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Thread: Prove some identities!

  1. #91
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    I can 'explain' how the substitution works $\displaystyle x = \frac{1-t}{1+t}$ but actually it is not the explanation , i am going to change the integral to a form which is more intuitive .

    Before this , let me ask you about another integral :

    $\displaystyle I = \int_0^{\pi/2} \frac{\sin(x)}{\sin(x) + \cos(x)} ~dx $

    Intuition or experiences or your friend tell you to subsitute $\displaystyle x = \frac{\pi}{2} - y $ so we have

    $\displaystyle I = \int_0^{\pi/2} \frac{\cos(y)}{\cos(y)+ \sin(y)} ~dy $

    $\displaystyle = \int_0^{\pi/2} \frac{\cos(y) + \sin(y) }{\cos(y)+ \sin(y)} ~dy - \int_0^{\pi/2} \frac{\sin(y)}{\cos(y)+ \sin(y)} ~dy $

    $\displaystyle = \frac{\pi}{2} - I $ so $\displaystyle I = \frac{\pi}{4} $

    Here is another one :

    $\displaystyle I = \int_0^{\pi} \frac{x\sin(x)}{1+ \cos^2(x) } ~dx $

    Again , they tell us to subsitute $\displaystyle x = \pi - y $

    $\displaystyle I = \int_0^{\pi} \frac{(\pi - y)\sin(y) }{1 + \cos^2(y) }~dy $

    $\displaystyle = \pi \int_0^{\pi} \frac{\sin(y)}{1 + \cos^2(y) }~dy - I $

    $\displaystyle = \pi \left[ - \tan^{-1}[\cos(y)] \right]_0^{\pi} - I $

    $\displaystyle = \frac{\pi^2}{2} - I $ so

    $\displaystyle I = \frac{\pi^2}{4} $

    The tricks of the above two integrals are very similar , what we do is to 'reflect' the integrand to obtain the answers . But we can't explain why one could think of , our intuition always cannot be explained .

    This one is very similar .

    Sub. $\displaystyle x = \tan(y) $

    $\displaystyle dx = \sec^2(y) ~dy $ we have

    $\displaystyle I = \int_0^{\pi/4} \ln[1 + \tan(y) ] ~dy $

    Again , sub $\displaystyle y = \frac{\pi}{4} - z $ You may find that it is equivalent to sub $\displaystyle x = \frac{1-t}{1+t} $ at the begining as $\displaystyle \tan( \frac{\pi}{4} - \theta ) = \frac{1 - \tan(\theta) }{1 + \tan(\theta) } $

    Therefore , $\displaystyle I = \int_0^{\pi/4} \ln{ \left( 1 + \frac{1 - \tan(z) }{1 + \tan(z) \right) }~dz $

    $\displaystyle = \int_0^{\pi/4} \ln(2) ~dz - I $

    $\displaystyle = \frac{\pi \ln(2)}{4} - I $ so

    $\displaystyle I = \frac{\pi \ln(2)}{8} $

    You see , these three integrals are not different at all so if you are asking how this works , let me ask you how this also works when applied in the first two integrals .
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  2. #92
    MHF Contributor chiph588@'s Avatar
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    Compute $\displaystyle \displaystyle \int_0^1 \frac{\log^ax}{x^2+1} dx $, for $\displaystyle a\geq0 $.
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  3. #93
    Super Member Random Variable's Avatar
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    The substitutions exploit the symmetries of the integrals. Is that what you're getting at?
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  4. #94
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Compute $\displaystyle \displaystyle \int_0^1 \frac{\log^ax}{x^2+1} dx $, for $\displaystyle a\geq0 $.
    Spoiler:


    Let


    $\displaystyle \displaystyle I(a)=\int_0^1\frac{\log^a(x)}{1+x^2}\text{ }dx$


    Note then that for $\displaystyle x\in[0,1]$ we have that


    $\displaystyle \displaystyle \frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^n x^{2n}$


    So that


    $\displaystyle \displaystyle \begin{aligned}I(a) &= \int_0^1\log^a(x)\sum_{n=0}^{\infty}(-1)^n x^{2n}\text{ }dx\\ &= \sum_{n=0}^{\infty}(-1)^n \int_0^1 \log^a(x) x^{2n}\text{ }dx\end{aligned}$


    But, let $\displaystyle y=-log(x)$ to get that


    $\displaystyle \displaystyle \int_0^1\log^a(x)x^{2n}\text{ }dx = (-1)^a\int_{0}^{\infty}y^a e^{-(2n+1)y}\text{ }dy$


    Letting $\displaystyle (2n+1)y=z$ gives that


    $\displaystyle \displaystyle \begin{aligned}(-1)^a\int_0^{\infty}y^a e^{-(2n+1)y}\text{ }dy &= \frac{(-1)^a}{(2n+1)^{a+1}}\int_0^{\infty}z^a e^{-z}\text{ }dz\\ &= \frac{(-1)^a \Gamma(a+1)}{(2n+1)^{a+1}}\end{aligned}$


