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I'm trying to verify a bound for the gamma function $$ \Gamma(z) = \int_0^\infty e^{-t}t^{z - 1}\;dt. $$

In particular, for real $m \geq 1$, I'd like to show that $$ \Gamma(m + 1) \leq 2\left(\frac{3m}{5}\right)^m. $$

Knowing that the bound should be attainable, my first instinct is to split the integral as $$ \Gamma(m + 1) = \int_0^{3m/5} e^{-t}t^{m}\;dt + \int_{3m/5}^\infty e^{-t}t^m\;dt \leq (1 - e^{-3m/5})\left(\frac{3m}{5}\right)^m + \int_{3m/5}^\infty e^{-t}t^m\;dt. $$

Using integration by parts, $$ \int_{3m/5}^\infty e^{-t}t^m\;dt = e^{-3m/5}\left(\frac{3m}{5}\right)^m + m\int_{3m/5}^\infty e^{-t}t^{m-1}\;dt.$$

So the problem has been reduced to showing $$ m\int_{3m/5}^\infty e^{-t}t^{m-1}\;dt \leq \left(\frac{3m}{5}\right)^m. $$

But this doesn't seem to have made the problem any easier.

Any help is appreciated, thanks.

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An idea: using induction on $m$ and $\Gamma(m+2)=(m+1)\Gamma(m+1)$. – vesszabo Oct 15 '12 at 20:46

Suppose $m = n + \alpha$ where $0 \le \alpha < 1$, so that $\alpha$ is the fractional part of $m$. Taking logarithms, the inequality becomes

$$ \sum_{k=1}^n \log (k + \alpha) + \log \Gamma (1+\alpha) < \log 2 + (n+\alpha) \log (n+\alpha) + (n+\alpha) \log \frac{3}{5} $$

Now using Riemann sums, we have

$$ \sum_{k=1}^n \log (k + \alpha) < \int_1^{n+1} \log(x+\alpha) dx$$ which is $$ (n+1+\alpha) \log (n+1+\alpha) - n - 1 - (1+\alpha) \log(1+\alpha) + 1.$$

This means we need to show that $$ (n+1+\alpha) \log (n+1+\alpha) - n - (1+\alpha) \log(1+\alpha) + \log \Gamma (1+\alpha) < \log 2 + (n+\alpha) \log (n+\alpha) + (n+\alpha) \log \frac{3}{5} $$

Rearranging terms, this is $$ (n+1+\alpha) \log (n+1+\alpha) - (n+\alpha) \log (n+\alpha) < \log 2 - \log \Gamma (1+\alpha) + (n+\alpha) \log \frac{3}{5} + n + (1+\alpha) \log(1+\alpha)$$

Now for the LHS we have $$ \log (n+1+\alpha) + \log \left( 1 + \frac{1}{n+\alpha}\right)^{n+\alpha} < \log (n+1+\alpha) + 1$$ because $ \log \left( 1 + \frac{1}{x}\right)^x < 1$ by a trivial calculation.

On the RHS we have $$ n \log\frac{3e}{5} + \log 2 - \log \Gamma (1+\alpha) + \alpha \log \frac{3}{5} + (1+\alpha) \log(1+\alpha) > \log (n+1+\alpha) + 1$$ for all $n > n_0$ for some $n_0$ because the coefficient $\log\frac{3e}{5}$ on $n$ is positive and the remaining terms are bounded by a constant. This concludes the proof. Note that the proof also goes through with a factor of $\frac{2}{5}$, just barely, and requiring $n_0 = 22.$ The original post has $n_0 = 1.$

I'm not sure of all the details but I hope it's a start.

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up vote 0 down vote accepted

I'll prove something that's close enough for my applications; in particular, that $$\Gamma(m + 1) \leq 3\left(\frac{3m}{5}\right)^m.$$

Let $0 < \alpha < 1$ be chosen later. We'll split $e^{-t}t^m$ as $(e^{-\alpha t}t^m)e^{-(1 - \alpha)t}$ and use this to bound the integral.

First, take a derivative to find a maximum for $e^{-\alpha t}t^m$.

$$\frac{d}{dt}e^{-\alpha t}t^m = -\alpha e^{-\alpha t}t^m + me^{-\alpha t}t^{m-1} = -\alpha e^{-\alpha t}t^{m - 1}\left(t - \frac{m}{\alpha}\right). $$

So $t = m / \alpha$ is a critical point, and in particular a maximum (increasing before and decreasing after, if you like).

Then we can bound the integral

$$ \Gamma(m + 1) = \int_0^\infty e^{-t}t^m\;dt \leq \left(\frac{m}{\alpha e}\right)^m \int_0^\infty e^{-(1 - \alpha)t}\;dt = \left(\frac{m}{\alpha e}\right)^m \left(\frac{1}{1 - \alpha}\right).$$

Choosing $\alpha = 5/(3e)$ and noting that $\frac{1}{1 - 5/(3e)} \leq 3$, we've proven

$$ \Gamma(m + 1) \leq 3\left(\frac{3m}{5}\right)^m. $$

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