Page 5 of 8 FirstFirst 12345678 LastLast
Results 61 to 75 of 106

Math Help - Prove some identities!

  1. #61
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by chiph588@ View Post
    I did the same thing except got  \displaystyle I(\alpha,\beta) = 2\arctanh^2\left(\frac{a-1}{\sqrt{a^2-1}}\right)-2\arctanh^2\left(\frac{b-1}{\sqrt{b^2-1}}\right) .
    I get that our integral is equal to \arccos(\beta)^2-\arccos(\alpha)^2.
    Follow Math Help Forum on Facebook and Google+

  2. #62
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
     2 \int^{\pi /2}_{0} \sec(x) \ln \big(\frac{1+ \alpha \cos x}{1+\beta \cos x}\big) \ dx =  2 \int_{0}^{\pi /2} \int^{\alpha}_{\beta} \frac{dt \ dx }{1+t \cos x }}
    Follow Math Help Forum on Facebook and Google+

  3. #63
    Member
    Joined
    Nov 2009
    Posts
    106
    Quote Originally Posted by Drexel28 View Post
    Yes, the obvious way is to use CA. One question, if you know CA how do you not know multiple integrals?
    We don't use multiple integrals in complex analysis (at least in the course in the university in which I study), and so it happened that I'll learn multiple integrals only after complex analysis.

    Quote Originally Posted by Drexel28 View Post
    Woah, your answer looks completely different than mine. Are you sure it's correct? Not doubting your, but it's just really different.
    I checked one value numerically with wolfram|alpha (for \alpha = 0.5\:\:\:\:,\:\:\:\:\beta=0. It can't do the integral in closed form ) and it seemed correct. Probably you can differentiate your answer w.r.t. alpha, my answer (although differentiating my answer looks like hell) and see they're the same.


    Quote Originally Posted by chiph588@ View Post
    I did magic differentiation and got  \displaystyle I(\alpha,\beta) = 4\left(\tanh^{-1}\left(\frac{a-1}{\sqrt{a^2-1}}\right)\right)^2-4\left(\tanh^{-1}\left(\frac{b-1}{\sqrt{b^2-1}}\right)\right)^2 .
    Your form looks much nicer

    Edit: Oh, that's even better:

    Quote Originally Posted by Drexel28 View Post
    I get that our integral is equal to \arccos(\beta)^2-\arccos(\alpha)^2.
    Follow Math Help Forum on Facebook and Google+

  4. #64
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by Drexel28 View Post
    I get that our integral is equal to \arccos(\beta)^2-\arccos(\alpha)^2.
    Hmm, it turns out  \displaystyle 4\left(\tanh^{-1}\left(\frac{a-1}{\sqrt{a^2-1}}\right)\right)^2 = -\arccos^2(a) , for  a>0 . I'd be interested to see what you did.
    Follow Math Help Forum on Facebook and Google+

  5. #65
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Random Variable View Post
     2 \int^{\pi /2}_{0} \sec(x) \ln \big(\frac{1+ \alpha \cos x}{1+\beta \cos x}\big) \ dx =  2 \int_{0}^{\pi /2} \int^{\alpha}_{\beta} \frac{dt \ dx }{1+t \cos x }}

    Yes, this is how I did it. And then interchanging th order of integration we get


    \displaystyle \begin{aligned}2\int_{\beta}^{\alpha}\int_0^{\frac  {\pi}{2}}\frac{1}{1+t\cos(x)}\text{ }dx\text{ }dt &= 2\int_{\beta}^{\alpha}\int_0^1\frac{2}{1+t\frac{1-u^2}{1+u^2}}\frac{2}{1+u^2}\text{ }du\text{ }dx\\ &= 4\int_{\beta}^{\alpha}\int_0^1\frac{1}{1-t}\frac{1}{\frac{1+t}{1-t}+u^2}\text{ }du\text{ }dt\\ &= 4\int_{\beta}^{\alpha}\arctan\left(\sqrt{\frac{1-t}{1+t}\right)\frac{1}{\sqrt{1-t^2}}\text{ }dt\\ &= 2\int_{\beta}^{\alpha}\frac{\arccos(w)}{\sqrt{1-w^2}}\text{ }dw\\ &= \arccos(\beta)^2-\arccos(\alpha)^2\end{aligned}
    Follow Math Help Forum on Facebook and Google+