    Thus,


    $\displaystyle \displaystyle \begin{aligned}I(a) &= \sum_{n=0}^{\infty}(-1)^n \int_0^1 \log^a(x) x^{2n}\text{ }dx\\ &=(-1)^a \Gamma(a+1)\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^{a+1}}\\ &= (-1)^a\Gamma(a+1)\beta(a+1)\end{aligned}$


    Which is now clear since letting $\displaystyle -\log(x)=y$ in our original integral we see that


    $\displaystyle \displaystyle \int_0^1\frac{\log^a(x)}{1+x^2}\text{ }dx=\frac{(-1)^a}{2}\int_0^{\infty}\frac{y^a}{\cosh(x)}\text{ }dx$

    Last edited by Drexel28; Dec 29th 2010 at 01:55 PM.
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  5. #95
    Super Member Random Variable's Avatar
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    How about showing that $\displaystyle \int^{1}_{0} \frac{\ln(x^2-x+1)}{x-1} \ dx = \frac{\pi^{2}}{18} $ ? I know of one approach, but I'm sure there are others.
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  6. #96
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    I find one , first put $\displaystyle x = 1 -t $ we have

    $\displaystyle \int_0^1 \frac{\ln(x^2 - x + 1 ) }{x-1}~dx $

    $\displaystyle = - \int_0^1 \frac{\ln(t^2 - t + 1 )}{ t}~dt $

    $\displaystyle = \int_0^1 \frac{\ln(t+1)}{t}~dt - \int_0^1 \frac{\ln(t^3+1)t^2}{t^3}~dt $

    Sub. $\displaystyle t^3 = u $ in the second integral ( Since the limit at zero of the integrand exists , i think i can make the substitution without changing to $\displaystyle \displaystyle \lim_{a\to 0 } \int_a^1 $ something . )

    $\displaystyle = \int_0^1 \frac{\ln(t+1)}{t}~dt - \frac{1}{3} \int_0^1 \frac{\ln(u+1)}{u}~du $

    $\displaystyle = ( 1 - \frac{1}{3} ) \int_0^1 \frac{\ln(t+1)}{t}~dt = \frac{2}{3} ~ \frac{\pi^2}{12} $

    $\displaystyle = \frac{\pi^2}{18} $
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  7. #97
    Super Member Random Variable's Avatar
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    You always find a better way.

    I let $\displaystyle I(a) = \int^{1}_{0} \frac{\ln\big(a(x^{2}-x)+1\big)}{x-1} \ dx $, differentiated with respect to $\displaystyle a $, integrated, and then used the fact that $\displaystyle I(0)=0.$

    EDIT: It may be be more correct to let $\displaystyle I(a) = \int^{1}_{0} \frac{\ln\Big(\frac{1}{a}(x^2-x)+1\Big)}{x-1} \ dx $ and use the fact that $\displaystyle I(\infty) = 0 $.

    Quote Originally Posted by simplependulum View Post
    I find one , first put $\displaystyle x = 1 -t $ we have

    $\displaystyle \int_0^1 \frac{\ln(x^2 - x + 1 ) }{x-1}~dx $

    $\displaystyle = - \int_0^1 \frac{\ln(t^2 - t + 1 )}{ t}~dt $

    $\displaystyle = \int_0^1 \frac{\ln(t+1)}{t}~dt - \int_0^1 \frac{\ln(t^3+1)t^2}{t^3}~dt $

    Sub. $\displaystyle t^3 = u $ in the second integral ( Since the limit at zero of the integrand exists , i think i can make the substitution without changing to $\displaystyle \displaystyle \lim_{a\to 0 } \int_a^1 $ something . )

    $\displaystyle = \int_0^1 \frac{\ln(t+1)}{t}~dt - \frac{1}{3} \int_0^1 \frac{\ln(u+1)}{u}~du $

    $\displaystyle = ( 1 - \frac{1}{3} ) \int_0^1 \frac{\ln(t+1)}{t}~dt = \frac{2}{3} ~ \frac{\pi^2}{12} $

    $\displaystyle = \frac{\pi^2}{18} $
    Last edited by Random Variable; Dec 29th 2010 at 08:29 PM.
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  8. #98
    MHF Contributor chiph588@'s Avatar
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    Here's one I just came up with. Compute $\displaystyle \displaystyle \sum_{k=0}^\infty (-1)^k \frac{2k+3}{(3k+4)(3k+5)} $.
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  9. #99
    MHF Contributor Drexel28's Avatar
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    And interesting corollary of this is that $\displaystyle \displaystyle \sum_{n=1}^{\infty}\frac{B(n+1,n)}{n}=\sum_{n=1}^{ \infty}\frac{(n!)^2}{n^2 (2n)!}=\frac{\pi^2}{18}$.
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  10. #100
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    And interesting corollary of this is that $\displaystyle \displaystyle \sum_{n=1}^{\infty}\frac{B(n+1,n)}{n}=\sum_{n=1}^{ \infty}\frac{(n!)^2}{n^2 (2n)!}=\frac{\pi^2}{18}$.
    I'm assuming you got to this by a substitution of some kind and then expanded into series. If so, what was your substitution?
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  11. #101
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    I'm assuming you got to this by a substitution of some kind and then expanded into series. If so, what was your substitution?
    No substitution,


