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This is my problem,I tried to use change of variable, but no result so far. Can anyone help me?

$$\int_0^1 {\frac{{\arcsin x}}{x}dx} $$

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    $\begingroup$ If you had to get a ball park figure fast, you could always use the fact that $\arcsin x \sim x$ to get that the integral is close to $1$, which indeed it is. $\endgroup$ – nbubis Jul 21 '13 at 4:58
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Replace $\arcsin x$ by $\theta$ Then the integral becomes $$I=\int_{0}^{\pi/2} \frac{\theta}{\sin \theta}\cos \theta d\theta$$ Then apply integration by parts to get $$I=-\int_{0}^{\pi/2}\ln\sin\theta d\theta=-\int_{0}^{\pi/2}\ln \cos\theta d\theta=-\frac{1}{2}\int_{0}^{\pi/2}\ln\sin 2\theta d\theta+\frac{1}{2}\int_{0}^{\pi/2}\ln2d\theta$$ The first integral can now be rewritten as below $$-\frac{1}{4}\int_{0}^{\pi}\ln \sin\theta d\theta=-\frac{1}{4}\int_{0}^{\pi/2}\ln \sin\theta d\theta-\frac{1}{4}\int_{\pi/2}^{\pi}\ln \sin\theta d\theta$$ after a change of variable from $2\theta $ to $\theta$ and then it can be rewritten as $$-\frac{1}{4}\int_{0}^{\pi/2}\ln \sin\theta d\theta-\frac{1}{4}\int_{0}^{\pi/2}\ln \cos\theta d\theta=-\frac{1}{2}\int_{0}^{\pi/2}\ln \sin\theta d\theta=\frac{I}{2}$$ Hence we get $$I=\frac{I}{2}+\frac{\pi}{4}\ln 2\\ \Rightarrow I=\frac{\pi}{2}\ln 2$$

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  • $\begingroup$ very clearly, best answear ! $\endgroup$ – Tien Quan Jul 21 '13 at 5:10
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Make the substitution $ x \mapsto \sin u $ to arrive at $ \int\limits_0^\frac{\pi}{2} u \cot u \ du $. Proceed with integration by parts to arrive at $ \left[u \ln \sin u \right] _0^{\frac{\pi}{2}} - \int\limits_0^\frac{\pi}{2} \ln \sin u \ du $. The latter integral is widel known to be $ -\frac{\pi \ln 2}{2} $ and hence the answer is $ \frac{\pi \ln 2}{2} $ as $ \left[u \ln \sin u \right] _0^{\frac{\pi}{2}} = 0 $.

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  • $\begingroup$ @JonClaus, I think we need to show how $ u \ln \sin u =0$ at $u=0$ $\endgroup$ – lab bhattacharjee Jul 21 '13 at 5:12
  • $\begingroup$ @labbhattacharjee That follows from $\sin u\sim u$ and $u\log u\to 0$. $\endgroup$ – Pedro Tamaroff Jul 21 '13 at 5:16
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Integrating by parts gives $$I = \int_0^1 {\frac{{\arcsin x}}{x}dx} = -\int_0^1 {\frac{\ln(x)}{\sqrt{1-x^2}}dx} .$$

Now, consider the integral

$$ F = \int_{0}^{1}\frac{x^{\alpha}}{\sqrt{1-x^2}}dx = \frac{1}{2}\,{\frac {\sqrt {\pi }\,\Gamma \left( \frac{\alpha}{2}+\frac{1}{2} \right) }{ \Gamma \left( \frac{\alpha}{2}\,+1 \right) }}, $$

which was evaluated using the beta function. Then our integral $I$ can be evaluate using $F$ as

$$ I = \lim_{\alpha \to 0} \frac{d{F}}{{d\alpha}} =\frac{\ln(2)\pi}{2}.$$

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  • $\begingroup$ And how does one derive such formula? $\endgroup$ – Pedro Tamaroff Jul 21 '13 at 6:33

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