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Thread: Prove some identities! (Round two)

  1. #31
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    It would appear that the answer is just a multiple of that impossible-looking integral.
    I'm a little busy now, but I'll think about it. The methodology is to note that $\displaystyle \displaystyle \sin\left(\pi\sqrt{z}\right)=\pi\sqrt{z}\prod_{n=1 }^{\infty}\left(1-\frac{z}{n^2}\right)$ and thus $\displaystyle \displaystyle \log\left(\sin\left(\pi\sqrt{z}\right)\right)-\log\left(\pi\sqrt{z}\right)=\sum_{n=1}^{\infty}\l og\left(1-\frac{z}{n^2}\right)$ but expanding the logarithm gives $\displaystyle \displaystyle \log\left(\sin\left(\pi\sqrt{z}\right)\right)-\log\left(\pi\sqrt{z}\right)\right)=-\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{z^m}{n ^{2m}m}$ doing the quick verification that the inner sum can be interchanged this implies that $\displaystyle \displaystyle \log\left(\sin\left(\pi\sqrt{z}\right)\right)-\log\left(\pi\sqrt{z}\right)\right)=\sum_{m=1}^{\i nfty}\frac{\zeta(2m)}{m}z^m$. Integrating both sides from $\displaystyle 0$ to $\displaystyle 1$ etc. etc
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  2. #32
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Random Variable View Post
    It would appear that the answer is just a multiple of that impossible-looking integral.
    $\displaystyle \displaystyle \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n} \implies \log(\pi \sqrt{x}) - \log(\sin(\pi \sqrt{x})) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{n} $ $\displaystyle \displaystyle \implies \sum_{n=1}^\infty \frac{\zeta(2n)}{n(n+1)} = \int_0^1 \log(\pi\sqrt{x}) dx - \int_0^1 \log(\sin(\pi \sqrt{x})) dx $

    For the second integral, let $\displaystyle t=\pi\sqrt{x} \implies dt = \frac\pi{2\sqrt{x}}dx \implies dx = \frac2{\pi^2} t dt $

    Thus the second integral equals $\displaystyle \displaystyle \frac2{\pi^2}\int_0^\pi t\log(\sin t) dt $.
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  3. #33
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    $\displaystyle \displaystyle \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n} \implies \log(\pi \sqrt{x}) - \log(\sin(\pi \sqrt{x})) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{n} $ $\displaystyle \displaystyle \implies \sum_{n=1}^\infty \frac{\zeta(2n)}{n(n+1)} = \int_0^1 \log(\pi\sqrt{x}) dx - \int_0^1 \log(\sin(\pi \sqrt{x})) dx $

    For the second integral, let $\displaystyle t=\pi\sqrt{x} \implies dt = \frac\pi{2\sqrt{x}}dx \implies dx = \frac2{\pi^2} t dt $

    Thus the second integral equals $\displaystyle \displaystyle \frac2{\pi^2}\int_0^\pi t\log(\sin t) dt $.
    Gah! I made a stupid Calc I error substituting! Haha! Anways, let's finish this

    Spoiler:

    Let $\displaystyle \displaystyle S=\sum_{n=1}^{\infty}\frac{\zeta(2n)}{n(n+1)}$ then we have shown that



    $\displaystyle \displaystyle S=\log(\pi)-\frac{1}{2}-I$



    Where


    $\displaystyle \displaystyle I=\frac{2}{\pi^2}\int_0^{\pi}t\log(\sin(t))\text{ }dt$


    Letting $\displaystyle \pi-y=t$ in this integral gives


    $\displaystyle \displaystyle I=\frac{2}{\pi^2}\int_0^\pi\left(\pi-y\right)\log\left(\sin(t)\right)\text{ }dt=\frac{2}{\pi}\int_0^{\pi}\log\left(\sin(t)\rig ht)\text{ }dt-I$


    And thus


    $\displaystyle \displaystyle \pi I=\int_0^{\pi}\log\left(\sin(t)\right)\text{ }dt$


