# Interesting integral

• August 26th 2009, 03:49 AM
simplependulum
Interesting integral
It should be in the sub-forum "Interesting Math Problem" but there is no such place for me to post (Rofl) It is not very hard but the answer is extremely elegant !!

Show that $I_{m} = \int_0^{\frac{\pi}{2}} \ln[\sin(x)] \sin^{2m}(x)~dx = \frac{\pi \binom{2m}{m} }{4^m} [( \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - ... + \frac{1}{2m-1} - \frac{1}{2m} )- \ln2 ]$

and $I_0 = -\frac{\pi}{2}\ln2$
• August 26th 2009, 02:13 PM
NonCommAlg
Quote:

Originally Posted by simplependulum

Show that $I_{m} = \int_0^{\frac{\pi}{2}} \ln[\sin(x)] \sin^{2m}(x)~dx = \frac{\pi \binom{2m}{m} }{4^m} [( \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - ... + \frac{1}{2m-1} - \frac{1}{2m} )- \ln2 ]$

are you sure this is correct? for example, according to may solution $I_1=\frac{\pi}{8} - \frac{\pi}{4} \ln 2,$ but your formula gives us: $I_1=\frac{\pi}{4} - \frac{\pi}{2} \ln 2.$ (Wondering) since i can't be wrong, so you must be wrong then! haha
• August 26th 2009, 04:24 PM
luobo
Quote:

Originally Posted by NonCommAlg
are you sure this is correct? for example, according to may solution $I_1=\frac{\pi}{8} - \frac{\pi}{4} \ln 2,$ but your formula gives us: $I_1=\frac{\pi}{4} - \frac{\pi}{2} \ln 2.$ (Wondering) since i can't be wrong, so you must be wrong then! haha

haha, I more and more tend to think this integral is a guess by some "experimental mathematicians".

Anyway, the correct answer is half of simplependulum's formula.
• August 26th 2009, 06:32 PM
NonCommAlg
let $m \geq 0$ be an integer. let $J_m = \int_0^{\frac{\pi}{2}} \sin^{2m} x \ dx = \frac{\pi}{4^m} \binom{2m-1}{m}.$ we have: $I_{m+1}=\int_0^{\frac{\pi}{2}}\ln(\sin x) (1 \ - \ \cos^2 x) \sin^{2m}x \ dx=I_m \ - \ \int_0^{\frac{\pi}{2}} \ln(\sin x) \cos^2 x \sin^{2m} x \ dx.$ call this $(1).$

now let $u=\cos x \sin^{2m} x, \ \ln(\sin x) \cos x \ dx = dv.$ then $v=\ln(\sin x) \sin x - \sin x$ and $du=(2m \sin^{2m-1} x - (2m+1)\sin^{2m+1}x) \ dx.$ therefore integration by parts gives us:

$\int_0^{\frac{\pi}{2}} \ln(\sin x) \cos^2 x \sin^{2m} x \ dx=(2m+1)I_{m+1} - 2mI_m - (2m+1)J_{m+1} + 2m J_m.$ put this in $(1)$ to get: $I_{m+1}=\frac{2m+1}{2m+2}I_m + \frac{\pi}{2(m+1)^24^{m+1}}\binom{2m}{m}, \ \ m \geq 0, \ \ I_0 = -\frac{\pi}{2} \ln 2.$

that's what i got and i think it's good enough! of course this recurrence relation will also give us a general formula for $I_{m+1}$ but i'm not going to do it! (Shake)
• August 26th 2009, 07:10 PM
simplependulum
Thanks for your corrections (Happy) , i think i need to have a double check after i finish a problem next time . (Crying)
• August 27th 2009, 04:31 AM
luobo
Quote:

Originally Posted by simplependulum
It should be in the sub-forum "Interesting Math Problem" but there is no such place for me to post (Rofl) It is not very hard but the answer is extremely elegant !!

Show that $I_{m} = \int_0^{\frac{\pi}{2}} \ln[\sin(x)] \sin^{2m}(x)~dx = \frac{\pi \binom{2m}{m} }{4^m} [( \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - ... + \frac{1}{2m-1} - \frac{1}{2m} )- \ln2 ]$

and $I_0 = -\frac{\pi}{2}\ln2$

The integral on the LHS is:
$
-\frac{1}{4} \sum_{n=1}^{\infty} \frac{\beta(m+\frac{1}{2},n+\frac{1}{2})}{n}
$
• August 27th 2009, 05:53 AM
simplependulum
Quote:

Originally Posted by luobo
The integral on the LHS is:
$
-\frac{1}{4} \sum_{n=1}^{\infty} \frac{\beta(m+\frac{1}{2},n+\frac{1}{2})}{n}
$

Oh i see you have made use of this identity (Happy) $\sin^2(x) = 1 - \cos^2(x)$

I wish to know how can we apply magic differentiation !

and here is another expression for L.H.S.

since
$\ln[\sin(x)] = \lim_{a\to0} \frac{1}{a} [ (\sin(x))^a - 1]$

the integral become $\lim_{a\to0}\frac{1}{2a} [ \beta( \frac{a}{2} + m + \frac{1}{2} , \frac{1}{2} ) - \beta( m + \frac{1}{2} , \frac{1}{2})] = \frac{1}{2} \lim_{a\to0} [ \beta( \frac{a}{2} + m + \frac{1}{2} , \frac{1}{2} ) ]'$
• August 27th 2009, 12:29 PM
luobo
Quote:

Originally Posted by simplependulum
It should be in the sub-forum "Interesting Math Problem" but there is no such place for me to post (Rofl) It is not very hard but the answer is extremely elegant !!

Show that $I_{m} = \int_0^{\frac{\pi}{2}} \ln[\sin(x)] \sin^{2m}(x)~dx = \frac{\pi \binom{2m}{m} }{4^m} [( \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - ... + \frac{1}{2m-1} - \frac{1}{2m} )- \ln2 ]$

and $I_0 = -\frac{\pi}{2}\ln2$

Here is another way, use this formula

$
\ln \sin x = -\ln 2 - \sum_{k=1}^{\infty} \frac{\cos(2kx)}{k}
$

Then use the following orthogonality
$
\int_0^{\frac{\pi}{2}} \cos(2kx) \cos(2nx) \ dx = \frac{\pi}{4} \ \delta_{kn}
$

Interested people can try, but I am just too busy to do so (Happy).