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I have the next integral: $$\int^{\pi/2}_0{\frac{\ln(\sin(x))}{\sqrt{x}}}dx$$ I have no clue how to start. At $x=0$ there is a clear discontinuity and I don't know how to solve the integral. The main problem is that I don't know how to solve $\int{\frac{\ln(\sin(x))}{\sqrt{x}}}dx$, Wolfram alpha says that the primite doesnt exist. Any idea?

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I don't think the problem is the discontinuity itself, but rather the fact that the integrand seems to blow up at $x = 0$. –  Orest Xherija Feb 24 '13 at 19:51
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The integrand behaves like $\log x/\sqrt{x}$ at $0$, which converges. So you should not worry about $0$ that much. –  1015 Feb 24 '13 at 19:55

3 Answers 3

Why don't you try the following:

$$\int^{\pi/2}_0{\frac{\ln(\sin(x))}{\sqrt{x}}}dx = \lim_{\epsilon \to 0+} \int^{\pi/2}_{\epsilon}{\frac{\ln(\sin(x))}{\sqrt{x}}}dx $$

You can try and calculate the latter integral and then take the proposed limit. I haven't tried the calculation myself so I don't know if it will work, but I think it's well worth an attempt.

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@julien is right. In fact the following limit says it converges: $$\lim_{x\to 0^+}x^{2/3}f(x)=0<+\infty$$

Note: If you set $\lim_{x\to a^+}(x-a)^pf(x)=A$ then $\int_a^bf(x)dx$ converges if $p<1$ and $A$ is finite.

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The y-axis is an asymptote for this function, the graph does not converge to 0 at all for x→0+ That ofcourse does not mean that the integral is divergent. No clue how to integrate that.

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