My book suggests $$\int_2^{\pi/2}\log\sin x \,dx$$ is an improper integral. But I think it is not for it is bounded on the respective interval....am I correct?
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2It is strange since $\sin x$ is strictly positive for $x \in [\frac{\pi}{2};2]$. – Martigan Feb 09 '15 at 09:57
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3may be the lower limit is 0 not 2, for then it would be improper – vnd Feb 09 '15 at 10:01
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$\displaystyle\int_0^\tfrac\pi2\ln\sin x~dx~=~\int_0^\tfrac\pi2\ln\cos x~dx~=-\frac\pi2~\ln2.~$ – Lucian Feb 09 '15 at 10:48