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Math Help - An Interesting Integral

  1. #1
    Member sbhatnagar's Avatar
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    An Interesting Integral

    Prove that

    \int_0^1 \ln \left( \Gamma(1+x)\right)dx = \frac{1}{2}\ln(2\pi)-1
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  2. #2
    Member sbhatnagar's Avatar
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    Re: An Interesting Integral

    This Problem is not as difficult as it looks. Use the property \Gamma(1+x)=x\Gamma(x):

    \begin{aligned} \int_0^1 \ln(\Gamma(1+x))dx &= \int_0^1 \ln(x\Gamma(x))dx \\ &= \int_0^1 \ln x dx+\int_0^1\ln(\Gamma(x))dx\end{aligned}

    The first integral equals -1.

    \begin{aligned} \int_0^1 \ln(\Gamma(1+x))dx &= \int_0^1\ln(\Gamma(x))dx-1\end{aligned}

    Let I=\int_0^1 \ln(\Gamma(x)) dx -------(1)

    A property of definite integral says that

    \int_0^a f(x)dx=\int_0^a f(a-x) dx

    therefore

    I=\int_0^1 \ln(\Gamma(1-x))dx --------(2)

    Add (1) and (2):

    2I = \int_0^1 \ln(\Gamma(x) \Gamma(1-x))dx

    = \int _0^1 \ln \left( \frac{\pi}{\sin(\pi x)}\right)dx (from Euler's Reflection Formula! )

    =\ln(\pi)\int_0^1 dx - \int_0^1 \ln(\sin (\pi x))dx

    Recall that \int_0^1 \ln(\sin(\pi x)) dx= -\ln(2).

    Therefore 2I= \ln(2\pi)

    I=\frac{1}{2}\ln(2\pi)

    \int_0^1 \ln(\Gamma(1+x))dx=\frac{1}{2}\ln(2\pi)-1
    Thanks from MarkFL and vincisonfire
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    Re: An Interesting Integral

    How would you prove the convergence of this integral ... ?
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    Re: An Interesting Integral

    Quote Originally Posted by zaidalyafey View Post
    How would you prove the convergence of this integral ... ?
    The integrand is real and finite in that interval.
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    Re: An Interesting Integral

    Here is another way to evaluate the integral. Write it out as a Riemann sum, the old-school way:

    \int_{a}^{a+1} f(x) dx = \lim_{n \to \infty}  \frac{1}{n} \left (f(a) + f(a + \frac{1}{n}) + f(a + \frac{2}{n}) + f(a + \frac{3}{n}) .... + f(a + \frac{n-1}{n}) \right )

    \int_0^1 \ln \left( \Gamma(1+x)\right)dx = \int_1^2 \ln \left( \Gamma(x)\right)dx

    So

    \int_1^2 \ln(\Gamma(x))dx = \lim_{n \to \infty} \frac{1}{n} \left (\ln(\Gamma(1)) + \ln(\Gamma(1 + \frac{1}{n})) + \ln(\Gamma(1 + \frac{2}{n})) + \ln(\Gamma(1 + \frac{3}{n})) .... + \ln(\Gamma(1 + \frac{n-1}{n})) \right )

    = \lim_{n \to \infty} \frac{1}{n} \ln\left ( \Gamma(1)\Gamma(1 + \frac{1}{n})\Gamma(1 + \frac{2}{n})\Gamma(1 + \frac{3}{n}) ....\Gamma(1 + \frac{n-1}{n}) \right )

    This exactly fits the format for the multiplication theorem for the gamma function! So

    I = \int_1^2 \ln \left( \Gamma(x)\right)dx = \lim_{n \to \infty} \frac{1}{n}\ln \left ((2\pi)^\frac{n-1}{2}\cdot n^{\frac{1}{2}-n}\cdot \Gamma(n))\right )

    e^I = \lim_{n \to \infty} \left ((2\pi)^\frac{n-1}{2}\cdot n^{\frac{1}{2}-n}\cdot \Gamma(n))\right )^\frac{1}{n}

    = \lim_{n \to \infty} \left ((2\pi)^\frac{n-1}{2n}\cdot n^{\frac{1}{2n}-1}\cdot ((n-1)!)^\frac{1}{n}\right )

    = \sqrt{2\pi }\lim_{n \to \infty} \left (\frac{n^{\frac{1}{2n}}}{n}\cdot ((n-1)!)^\frac{1}{n}\right )

    = \sqrt{2\pi } \cdot \lim_{n \to \infty} \left ( n^{\frac{1}{2n}} \right ) \cdot \lim_{n \to \infty} \left ( \frac{((n-1)!)^{\frac{1}{n}}}{n} \right )

    = \frac{\sqrt{2\pi}}e

    I = \ln{\frac{\sqrt{2\pi}}e}
    = \frac{1}{2}\ln(2\pi) - 1
    Last edited by SworD; January 19th 2013 at 04:33 PM.
    Thanks from sbhatnagar
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