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Math Help - Interesting integral

  1. #1
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    Interesting integral

    It should be in the sub-forum "Interesting Math Problem" but there is no such place for me to post 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
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  2. #2
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    Quote Originally Posted by simplependulum View Post

    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. since i can't be wrong, so you must be wrong then! haha
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    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. 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.
    Last edited by mr fantastic; September 18th 2009 at 08:42 AM. Reason: Restored original reply
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  4. #4
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    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!
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  5. #5
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    Thanks for your corrections , i think i need to have a double check after i finish a problem next time .
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  6. #6
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    Quote Originally Posted by simplependulum View Post
    It should be in the sub-forum "Interesting Math Problem" but there is no such place for me to post 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:
    <br />
-\frac{1}{4} \sum_{n=1}^{\infty} \frac{\beta(m+\frac{1}{2},n+\frac{1}{2})}{n}<br />
    Last edited by mr fantastic; September 18th 2009 at 08:42 AM. Reason: Restored original reply
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  7. #7
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    Quote Originally Posted by luobo View Post
    The integral on the LHS is:
    <br />
-\frac{1}{4} \sum_{n=1}^{\infty} \frac{\beta(m+\frac{1}{2},n+\frac{1}{2})}{n}<br />
    Oh i see you have made use of this identity  \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} )  ]'
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  8. #8
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    Quote Originally Posted by simplependulum View Post
    It should be in the sub-forum "Interesting Math Problem" but there is no such place for me to post 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

    <br />
\ln \sin x = -\ln 2 - \sum_{k=1}^{\infty} \frac{\cos(2kx)}{k}<br />

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

    Interested people can try, but I am just too busy to do so .
    Last edited by mr fantastic; September 18th 2009 at 08:43 AM. Reason: Restored original reply
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