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$\def\cosec{\operatorname{csc}}$ Calculate the value of the improper integral

$$\int^{\pi/2}_0\left (\cosec x - \dfrac{1}{x}\right)\,\mathrm dx.$$

You may use the standard integral $\int \cosec x \,\mathrm dx = -\ln|\cosec x + \cot x| + c $.

Please note this is a 3 point question. Here is my solution:

This definite integral is improper at $x=0$, so we have: \begin{align*} \lim_{b\to0}\int^{\pi/2}_b\cosec x - \dfrac{1}{x}\,\mathrm dx &= -\ln|\cosec x + \cot x|- \ln|x| \biggr|^{\pi/2}_b\\ &= \lim_{b\to0} -\left(\ln\left|\cosec \frac{\pi}{2} + \cot \frac{\pi}{2}\right|- \ln\left|\frac{\pi}{2}\right| \right) - (-\ln|\cosec b + \cot b|- \ln|b| ) \end{align*}

For $\lim\limits_{b\to0}-(\ln|\cosec b + \cot b|+ \ln|b|)$, using the properties of logarithms, I have $-\lim\limits_{b\to0} \ln|b\cosec b + b\cot b|$.

The limits are both solved similarly by L'Hôpital's rule, giving:

$$-\lim_{b\to0}\ln|b\cosec b + b\cot b| = -\lim_{b\to0}\ln|1 + 1| = -\ln|2|$$

Therefore, I have:

$$\lim_{b\to0} \left(-\ln|1+0|- \ln\left|\frac{\pi}{2}\right| \right) - (-\ln|2| ) = \ln\left|\frac{2}{\pi/2}\right| = \ln\left|\dfrac{4}{\pi}\right|$$

Is this correct and answered sufficiently? Have I made any errors?

Thanks so much!

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What is csc? Maybe $1/\sin$? –  Siminore Sep 26 '12 at 9:18
    
It looks fine to me. Just please delete the absolute values for the positive arguments of the logarithm: it just makes the reading more cumbersome. –  DonAntonio Sep 26 '12 at 9:22
    
@Avatar - $cot \dfrac{\pi}{2}$ evaluates to 0 –  JackReacher Sep 26 '12 at 9:48
    
@Siminore - yes csc is cosec, 1/sin –  JackReacher Sep 26 '12 at 9:49
2  
Your answer is correct. –  Mhenni Benghorbal Sep 26 '12 at 13:22
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1 Answer

This CW answer is intended to remove this question from the Unanswered queue.


As Mhenni Benghorbal remarks, your answer is entirely correct. Also, the derivation is nicely written out.

Cheers!

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