    Let $\displaystyle \displaystyle t=\frac{\pi}{2}-y$ to get that


    $\displaystyle \displaystyle \pi I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\log\left(\cos(y)\right)\t ext{ }dy$


    But, since $\displaystyle \cos(y)$ is symmetric on $\displaystyle \displaystyle \left[\frac{-\pi}{2},\frac{\pi}{2}\right]$ we get that


    $\displaystyle \displaystyle \frac{\pi}{2}I=\int_0^{\frac{\pi}{2}}\log\left(\co s(y)\right)\text{ }dy$


    Letting $\displaystyle \displaystyle y=\frac{\pi}{2}-u$ gives that


    $\displaystyle \displaystyle \frac{\pi}{2}I=\int_0^{\frac{\pi}{2}}\log\left(\si n(u)\right)\text{ }du$



    This integral is well-known (see here fore example) and so we get


    $\displaystyle \displaystyle I=\frac{\frac{-\pi\log(2)}{2}}{\frac{\pi}{2}}=-\log(2)$


    So, finally we may conclude that


    $\displaystyle \displaystyle S=\log(2\pi)-\frac{1}{2}$

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  4. #34
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    Gah! I made a stupid Calc I error substituting! Haha! Anways, let's finish this

    Spoiler:

    Let $\displaystyle \displaystyle S=\sum_{n=1}^{\infty}\frac{\zeta(2n)}{n(n+1)}$ then we have shown that



    $\displaystyle \displaystyle S=\log(\pi)-\frac{1}{2}-I$



    Where


    $\displaystyle \displaystyle I=\frac{2}{\pi^2}\int_0^{\pi}t\log(\sin(t))\text{ }dt$


    Letting $\displaystyle \pi-y=t$ in this integral gives


    $\displaystyle \displaystyle I=\frac{2}{\pi^2}\int_0^\pi\left(\pi-y\right)\log\left(\sin(t)\right)\text{ }dt=\frac{2}{\pi}\int_0^{\pi}\log\left(\sin(t)\rig ht)\text{ }dt-I$


    And thus


    $\displaystyle \displaystyle \pi I=\int_0^{\pi}\log\left(\sin(t)\right)\text{ }dt$


    Let $\displaystyle \displaystyle t=\frac{\pi}{2}-y$ to get that


    $\displaystyle \displaystyle \pi I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\log\left(\cos(y)\right)\t ext{ }dy$


    But, since $\displaystyle \cos(y)$ is symmetric on $\displaystyle \displaystyle \left[\frac{-\pi}{2},\frac{\pi}{2}\right]$ we get that


    $\displaystyle \displaystyle \frac{\pi}{2}I=\int_0^{\frac{\pi}{2}}\log\left(\co s(y)\right)\text{ }dy$


    Letting $\displaystyle \displaystyle y=\frac{\pi}{2}-u$ gives that


    $\displaystyle \displaystyle \frac{\pi}{2}I=\int_0^{\frac{\pi}{2}}\log\left(\si n(u)\right)\text{ }du$



    This integral is well-known (see here fore example) and so we get


    $\displaystyle \displaystyle I=\frac{\frac{-\pi\log(2)}{2}}{\frac{\pi}{2}}=-\log(2)$


    So, finally we may conclude that


    $\displaystyle \displaystyle S=\log(2\pi)-\frac{1}{2}$

    Wow! You made that integral look easy!
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  5. #35
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Wow! You made that integral look easy!
    I actually found a much better way.
    Spoiler:



    Looking at my last solution see that we have


    $\displaystyle \displaystyle \pi I=\int_0^\pi \log(\sin(t))\text{ }dt$


    and


    $\displaystyle \displaystyle \frac{\pi}{2}I=\int_0^{\frac{\pi}{2}}\log\left(\si n(t)\right)\text{ }dt=\int_0^{\frac{\pi}{2}}\log\left(\cos(t)\right) \text{ }dt$


