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Math Help - complex analysis method to solve.

  1. #1
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    complex analysis method to solve.

    This may be aimed at PH(or whomever), since he is up on CA and I am just learning it a little as I find time.

    How could we go about using complex integration to solve this integral from a previous post?. Assuming it's worthwhile in this case. Are cases where there are infinite poles more difficult?.

    \int_{0}^{\infty}\frac{x}{e^{x}-1}dx

    I see it has poles at 2c{\pi}i and a double at 0, I think.

    Should we end up with something like 2{\pi}i(\frac{\pi}{12i})=\frac{{\pi}^{2}}{6}

    Or maybe even 2{\pi}i(-\frac{\pi}{12}i)=\frac{{\pi}^{2}}{6}
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    Quote Originally Posted by galactus View Post
    \int_{0}^{\infty}\frac{x}{e^{x}-1}dx
    I would not do this problem with complex analysis for two reasons:
    1)The function is odd, usually we find the integral from -\infty to \infty which is 1/2 of your desired integral if the function is even.
    2)There are infinitely many poles for this function. And one of them is right on 0. Which means we have to avoid the origin somehow. Which is not looking good at all.

    Maybe if the integral was \int_0^{\infty} \frac{x}{e^x + 1} dx we can do something nicer. Like for example take the contour with vertices at 0,R,\frac{1}{2}\pi i, \frac{1}{2}\pi i + R which has no poles and work from there.
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    Thank you PH. I like this CA stuff. You're right, it's an interesting branch of mathematics.

    Here's something you might find of interest. It involves the gamma function.

    {\Gamma}(x)=\int_{0}^{\infty}t^{x-1}e^{-t}dt, for x>0.

    Make a change of variables, t=ru:

    {\Gamma}(x)=\int_{0}^{\infty}(ru)^{x-1}e^{-ru}rdu=r^{x}\int_{0}^{\infty}u^{x-1}e^{-ru}du

    \frac{1}{r^{x}}=\frac{1}{{\Gamma}(x)}\int_{0}^{\in  fty}u^{x-1}e^{-ru}du

    {\zeta}(x)=\sum_{r=1}^{\infty}\frac{1}{r^{x}}=\fra  c{1}{{\Gamma}(x)}\sum_{r=1}^{\infty}\int_{0}^{\inf  ty}u^{x-1}e^{-ru}du

    = \frac{1}{{\Gamma}(x)}\int_{0}^{\infty}u^{x-1}\sum_{r=1}^{\infty}e^{-ru}du

    having been pushed through the integral, and summing the infinite geometric series gives us:

    {\zeta}(x)=\frac{1}{{\Gamma}(x)}\int_{0}^{\infty}u  ^{x-1}\frac{e^{-u}}{1-e^{-u}}du

    And we get a marvelous formula:

    \boxed{{\zeta}(x){\Gamma}(x)=\int_{0}^{\infty}\fra  c{u^{x-1}}{e^{u}-1}du}

    Since {\zeta}(x)=\sum_{r=1}^{\infty}\frac{1}{r^{x}}

    This gives us, for our problem, {\zeta}(2){\Gamma}(2)=\int_{0}^{\infty}\frac{u}{e^  {u}-1}du=\frac{{\pi}^{2}}{6}

    I think this is really cool.
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    Quote Originally Posted by galactus View Post
    I think this is really cool.
    Sweet! I like it. I happen to know that identity with the Gamma-Zeta function, I cannot believe I did not see it.
    You come up with it yourself?

    By the way \zeta (4) = \frac{\pi^4}{90} so you can do the same trick. But not with \zeta (3) .
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  5. #5
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    You come up with it yourself?
    No, PH. I didn't have to. It's in the Gamma book by Havil. I recommended this book to you a while back. If you don't have it, try to find it and give a look. It's your kinda stuff.


    By the way \zeta (4) = \frac{\pi^4}{90} so you can do the same trick. But not with \zeta (3) .
    Yes, I know. It's cool stuff. Also, how these are tied in with the Bernoulli numbers is another thing that's interesting.

    You probably know it already, but here is the general formula for the sum of the reciprocals of the even powers.

    \sum_{n=1}^{\infty}\frac{1}{n^{2k}}=\boxed{\frac{(-1)^{k-1}B_{2k}(2\pi)^{2k}}{2(2k)!}}, where B_{2k} is a Bernoulli number. For instance, if we wanted the sum of the reciprocals of the powers of 10, we would use k=5 and find B_{10}=\frac{5}{66} and find \sum_{n=1}^{\infty}\frac{1}{n^{10}}=\frac{{\pi}^{1  0}}{93555}. Of course, all Bernoulli number with odd subscripts are 0.

    Kind of off topic, but another formula you may like is the remarkable formula for the Bell numbers. The nth Bell number, b_{n}, is the number of ways of partitioning a set of n objects into subsets. For instance, b_{3}=5 since the partitions of {1,2,3} are

    {1,2,3}, \;\ (1,2)\cup(2), \;\ (1,3)\cup(2), \;\ (2,3)\cup(1), \;\ 1\cup{2}\cup{3}

    We define b_{0}=1

    They go 1,1,2,5,15,52,..........

    The general formula is b_{n}=\frac{1}{e}\sum_{k=1}^{\infty}\frac{k^{n-1}}{(k-1)!}

    How many ways can 10 objects be partitioned into subsets?.

    Using the formula we see it is 115975.

    I just thought this was interesting. Thought you may too.
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