Question

show that integral (ln (sin(x))=-piln(2) from 0..pi.

Answer 1

\[\int\limits_{0}^{\pi} \ln \sin(x)=-\pi \ln(2)\]

Answer 2

interesting

Answer 3

ahh this classic integral again...!

Answer 4

i have solved Numerically . but want to know is there analytical way .?

Answer 5

yeah there is...

Answer 6

how ? can you guide me ?

Answer 7

i remember solving it....will take some time to type the reply....

Answer 8

ok,first since sin is symmetric in the range [0,pi],we can write that as 2*integral from 0 to pi/2,right?.....because there should be mod sign inside ln,as ln cannot take negative values.

Answer 9

now we can use the property that
\[\int\limits_{a}^{b}f(x)dx=\int\limits_{a}^{b}f(a+b-x)dx\]
to get \[\int\limits_{0}^{\pi/2}\ln|\cos x| dx\]
with me till here?

Answer 10

yes.
how did you get cos(x)?

Answer 11

sin(0+pi/2-x)=cos x

Answer 12

ok

Answer 13

now ?

Answer 14

now let the original integral be
\[I=2\int\limits_{0}^{\pi/2}\ln|\sin x|=2\int\limits_{0}^{\pi/2}\ln|\cos x|\]
so can i write that as
\[I=\int\limits_{0}^{\pi/2}\ln|sinx|dx+\int\limits_{0}^{\pi/2}\ln|\cos x|dx\]
be careful,this is the main step

Answer 15

ok .

Answer 16

so now,i would use definite integral and ln properties,like this
\[I=\int\limits_{0}^{\pi/2}\ln|\sin x \cos x|=\int\limits_{0}^{\pi/2}\ln|2\sin x \cos x|-\ln 2\]
got this?
then u can write 2 sin x cos x as sin 2x and make the substitution of 2x=t,try it and tell me what u get..

Answer 17

sorry,i keep on forgetting dx.....

Answer 18

that's a beautiful proof.

Answer 19

i am getting
\[\int\limits_{0}^{\pi/2}\ln|\sin(2x)-\ln(2)\]
\[\int\limits_{0}^{\pi}[\frac{1}{2}\ln|\sin(t)-ln(2)]dt\]

Answer 20

absolutely correct,now separate those two integrals,solve 2nd one(ln 2)
and in 1st integral,write
\[I=\int\limits_{0}^{\pi}\ln|\sin x|\]
and don't forget to write ,I=..............
then u can solve for I to get to your answer
tell me if this was confusing....

Answer 21

should i solve them separately ?

Answer 22

the 1st integral need not be solved,u get 1/2 I ,and 2nd integral when solved will just give pi/2*ln2 ,right?
so now u have
I=I/2-pi/2*ln 2
solve for I

Answer 23

ohh thanks .. i got it now.

Answer 24

welcome :) its good to remember such beautiful proofs......

Answer 25

there is a lot of beautiful integrals in this putnam competitions
http://mks.mff.cuni.cz/kalva/putnam/putn53.html

Answer 26

http://mks.mff.cuni.cz/kalva/putnam.html

Answer 27

and u have solved them all??

Answer 28

lol...most of them