    Thus, from the equality immediately preceding this sentence


    $\displaystyle \displaystyle \begin{aligned}\pi I &= \int_0^{\frac{\pi}{2}}\left(\log(\sin(t)\right)+\l og\left(\cos(t)\right)\right)\text{ }dt\\ &= \int_0^{\frac{\pi}{2}}\log\left(\frac{1}{2}\sin\le ft(2t\right)\right)\\ &= \frac{-\pi\log(2)}{2}+\int_0^{\frac{\pi}{2}}\log\left(\si n(2t)\right)\text{ }dt\\ &= \frac{-\pi\log(2)}{2}+\frac{1}{2}\int_0^\pi \log(\sin(w))\text{ }dw\quad\quad 2t\mapsto w\\ &= \frac{-\pi\log(2)}{2}+\frac{\pi}{2}I\end{aligned}$


    Solving we find that $\displaystyle \displaystyle I=-\log(2)$ as desired. Thus, we've completely circumvented the need to calculate$\displaystyle \displaystyle \int_0^{\frac{\pi}{2}}\log\left(\sin(x)\right)\tex t{ }dx$ directly, and have even found an alternate way of computing it.


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  6. #36
    MHF Contributor chiph588@'s Avatar
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    This isn't too hard to calculate, but I think it's a good integral to know. Compute $\displaystyle \displaystyle \int_0^\infty \frac{x^n}{(ax+b)^m} dx $.
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  7. #37
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    This isn't too hard to calculate, but I think it's a good integral to know. Compute $\displaystyle \displaystyle \int_0^\infty \frac{x^n}{(ax+b)^m} dx $.
    Of course assuming $\displaystyle m>n+1$. I'll assume that $\displaystyle b\ne 0$ since this is trivial otherwise. Similarly, I'll assume that $\displaystyle \frac{a}{b}>0$.

    Spoiler:



    Let


    $\displaystyle \displaystyle I=\int_0^{\infty}\frac{x^n}{(ax+b)^m}\text{ }dx$


    Then,


    $\displaystyle \displaystyle b^mI=\int_0^{\infty}\frac{x^n}{(1+cx)^m}\text{ }dx$


    where $\displaystyle \displaystyle c=\frac{a}{b}$. Let $\displaystyle 1+cx=z$ so that



    $\displaystyle \displaystyle c^{n+1}b^mI=\int_1^{\infty}\frac{(z-1)^n}{z^m}\text{ }dz$


    Let $\displaystyle \displaystyle z=\frac{1}{u}$ to get that



    $\displaystyle \displaystyle \begin{aligned}c^{n+1}b^m I &=\int_0^1 u^{m-2}\left(\frac{1}{u}-1\right)^n\text{ }du\\ &=\int_0^1 u^{m-2-n}(1-u)^n\text{ }du\\ &=B\left(m-1-n,n+1\right)\end{aligned}$


    and thus


    $\displaystyle \displaystyle I=\frac{b^{m-n-1}}{a^{n+1}}B\left(m-1-n,m+1\right)$

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  8. #38
    MHF Contributor Drexel28's Avatar
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    This one's not too bad.

    Compute:


    $\displaystyle \displaystyle \int_0^{\infty}x\sin\left(ax^2\right)\sin\left(2bx \right)\text{ }dx$
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  9. #39
    Super Member Random Variable's Avatar
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    Are you sure that integral converges?
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  10. #40
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    Are you sure that integral converges?
    Assuming I didn't miscalculate something? Anyways, try this integral. It's much more surprising (albeit not too hard)!


    $\displaystyle \displaystyle \int_0^1\log\left(\Gamma(x)\right)\cos\left(2\pi n x\right)\text{ }dx,\quad n\in\mathbb{N}$
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  11. #41
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    Quote Originally Posted by Drexel28 View Post

    $\displaystyle \displaystyle \int_0^1\log\left(\Gamma(x)\right)\cos\left(2\pi n x\right)\text{ }dx,\quad n\in\mathbb{N}$


