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This time I want to prove that $\displaystyle \Gamma(x) = \lim_{\varepsilon\rightarrow 0} \int_\varepsilon ^{1/\varepsilon} t^{x-1} e^{-t}$, I know this is true because we have defined $\displaystyle \Gamma(x) = \int_{\mathbb R_+}t^{x-1}e^{-t} dt$ but the thing is how can I say this formally and using the fact that the function $t\mapsto t^{x-1}e^{-t} $ is integrable on $\mathbb R_+ = (0,\infty)$, because I have the following result : Prove that $ f:(a,b)\to\mathbb{R}$ is integrable iff $\lim_{\epsilon\to0} \int_{[a+\epsilon,b-\epsilon]}f$ exists and I think this could help, Can you help me with this ?

Thanks a lot in advance.

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user162343
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1 Answers1

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Substitute $\varepsilon = 1/n$, so we are looking at $$ \lim_{n \to \infty} \int_{1/n}^n t^{x-1} e^{-t} \, dt $$

You probably want to show that you have a Cauchy sequence, viz., given $\delta>0$, there is an $M$ so that $$ \left\lvert \int_{1/n}^{n} t^{x-1} e^{-t} \, dt - \int_{1/m}^{m} t^{x-1} e^{-t} \, dt \right\rvert < \delta $$ whenever $m,n>M$. Without loss of generality, take $n>m>M$. Then the left-hand side reduces to $$ \left\lvert \int_{m}^{n} t^{x-1} e^{-t} \, dt + \int_{1/n}^{1/m} t^{x-1} e^{-t} \, dt \right\rvert \leqslant \left\lvert \int_{m}^{n} t^{x-1} e^{-t} \, dt \right\rvert + \left\lvert \int_{1/n}^{1/m} t^{x-1} e^{-t} \, dt \right\rvert \\ = \int_{m}^{n} t^{x-1} e^{-t} \, dt + \int_{1/n}^{1/m} t^{x-1} e^{-t} \, dt, $$ and now we have to show these can be made small.

At this point it is sensible to integrate by parts, just to make $t^x$ an increasing function. This gives you $$ \int_a^b t^{x-1} e^{-t} \, dt = \frac{1}{x}[t^x e^{-t}]_a^b + \frac{1}{x} \int_a^b t^x e^{-t} \, dt = \frac{1}{x}(b^x e^{-b}-a^x e^{-a}) + \frac{1}{x} \int_a^b t^x e^{-t} \, dt $$

At this point I'm going to forget about the $1/x$, since it is just a constant. For the interval $[1/n,1/m]$, $$ \int_{1/n}^{1/m} x t^{x-1} e^{-t} \, dt = m^{-x} e^{-1/m} - n^{-x} e^{-1/n} + \int_{1/n}^{1/m} t^x e^{-t} \, dt \\ \leqslant m^{-x} + \int_0^{1/m} t^x \, dt, $$ and it is obvious that both of these can be made as small as we like by choosing $m$ large enough.

Now, the other integral requires a different idea. My suggestion is to integrate by parts the other way: this trick only requires that we know that $$ y^s e^{-y} \to 0 $$ as $y \to +\infty$, for any constant $s$. (There are plenty of places this is proved, including on this site.)

So what happens is: $$ \int_m^n t^{x-1} e^{-t} \, dt = [-t^{x-1}e^{-t}]_n^m + (x-1)\int_n^m t^{x-2} e^{-t} \, dt \\ \leq n^{x-1} e^{-n} + (x-1)n^{x-2} e^{-n} + \dotsb + (x-1)(x-2) \dotsm (x-k)\int_n^m t^{x-k} e^{-t} \, dt, $$ repeating this $k$(ish) times, where $k=\lfloor x \rfloor +1$ The first set of terms is basically a polynomial in $n$ of degree $k-2$, multiplied by $t^{x-k} e^{-n}$, and using the limit I mentioned, this can be made as small as we like by choosing $n$ large enough. The last term is the integral of a decreasing function since $x-k<0$, and hence is bounded by $ n^{x-k} \int_n^{\infty} e^{-t} \, dt $, which again can be made as small as we like.

Hence the improper integral is the limit of a Cauchy sequence, so it exists.

Chappers
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