function. – Mark Viola Jul 03 '15 at 21:05
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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.

Lucian
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  • Well, it is not «non-integrable»... – Mariano Suárez-Álvarez Jul 03 '15 at 20:56
  • @MarianoSuárez-Alvarez: There's no point making the discussion more complicated than it already is. – Lucian Jul 03 '15 at 21:01
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    There is a difference between making a discussion complicated and asserting what is a false statement under the usual definitions of the terms involved. You could expand a bit in what exactly is meant by «integrability in elementary terms» as opposed to good ol' «intregrability». As it stands, I am afraid this is more confusing than helpful. – Mariano Suárez-Álvarez Jul 03 '15 at 21:05
  • @MarianoSuárez-Alvarez: Not to the OP. If you want to rub the man's head in that “being integrable” and “possessing a closed form anti-derivative” are two completely different technical terms belonging to the most basic mathematical jargon, be my guest... You can do so by leaving either a comment to main post, or just answering the question yourself, but persecuting me in particular serves no purpose. – Lucian Jul 03 '15 at 21:15
  • I am using comments for one of its intended purposes: to signal an error. – Mariano Suárez-Álvarez Jul 03 '15 at 21:48
  • @MarianoSuárez-Alvarez: Then signal it to the one who made it, and leave me out of it. – Lucian Jul 03 '15 at 21:55