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$$\int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{x}{\sin x} dx$$ How to calculate this integral? I tried using integration by part without too much success. The indefinite integral of it seemed to be to complicated, but the answer should probably be something like $\frac{{\pi}\ln 3}2$ . I only know that $$\int\frac1{\sin x}dx=\ln\left|\frac1{\sin x}-\frac1{\tan x }\right|+C$$, and was told that it has something to do with the integral. I hope someone can answer this question. Thanks.

EDIT: sorry, but I mistakenly got the lower limit wrong. It should be $\frac{\pi}3$

Solution: (inspired by @Claude Leibovici)

Sub $$x=\frac{\pi}2+y$$ Then $$\int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{x}{\sin x} dx$$ $$=\int_{\frac{-\pi}{6}}^{\frac{\pi}{6}} {\frac{{\frac{\pi}2}+y}{\cos y}} dy$$ $$=\int_{\frac{-\pi}{6}}^{\frac{\pi}{6}} {\frac{{\frac{\pi}2}}{\cos y}} dy + \int_{\frac{-\pi}{6}}^{\frac{\pi}{6}} {\frac y{\cos y}}dy$$ $$=\frac{{\pi}\ln3}2+0$$ $$=\frac{{\pi}\ln3}2$$ Again, thanks for those who put efforts to solve this problem.

Brian61354270
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ph-chan
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3 Answers3

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Robert Israel gave the answer : no solution except using polylogarithm functions and, more than likely, numerical method would be required.

What you could do is first set $x=\frac \pi 2+y$ which would give $$I=\int_{\frac{\pi}{2}}^{\frac{2\pi}{3}} \frac{x}{\sin x} dx=\int_0^{\frac \pi 6}\frac{\frac \pi 2+y}{\cos(y)}\,dy=\frac \pi 2\int_0^{\frac \pi 6}\frac{dy}{\cos(y)}+\int_0^{\frac \pi 6}\frac{y}{\cos(y)}\,dy$$ The first integral is simple (using the tangent half-angle subsitution).

For the second one, we could expand the integrand as a Taylor series around $y=0$. This would give $$\frac{y}{\cos(y)}=y+\frac{y^3}{2}+\frac{5 y^5}{24}+\frac{61 y^7}{720}+O\left(y^9\right)$$ $$\int\frac{y}{\cos(y)}\,dy=\frac{y^2}{2}+\frac{y^4}{8}+\frac{5 y^6}{144}+\frac{61 y^8}{5760}+O\left(y^{10}\right)$$ $$\int_0^{\frac \pi 6}\frac{y}{\cos(y)}\,dy=\frac{\pi ^2}{72}+\frac{\pi ^4}{10368}+\frac{5 \pi ^6}{6718464}+\frac{61 \pi ^8}{9674588160}+\cdots$$ Finally, we end with $$I=\frac \pi 2 \frac{\log (3)}{2}+\frac{\pi ^2}{72}+\frac{\pi ^4}{10368}+\frac{5 \pi ^6}{6718464}+\frac{61 \pi ^8}{9674588160}+\cdots\approx 1.010096$$ while numerical integration would lead to $\approx 1.010102$.

Edit

For the second integral, instead of a Taylor expansion, we could use a Padé approximant such as $$\frac{y}{\cos(y)}\approx\frac{y+\frac{7 }{75}y^3+\frac{1}{200} y^5} { 1-\frac{61 }{150}y^2}$$ which would make $$\int_0^{\frac \pi 6} \frac{y}{\cos(y)}\,dy\approx \frac{703125 \log \left(\frac{5400}{5400-61 \pi ^2}\right)}{453962}-\frac{\pi ^2 \left(92784+61 \pi ^2\right)}{25719552}$$ and then $$I\approx 1.0101022740$$ while numerical integration would lead to $\approx 1.0101022541$.

Claude Leibovici
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The antiderivative $\int \frac{x}{\sin(x)}\; dx$ is not an elementary function. It can be expressed in terms of the dilog function. I do not expect your definite integral to have an elementary expression either.

Robert Israel
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  • Hmmm, but I think there might be another way to calculate this definite integral without directly calculating it – ph-chan May 08 '17 at 06:46
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Solution: (inspired by @Claude Leibovici)

Sub $$x=\frac{\pi}2+y$$ Then $$\int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{x}{\sin x} dx$$ $$=\int_{\frac{-\pi}{6}}^{\frac{\pi}{6}} {\frac{{\frac{\pi}2}+y}{\cos y}} dy$$ $$=\int_{\frac{-\pi}{6}}^{\frac{\pi}{6}} {\frac{{\frac{\pi}2}}{\cos y}} dy + \int_{\frac{-\pi}{6}}^{\frac{\pi}{6}} {\frac y{\cos y}}dy$$ $$=\frac{{\pi}\ln3}2+0$$ $$=\frac{{\pi}\ln3}2$$

ph-chan
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