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$$\displaystyle\begin{align} & \int_{0}^{\infty }{{{\text{e}}^{-x}}\left| \sin x \right|}\text{d}x \\ & \int_{0}^{\pi }{\cos \left( nx \right)\ln \left( 2\sin \frac{x}{2} \right)\text{d}x} \\ \end{align} $$

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put on hold as unclear what you're asking by Mice Elf, Adam Hughes, avid19, Harish Chandra Rajpoot, Hurkyl 2 days ago

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

Do you want to know about the convergent or divergent? Or you wanna find the probable values? –  Babak S. Jan 26 '13 at 9:50
If you split the first integral into intervals where $\sin$ has constant sign, you will end up with a geometric series that you can compute. –  mrf Jan 26 '13 at 10:08
@mrf : Thx mrf, I will hav a try –  Ryan Jan 26 '13 at 10:22
@Ryan As implied by Babak's comment, you should clarify what is it exactly you want on your question. –  Git Gud Jan 26 '13 at 10:56

1 Answer 1

up vote 2 down vote accepted

For 2nd integral, we have
$\begin{align*} \int_{0}^{\pi}{\cos\left(nx\right)\ln\left(2\sin\frac{x}{2}\right)\text{d}x}&=\int_{0}^{\pi}{\ln\left(2\sin\frac{x}{2}\right)\text{d}\dfrac{\sin\left(nx\right)}{n}} \\&=\left[\dfrac{\sin\left(nx\right)}{n}\ln\left(2\sin\frac{x}{2}\right)\right]_{0+}^\pi-\dfrac{1}{2n}\int_0^\pi \sin\left(nx\right)\cot\dfrac{x}{2}\text{d}x \\&=-\dfrac{1}{n}\int_0^\frac{\pi}{2} \sin\left(2nx\right)\cot x\text{d}x \end{align*}$
notice that, for $n\geq 1$
$\begin{align*} \int_0^\frac{\pi}{2} \sin\left(2(n+1)x\right)\cot x\text{d}x-\int_0^\frac{\pi}{2}\sin\left(2nx\right)\cot x\text{d}x&=2\int_0^\frac{\pi}{2} \cos\left((2n+1)x\right)\cos x\text{d}x \\&=\int_0^\frac{\pi}{2} \left\{\cos\left((2n+2)x\right)+\cos\left(2nx\right)\right\}\text{d}x \\&=0 \end{align*}$
$\displaystyle \int_0^\frac{\pi}{2}\sin\left(2x\right)\cot x\text{d}x=2\int_0^\frac{\pi}{2}\cos^2 x\text{d}x=\frac{\pi}{2}$.
Hence, $$\int_{0}^{\pi}{\cos\left(nx\right)\ln\left(2\sin\frac{x}{2}\right)\text{d}x}=-\frac{\pi}{2n}$$

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