    $\displaystyle \displaystyle I= \int_0^1 \ln[\Gamma(x)]\cos(2\pi nx)~dx $



    $\displaystyle \displaystyle = \frac{1}{2} [ \int_0^1 \ln[\Gamma(x)]\cos(2\pi nx)~dx + \int_0^1 \ln[\Gamma(x)]\cos(2\pi nx)~dx ] $



    Sub. $\displaystyle x = 1-t $ on the second integral and change the dummy variable , we have



    $\displaystyle \displaystyle I= \frac{1}{2} [ \int_0^1 \ln[\Gamma(x)]\cos(2\pi nx)~dx + \int_0^1 \ln[\Gamma(1-x)]\cos(2\pi nx)~dx$



    $\displaystyle \displaystyle = \frac{1}{2} \int_0^1 \ln[ \Gamma(x) \Gamma(1-x) ] \cos(2\pi nx ) ~dx $



    $\displaystyle \displaystyle = \frac{1}{2} \int_0^1 [ \ln(\pi) - \ln[\sin(\pi x)] ] \cos(2\pi nx ) ~dx $




    $\displaystyle \displaystyle = - \frac{1}{2\pi} \int_0^{\pi} \ln[\sin(x)] \cos(2nx)~dx $



    Integration by parts



    $\displaystyle \displaystyle = - \frac{1}{4n\pi} \left[ \ln[\sin(x)] \sin(2nx)\right]_0^\pi + \frac{1}{4n\pi} \int_0^{\pi} \frac{\sin(2nx)\cos(x)}{\sin(x)}~dx $



    $\displaystyle \displaystyle = \frac{1}{8n\pi} \int_0^{\pi} \frac{\sin(2n+1)x + \sin(2n-1)x}{\sin(x)}~dx $



    $\displaystyle \displaystyle = \frac{1}{8n\pi} (2\pi) = \frac{1}{4n} $



    Problem :

    How do you prove this

    $\displaystyle \displaystyle \frac{\Gamma(x) \Gamma(y) }{\Gamma(x+y)} = \int_0^1 t^{x-1} (1-t)^{y-1} ~dt $

    My solution is to make a substitution , followed by changing the order of a double integral .
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  12. #42
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by simplependulum View Post

    Problem :

    How do you prove this

    $\displaystyle \displaystyle \frac{\Gamma(x) \Gamma(y) }{\Gamma(x+y)} = \int_0^1 t^{x-1} (1-t)^{y-1} ~dt $

    My solution is to make a substitution , followed by changing the order of a double integral .
    Let $\displaystyle \displaystyle B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\text{ }dt$. It's easy to show that $\displaystyle \displaystyle B(x,y)=\frac{x+y}{y}B(x,y+1)$ and more so by induction that $\displaystyle \displaystyle B(x,y)=\frac{(x+y)_n}{n! n^{x+y-1}}\frac{n! n^{y-1}}{(y)_n}\int_0^n t^{x-1}\left(1-\frac{t}{n}\right)^{n+y-1}\text{ }dt$. But, by definition (I guess it really depends upon your definition of the Gamma function, but this is a common one) it follows that $\displaystyle \displaystyle B(x,y)=\frac{\Gamma(y)}{\Gamma(x+y)}\lim_{n\to\inf ty}\int_0^n t^{x-1}\left(1-\frac{t}{n}\right)^{y+n-1}\text{ }dt$ but one can verify (using the LDC theorem for example) that this integral tends to $\displaystyle \displaystyle \int_0^{\infty}t^{x-1}e^{-t}\text{ }dt$ from where the conclusion follows.
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  13. #43
    Super Member Random Variable's Avatar
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    $\displaystyle \int^{1}_{0} t^{x-1} (1-t)^{y-1} \ dt $


    let $\displaystyle t = \sin^{2} \theta $

    $\displaystyle = 2 \int^{\pi /2}_{0} (\sin^{2} \theta)^{r-1} (1-\sin^{2} \theta)^{s-1} \sin \theta \cos \theta \ d \theta $

    $\displaystyle = 2 \int^{\pi /2}_{0} (\sin \theta)^{2r-1} (\cos \theta)^{2s-1} \ d \theta $


