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Convergence of the gamma function

April 26, 2012

I will write up the proof of the convergence of the gamma function as a follow-up to this post, where I show why the exponential grows faster than any polynomial. The gamma function is defined by

\Gamma(s) = \displaystyle\int_{0}^{\infty} e^{-t}t^{s-1}dt.

Proposition: This integral converges for s > 0.

Proof: Let’s divide the integral in a sum of two terms,

\Gamma(s) = \displaystyle\int_0^1 e^{-t}t^{s-1}dt + \int_1^{\infty}e^{-t}t^{s-1}dt.

For the first term, since the function e^{-t} is decreasing, it’s maximum on the interval [0,1] is attained at t = 0 so

\displaystyle\int_0^1 e^{-t}t^{s-1}dt < \int_0^1t^{s-1}dt.

But for s > 0, this last integral converges to 1/s.

For the second term, we use what we showed in this post: since the exponential grows faster than any polynomial, for every s we can take N \in \textbf{N} so big that t \geq N \Rightarrow e^{t/2} > t^{s-1}. So

\displaystyle\int_1^{\infty}e^{-t}t^{s-1}dt = \int_1^Ne^{-t}t^{s-1}dt + \int_N^{\infty}e^{-t}t^{s-1}dt

< \displaystyle\int_1^Ne^{-t}t^{s-1}dt + \int_N^{\infty} e^{-t}e^{t/2}dt

= \displaystyle\int_1^Ne^{-t}t^{s-1}dt + \int_N^{\infty}e^{-t/2}dt

< \infty

which completes the proof.

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3 Comments
  1. Erica Liu permalink

    May I ask why e^(t/2) was chosen not e^(t/3) for instance for the second term? Much appreciated.

  2. Madeline permalink

    I don’t think it makes a difference what is used (e^(t/2) or something else) so long as it is in fact > x^(p-1). For instance what if we used e^x? e^x > x^(p-1). Using e^x would make it even more obvious.

    • It is true that we could also use e^{t/3} because it’s greater than t^{s-1} for any s, but also because e^{-t}e^{t/3} is integrable. What you suggest Madeline doesn’t work, if we take e^t instead of e^{t/2}, then we get \int_N^{\infty}e^{-t}t^{s-1} < \int_N^{\infty}e^{-t}e^{t} = \int_N^{\infty}1 = +\infty and we can’t conclude anything.

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