  6. #66
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by chiph588@ View Post
    Hmm, it turns out  \displaystyle 4\left(\tanh^{-1}\left(\frac{a-1}{\sqrt{a^2-1}}\right)\right)^2 = -\arccos^2(a) , for  a>0 . I'd be interested to see what you did.
    Well, it's commonly known (I should have noticed this in Unbeatable's answer) that \displaystyle \arccos(x)=\frac{1}{2}\arctan\left(\frac{\sqrt{1-x}}{\sqrt{1+x}}\right). Maybe there's a way to use this to derive your result. Using this one can see that Unbeatable's answer is totally correct.
    Follow Math Help Forum on Facebook and Google+

  7. #67
    Member
    Joined
    Nov 2009
    Posts
    106
    Quote Originally Posted by Drexel28 View Post
    Well, it's commonly known (I should have noticed this in Unbeatable's answer) that \displaystyle \arccos(x)=\frac{1}{2}\arctan\left(\frac{\sqrt{1-x}}{\sqrt{1+x}}\right). Maybe there's a way to use this to derive your result. Using this one can see that Unbeatable's answer is totally correct.
    The \frac{1}{2} should have been on the other side of the equality.

    It follows that:

    \tanh^{-1}\frac{a-1}{\sqrt{a^2-1}} = -i\tan^{-1}\sqrt\frac{1-a}{1+a} = -\frac{1}{2}i\arccos(a), deriving Chiph's result
    Last edited by Unbeatable0; December 27th 2010 at 02:19 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #68
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Is the answer still valid if  \alpha, \beta >1 ?

    Quote Originally Posted by Drexel28 View Post
    Yes, this is how I did it. And then interchanging th order of integration we get


    \displaystyle \begin{aligned}2\int_{\beta}^{\alpha}\int_0^{\frac  {\pi}{2}}\frac{1}{1+t\cos(x)}\text{ }dx\text{ }dt &= 2\int_{\beta}^{\alpha}\int_0^1\frac{2}{1+t\frac{1-u^2}{1+u^2}}\frac{2}{1+u^2}\text{ }du\text{ }dx\\ &= 4\int_{\beta}^{\alpha}\int_0^1\frac{1}{1-t}\frac{1}{\frac{1+t}{1-t}+u^2}\text{ }du\text{ }dt\\ &= 4\int_{\beta}^{\alpha}\arctan\left(\sqrt{\frac{1-t}{1+t}\right)\frac{1}{\sqrt{1-t^2}}\text{ }dt\\ &= 2\int_{\beta}^{\alpha}\frac{\arccos(w)}{\sqrt{1-w^2}}\text{ }dw\\ &= \arccos(\beta)^2-\arccos(\alpha)^2\end{aligned}
    Follow Math Help Forum on Facebook and Google+

  9. #69
    Member
    Joined
    Nov 2009
    Posts
    106
    Quote Originally Posted by Random Variable View Post
    Is the answer still valid if  \alpha, \beta >1 ?
    if a>1 you can use \arccos(a) = -i \ln(a+\sqrt{a^2-1})
    Follow Math Help Forum on Facebook and Google+

  10. #70
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    But the integrand is real-valued when  \alpha, \beta > 0. That doesn't make any sense, does it?
    Follow Math Help Forum on Facebook and Google+

  11. #71
    Member
    Joined
    Nov 2009
    Posts
    106
    Quote Originally Posted by Random Variable View Post
    But the integrand is real-valued when  \alpha, \beta > 0. That doesn't make any sense, does it?
    But you're squaring, and thus the i disappears
    Follow Math Help Forum on Facebook and Google+

  12. #72
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Oh, I thought a complex number was being squared, not necessarily a purely imaginary number. My bad.
    Follow Math Help Forum on Facebook and Google+

  13. #73
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by Drexel28 View Post
    Problem \mathbf{(\star\star)}: Compute

    \displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2+a^2}
    Spoiler:

    We have  \displaystyle \frac{\sin\pi a}{\pi a} = \prod_{n=1}^\infty \left(1-\frac{a^2}{n^2}\right) \implies \frac{\sinh \pi a}{\pi a} = \frac{\sin\pi ia}{\pi ia} = \prod_{n=1}^\infty \left(1+\frac{a^2}{n^2}\right) .

