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I have to find out the convergence of the next integral: $$\int^{\pi/2}_0{\frac{\ln(\sin(x))}{\sqrt{x}}}dx$$ Any help? Thanks

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I've already seen this one today...but I'm very bad at finding duplicates. –  1015 Feb 24 '13 at 23:00
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The only problem is at $0$.

We have, as $x$ approaches $0$, $$ \sin x\sim x $$ so $$ \frac{\log(\sin x)}{\sqrt{x}}\sim\frac{\log x}{\sqrt{x}} $$ and the integral of the latter converges at $0$.

If you did not know that, compare for instance with $\frac{1}{x^{2/3}}$. Since $$ x^{2/3}\frac{\log x}{\sqrt{x}}\longrightarrow 0 $$ we have $$ \lvert \frac{\log x}{\sqrt{x}}\rvert\leq \frac{C}{x^{2/3}}. $$ Now it is easily seen that the integral of $1/x^{2/3}$ converges at $0$.

So your integral converges.

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true, but not a trivial statement. –  Ron Gordon Feb 24 '13 at 23:09
    
@rlgordonma Not that difficult either. Like with Bertrand series, the behavior of $\int_0 x^\alpha(\log x)^\beta dx$ is standard. –  1015 Feb 24 '13 at 23:10
    
To you, sure; to the OP, who knows. –  Ron Gordon Feb 24 '13 at 23:12
    
@rlgordonma Sure. But we can always add details, no? By the way, I did. –  1015 Feb 24 '13 at 23:12
    
And good details they are! Seriously, I wasn't picking on you, but it was that you labeled something as self-evident which I felt it took pains to show. –  Ron Gordon Feb 24 '13 at 23:25
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If you integrate by parts, then you find $$\int_0^{\pi/2}dx\,\frac{\ln\sin x}{\sqrt{x}}=2\sqrt{x}\ln\sin x\Bigg|_0^{\pi/2}+2\int_0^{\pi/2}dx\,\sqrt{x}\frac{\cos x}{\sin x}.$$ It is fairly clear at this point that the integral converges because $\sqrt{x}/\sin x\sim x^{-1/2}$ at $x=0$.

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The tricky part of the integral is near $x=0$. There, note that $\sin{x} \sim x$, and consider

$$\int dx \frac{\log{x}}{\sqrt{x}}$$

Substitute $x=u^2$, $dx=2 u du$ and this integral is equal to

$$2 \int du u \frac{1}{u} \log{u^2} = 4\int du \log{u} = 4 (u \log{u}-u) = 2 (\sqrt{x} \log{x} - 2 \sqrt{x}) $$

Thus, near $x=0$, the singularity is integrable, and the integral converges.

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