Why doesn't this approach work for integral of $\log(\sin(x))$?

Asked Dec 28 '14 at 11:52

Active Dec 28 '14 at 11:52

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Evaluation of:

$$I = \int_{0}^{\pi} \log(\sin(x)) dx$$

Over closed rectangular contour $ABCD$ complex analysis.

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BUT INSTEAD the right bound will be $\pi$ instead of $\pi/2$

Here is what I do not understand.

I will use the fact that

$$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$

$$I = \int_{0}^{\pi} \log(\frac{e^{ix} - e^{-ix}}{2i}) dx$$

$$I = \int_{0}^{\pi} \log(e^{ix} - e^{-ix}) dx - \left( \pi\log(2) + \frac{i\pi^2}{2} \right)$$

All we need to figure out now is:

$$J = \int_{0}^{\pi} \log(e^{ix} - e^{-ix}) dx$$

Using that contour.

How can we do so?

edited Apr 13 '17 at 12:20

Community

asked Dec 28 '14 at 11:52

Amad27

Have you tried simplifying? $y=e^{ix}$ you have $e^{ix}-e^{-ix}=y-1/y=(1-y)/y$ and splitting up. – Ali Caglayan Dec 28 '14 at 11:55
1

On another note, this integral is done by exploiting symmetrical properties of the sin function. No complex analysis needed. – Ali Caglayan Dec 28 '14 at 11:57
@Alizter, I know it can be done without complex analysis. But I am interested in complex analysis. Which is why I posted this. – Amad27 Dec 28 '14 at 11:58
see here http://math.stackexchange.com/questions/37829/computing-the-integral-of-log-sin-x – Dr. Sonnhard Graubner Dec 28 '14 at 15:22
@Amad27, see here http://math.stackexchange.com/questions/1080710/integral-of-log-sinx-using-contour-integrals. – xpaul Jan 02 '15 at 15:00

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