    Taking logs and differentiating, we get  \displaystyle \left(\log\frac{\sinh \pi a}{\pi a}\right)' = \sum_{n=1}^\infty \left(\log\left\{1+\frac{a^2}{n^2}\right\}\right)' = \sum_{n=1}^\infty \frac{2a}{n^2+a^2} .

    Thus  \displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2+a^2} = \frac1{2a}\left[\pi\coth\pi a-\frac1a\right] .
    Follow Math Help Forum on Facebook and Google+

  14. #74
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Problem (\star\star\star\star): For  |x|<1 , compute

     \displaystyle \sum_{s=2}^\infty \frac{\zeta(s)}sx^s
    Last edited by chiph588@; December 27th 2010 at 06:09 PM.
    Follow Math Help Forum on Facebook and Google+

  15. #75
    Super Member
    Joined
    Jan 2009
    Posts
    715
    The answer is  \displaystyle \ln[ \Gamma(1-x) ] - \gamma x But the way I evaluate is to use the infinite product of Gamma function which makes me confused .

    For this reason , i attempt to solve it without using the infinite product :

    First , I need to obtain the expression of Digamma function :


     \displaystyle   \Gamma'(z) = \int_0^{\infty} e^{-t} t^{z-1} \ln(t) ~dt

    Since  \displaystyle   \ln(t) = \int_0^{\infty} \frac{e^{-x} - e^{-xt} }{x}~dx substituting it we obtain :

     \displaystyle   \Gamma'(z) = \int_0^{\infty}\int_0^{\infty}(e^{-x} - e^{-xt})e^{-t} t^{z-1} ~dt~\frac{dx}{x}

    \displaystyle   = \Gamma(z)  \int_0^{\infty} ( e^{-x} - \frac{1}{(x+1)^z} ) ~\frac{dx}{x}

    Then \displaystyle    \frac{ \Gamma'(1)}{\Gamma(1)}  - \frac{ \Gamma'(z)}{\Gamma(z)}  = \int_0^{\infty} \left( \frac{1}{(x+1)^z} - \frac{1}{x+1} \right) ~ \frac{dx}{x}

    Sub.  x+1 = e^{y} we have

     \displaystyle  \frac{ \Gamma'(1)}{\Gamma(1)}  - \frac{ \Gamma'(z)}{\Gamma(z)}  = \int_0^{\infty} \frac{e^{(1-z)y} - 1 }{ e^y -1 }~dy

    Therefore , \displaystyle   \int_0^{\infty} \frac{e^{(1-z)y} - 1 }{ e^y -1 }~dy = -\gamma  -  \frac{ \Gamma'(z)}{\Gamma(z)}


    Let's get back to your problem :

    Let \displaystyle   F(x) = \sum_{s=2}^{\infty} \zeta(s) x^{s-1}

     \displaystyle  = \int_0^{\infty} \sum_{s=2}^{\infty} \frac{(xt)^{s-1}}{(s-1)!} ~\frac{dt}{e^t - 1 }

     \displaystyle  = \int_0^{\infty} \frac{ e^{xt} - 1 }{e^t - 1 } ~dt

     \displaystyle  = - \gamma  -  \frac{ \Gamma'(1-x)}{\Gamma(1-x)}

    Integrating it to obtain the result :

      \displaystyle  \sum_{s=2}^{\infty} \frac{\zeta(s)}{s} x^s = \ln[\Gamma(1-x)] - \gamma x


    Using the similar method , Prove this formula ( Gauss's Multiplication Formula ) :

     \displaystyle   \Gamma(nz) = n^{nz - \frac{1}{2} } (2\pi)^{\frac{1-n}{2}  \prod_{k=0}^{n-1} \Gamma{\left( z + \frac{k}{n} \right)} ~ n \in \mathbb{N}

    ps I don't know how many stars it should be given ...
    Last edited by simplependulum; December 27th 2010 at 08:20 PM.
    Follow Math Help Forum on Facebook and Google+

Page 5 of 8 FirstFirst 12345678 LastLast

Similar Math Help Forum Discussions

  1. Can't prove these identities?
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: November 20th 2010, 08:07 AM
  2. prove the following identities.
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: June 23rd 2010, 12:15 AM
  3. Can someone help me prove these two identities?
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: April 7th 2010, 01:48 PM
  4. prove identities
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: June 26th 2009, 12:08 AM
  5. Prove identities, can anyone help?
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: February 14th 2009, 02:04 PM

Search Tags


/mathhelpforum @mathhelpforum