# Improper integral $\int^{\pi/2}_0 (\operatorname{csc} x - \frac{1}{x})\,\mathrm dx$

$\def\cosec{\operatorname{csc}}$ 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:

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

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
Your answer is correct. –  Mhenni Benghorbal Sep 26 '12 at 13:22