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I got this doubt while evaluating the integrals:

$$I=\int_{0}^{\frac{\pi}{2}}\ln(\sin x)\sin xdx$$ and

$$J=\int_{0}^{\frac{\pi}{4}}\csc xdx$$

Now even though the integrand $f(x)=\ln(\sin x)\sin x$ is not defined at $x=0$ which is the lower limit, still it has a finite answer.

But integrand in $J$ is not defined at $x=0$ and integral is infinite.

So how to identify without explicitly evaluating?

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  • $\begingroup$ Comparison test is definite a way to go. $\endgroup$ – Sangchul Lee Mar 24 at 17:57
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Notice, however, that

\begin{eqnarray} \lim_{x\to0^+}\ln(\sin x)\sin x&=&\lim_{x\to0^+}\frac{\ln(\sin x)}{\csc x}\\ &=&\lim_{x\to0^+}\frac{\cot x}{(-\csc x\cot x)}\\ &=&\lim_{x\to0^+}(-\sin x)\\ &=&0 \end{eqnarray}

Here is the graph of $y=\ln(\sin x)\sin x$

graph of ln(sin x)(sin x)

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  • $\begingroup$ So does it mean if limit is finite at the end points, integral exists? $\endgroup$ – Umesh shankar Mar 24 at 18:08
  • $\begingroup$ In this case, the discontinuity at $x=0$ was a removable discontinuity. But such is not the case at $x=0$ for $\csc x$ because of the vertical asymptote. However, even in the case where there is a vertical asymptote the integral may still converge. For example, $\int_0^1\dfrac{1}{\sqrt{x}}\,dx$. $\endgroup$ – John Wayland Bales Mar 24 at 18:20
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For the second integral $J$, the integrand goes to $\infty$ as $x\to 0$. Roughly $\frac{1}{\sin x} = \frac{1}{x}$. From the examples $\int_0^1 x^{-a} dx$ we know that $a=1$ is divergent, albeit borderline so. Since $$ \sin x = x - \frac{1}{3!}x^3 \pm ... $$ we have $\sin x \leq x$ for $x>0$ small (from elementary trigonometry, or since the tail of a (edit: convergent) alternating series with terms of decreasing value is dominated by any previous term). Thus $\frac{1}{\sin x} \geq \frac{1}{x}$ and the integral $J$ is divergent.

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