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$$\displaystyle\int_0^{\pi/2} x\cot x\,dx$$

By integration by parts I get a part $[x \ln|\sin x|]_0^{\pi/2}$ where $(\pi/2)\ln|\sin(\pi/2)|=0$ but, does $0\cdot \ln|\sin0|=0$ or is it indeterminate form of $0\cdot\infty$? Is there any way to find the above integral other than the integration by parts? I prefer if it is in elementary functions as I'm still a new student to Integration.

Mike
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  • http://math.stackexchange.com/questions/692203/integral-int-0-pi-2-x-cotxdx-differntiation-wrt-parameter-only – lab bhattacharjee Mar 06 '14 at 07:42
  • $$\lim\limits_{x\to0}x\log\sin x=-\lim\limits_{x\to0}x^2\cot x=-\lim\limits_{x\to0}x\cos x=0$$Where I have used$$\lim\limits_{x\to0}\frac {x}{\sin x}=1$$ – Frank W Jun 27 '22 at 19:20

2 Answers2

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Your integral is an improper integral because $\cot x$ is not continuous at $x=0$. To correct this issue, you should set up the problem as $$ \int_0^{\pi/2}x\cot x\,dx=\lim_{a\to 0^+}\int_a^{\pi/2}x\cot x\,dx $$ Then evaluate the integral $$ \int_a^{\pi/2}x\cot x\,dx $$ This will give you an expression in $a$. Once you have this expression take $a\to 0$.

Brian Fitzpatrick
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The function you have to integrate is not defined at $0$, but it is bounded on $(0,\frac{\pi}{2}]$, so this is a regular Riemann Integral, not an improper one. For $0<h<\frac{\pi}{2}$ the function is continuous on $[h,\frac{\pi}{2}]$, so you can apply the Fundamental Theorem of Calculus, and then take the limit for $h\rightarrow 0^+$ of the function you get.

alex
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