    $\displaystyle \Gamma(x) = \int^{\infty}_{0} t^{x-1} e^{-t} \ dt $

    let $\displaystyle t = u^{2} $

    $\displaystyle \Gamma (x) = \int^{\infty}_{0} u^{2x-2} e^{-u^{2}} (2 u \ du)$

    $\displaystyle = 2 \int^{\infty}_{0} u^{2x-1} e^{-u^{2}} \ du $


    then $\displaystyle \Gamma(x) \Gamma(y) = 4 \int^{\infty}_{0} u^{2x-1} e^{-u^{2}} \ du \int^{\infty}_{0} v^{2y-1} e^{-v^{2}} \ dv $

    $\displaystyle = 4 \int^{\infty}_{0} \int^{\infty}_{0} u^{2x-1} v^{2y-1} e^{-(u^{2}+v^{2})} \ du \ dv $

    change to polar coordinates

    $\displaystyle \Gamma(x) \Gamma (y) = 4 \int^{\pi / 2}_{0} \int^{\infty}_{0} (r \cos \theta)^{2x-1} \ (r \sin \theta)^{2y-1} e^{-r^{2}} r \ dr \ d \theta $

    $\displaystyle = 4 \int^{\pi /2}_{0} \int^{\infty}_{0} (cos \theta)^{2x-1} \ (\sin \theta)^{2y-1} \ r^{2(x+y)-1} e^{-r^{2}} \ dr \ d \theta $

    $\displaystyle = 2 \Gamma(x+y) \int^{\pi /2}_{0} (cos \theta)^{2x-1} \ (\sin \theta)^{2y-1} \ d \theta $

    $\displaystyle = \Gamma (x+y) \int^{1}_{0} t^{x-1} (1-t)^{y-1} \ dt $
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  14. #44
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    This one's not too bad.

    Compute:


    $\displaystyle \displaystyle \int_0^{\infty}x\sin\left(ax^2\right)\sin\left(2bx \right)\text{ }dx$
    Here's a thought: $\displaystyle \displaystyle x\sin(ax^2)\sin(2bx) = \frac{x}2 (\cos(a x^2-2 b x)-\cos(a x^2+2 b x)) $

    $\displaystyle = \frac{x}2\left(\cos\left(\frac{(a x+b)^2}a-\frac{b^2}a\right)-\cos\left(\frac{(a x+b)^2}a-\frac{b^2}a\right)\right) $

    Now split the cosines up into its exponentials.

    I could be wrong, and won't have time to work it out... I'll be away from a computer for a week.
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  15. #45
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    Here is my method :

    $\displaystyle \displaystyle I = \int_0^1 t^{x-1} (1-t)^{y-1}~dt $


    Sub. $\displaystyle \displaystyle t = \frac{1}{u+1} $

    $\displaystyle \displaystyle I = \int_0^{\infty} \frac{u^{y-1}}{(u+1)^{x+y}}~du $


    Consider $\displaystyle \displaystyle \Gamma(x+y) = \int_0^{\infty} e^{-z} z^{x+y-1}~dz $

    Sub. $\displaystyle z = (u+1)s $ , $\displaystyle u $ as parameter , we have

    $\displaystyle \displaystyle \frac{\Gamma(x+y)}{(u+1)^{x+y}} = \int_0^{\infty} e^{-(u+1)s} s^{x+y-1}~ds$


    Let's come back to the original problem , we thus have

    $\displaystyle \displaystyle I = \frac{1}{\Gamma(x+y)}~ \int_0^{\infty} \int_0^{\infty} e^{-(u+1)s} u^{y-1} s^{x+y-1}~dsdu$

    $\displaystyle \displaystyle = \frac{1}{\Gamma(x+y)}~ \int_0^{\infty} \left[ \frac{\Gamma(y)}{s^y}~ e^{-s} s^{x+y-1} \right] ~ds $

    $\displaystyle \displaystyle = \frac{\Gamma(y)}{\Gamma(x+y)}~\Gamma(x) $

    $\displaystyle \displaystyle = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} $
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