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Problem: Evaluate $$\int \Gamma(x)\, \mathrm{d}x$$

In asymptotic analysis, functions are compared with each other in terms of growth. I want to know how much better one function is than another.

To do this, one idea I had was to compute the average values of two functions and then compare them. This involves integration.

$n!$ is a common function in asymptotic analysis for describing the functions of higher growth rates. Of course, being a discrete function, you cannot integrate the factorial function, but the Gamma Function is its continuous analog.

So how do you integrate the gamma function? In other words, evaluate: $\int \Gamma(x)\, \mathrm{d}x$

Mikasa
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Flyster McFlyster
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2 Answers2

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If you want to compare the asymptotic behaviour of two functions, average values don't turn out to be an efficient tool.

Let $f,g$ be those two functions and let's assume they are positive for the sake of simplicity. The simplest way to compare them would be to compute the ratio $\displaystyle\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}$. If the result is above 1, then $f$ grows asymptotically faster than $g$; if the result is under 1, then $g$ grows faster; and if it is equal to 1, the two functions are asymptotically equal.

If you want to study the $f,g$ individually, you can determine their leading terms (using the big-O notation) and then compare them. You can also compare their growth rates for large values, i.e. the asymptotic behaviour of their derivatives.

The problem with average values is that you must compute them on finite intervals, otherwise the integrals won't converge, unless the functions are $L^1$, but then it means that $\displaystyle \lim_{x\rightarrow\infty}f(x) = \lim_{x\rightarrow\infty}g(x) = 0$, hence a trivial asymptotic behaviour. One way to circumvent this problem would be a continuous version of moving average, but that seems troublesome, since the integral of the gamma function doesn't possess a closed form anyway.

In conclusion, I would recommend you the first method I mentioned, taking into account the asymptotic behaviour of the factorial given by Stirling's formula, namely $n! \sim \sqrt{2\pi} \,n^{n+1/2}e^{-n}$.

Abezhiko
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Here goes nothing, looking at $$\int\Gamma(1+x)\mathrm{d}x=\int x\Gamma(x)\mathrm{d}x$$ we consider the integral form of $\Gamma(1+x)$, giving us, $$\int_0^{\infty}e^{-t}t^x\mathrm{d}t$$ We can rewrite the original integral as, $$\int x\left(\int_0^{\infty}e^{-t}t^{x-1}\mathrm{d}t\right)\mathrm{d}x$$ $$\int_0^{\infty}e^{-t}\left(\int xt^{x-1}\mathrm{d}x\right)\mathrm{d}t$$ Now, computing, $$\int xt^{x-1}\mathrm{d}x$$ we can use integration by parts and easily get that it simplifies to $$\frac{t^{x-1}(x\ln(t)-1)}{\ln^2t}+C$$ where $C$ is the constant of integration. Substituting this back in, we get; $$C+\int_0^{\infty}\frac{t^{x-1}(x\ln(t)-1)}{e^t\ln^2t}\mathrm{d}t$$ By inputting a numerical value of $x$, we can calculate this integral.

#1: To avoid the constant of integration, I suggest you to consider the function, $$\int_a^b\Gamma(1+x)\mathrm{d}x$$ That is, integrate $\Gamma(x)$ on a specified domain $[a,b]$. As usually, that is what you are going to be finding the indefinite for.

wolfqz
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