    $\displaystyle \displaystyle \begin{aligned}\frac{\pi^2}{18} &= \int_0^1\frac{\log\left(1-x(x-1)\right)}{x-1}\text{ }dx\\ &= \int_0^1\frac{1}{x-1}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}x^n(1-x)^n\\ &= \sum_{n=1}^{\infty}\frac{1}{n}\int_0^1 x^n(1-x)^{n-1}\text{ }dx\\ &= \sum_{n=1}^{\infty}\frac{B(n+1,n)}{n}\\ &= \sum_{n=1}^{\infty}\frac{\Gamma(n+1)\Gamma(n)}{n\G amma(2n+1)}\\ &= \sum_{n=1}^{\infty}\frac{(n!)^2}{n^2 (2n)!}\end{aligned}$


    I believe that every lover of series and integrals should do the following four at least once in their life


    $\displaystyle \displaystyle \sum_{n=1}^{\infty}\frac{n!}{n^k(2n)!},\quad k=0,1,2,4$


    and compute


    $\displaystyle \displaystyle \lim_{k\to\infty}\sum_{n=1}^{\infty}\frac{(n!)^2}{ n^k (2n)!}$
    Last edited by Drexel28; Dec 29th 2010 at 09:33 PM.
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  12. #102
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Here's one I just came up with. Compute $\displaystyle \displaystyle \sum_{k=0}^\infty (-1)^k \frac{2k+3}{(3k+4)(3k+5)} $.
    Not to be a problem spoiler, but is there some trick to compute this? There is a really, really ugly way. But I don't feel like carrying that out haha.
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  13. #103
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    Not to be a problem spoiler, but is there some trick to compute this? There is a really, really ugly way. But I don't feel like carrying that out haha.
    Let's start off with a small hint and see where that gets you. I solved this one by accident when trying to solve the last integral Random Variable posted.
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  14. #104
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    $\displaystyle \displaystyle \sum_{n=1}^{\infty}\frac{n!}{n^k(2n)!},\quad k=0$
    $\displaystyle \displaystyle S_0 = \sum_{n=0}^\infty \frac{n!}{(2n)!} = \sum_{n=0}^\infty \frac1{\binom{2n}{n}n!} $

    Now notice $\displaystyle \displaystyle \binom{2n}{n}^{-1} = (2n+1)\int_0^1t^n(1-t)^n dt $. Thus $\displaystyle \displaystyle S_0 = \int_0^1 \sum_{n=0}^\infty \frac{(2n+1)(t-t^2)^n}{n!} dt = \int_0^1 (-2t^2+2t+1)e^{t-t^2} dt $.

    Unfortunately this has no closed integral but can be whittled down to $\displaystyle \displaystyle S_0 = 1+e^{1/4}\int_0^{1/2}e^{-t^2} dt $.
    Last edited by chiph588@; Dec 29th 2010 at 10:12 PM.
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  15. #105
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    $\displaystyle \displaystyle S_0 = \sum_{n=0}^\infty \frac{n!}{(2n)!} = \sum_{n=0}^\infty \frac1{\binom{2n}{n}n!} $

    Now notice $\displaystyle \displaystyle \binom{2n}{n}^{-1} = (2n+1)\int_0^1t^n(1-t)^n dt $. Thus $\displaystyle \displaystyle S_0 = \int_0^1 \sum_{n=0}^\infty \frac{(2n+1)(t-t^2)^n}{n!} dt = \int_0^1 (-2t^2+2t+1)e^{t-t^2} dt $.

    Unfortunately this has no closed integral but can be whittled down a bit to $\displaystyle \displaystyle S_0 = e^{1/4}\int_0^{1/2}e^{-t^2} dt $.
    Spoiler:



    Let


    $\displaystyle \displaystyle S=\sum_{n=1}^{\infty}\frac{(n!)^2}{(2n)!}$


    Notice then that,


    $\displaystyle \displaystyle \begin{aligned} S &= \sum_{n=1}^{\infty}\frac{n n! (n-1)!}{(2n)!}\\ &= \sum_{n=1}^{\infty}\frac{n \Gamma(n+1)\Gamma(n)}{\Gamma(2n+1)}\\ &= \sum_{n=1}^{\infty}n B(n+1,n)\\ &= \sum_{n=1}^{\infty}n\int_0^1 x^{n-1}(1-x)^n\text{ }dx\\ &= \int_0^1 \frac{1}{x}\sum_{n=1}^{\infty}n\left(x(1-x)\right)^n\\ &= \int_0^1\frac{1-x}{(1-x+x^2)^2}\text{ }dx\\ &= \frac{1}{27}\left(9+2\sqrt{3}\pi\right)\end{aligne d}$


    I'll leave you to verify this last integral.

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