1. Show that $\Gamma(y) = \int_0^{\infty}{e^{-x}x^{y-1}\,dx}$ is finite for $y>0$ both as an improper Riemann integral and as a Lebesgue integral.

  2. Show $\Gamma'(y) = \int_0^{\infty}{e^{-x}x^{y-1}\ln{x}\,dx}$ for $y>0$.

For one: I've tried simply integrating it as an improper Riemann integral, but you always end up with another integral of the "same type" (which is how you eventually show $\Gamma(y+1)=y\Gamma(y)$ ). How do I get around this? As for the Lebesgue integral, I think it'd be easiest to compare the integrand to a larger function whose integral converges, but I haven't come up with a good candidate.

For two: Fix $y_0>0$. Write $$\Gamma'(y) = \lim_{y\to y_0}{\int_0^{\infty}{\frac{e^{-x}x^{y-1}-e^{-x}x^{y_0-1}}{y-y_0}\,dx}}\,.$$ By the MVT, there exists $\eta$ between $y$ and $y_0$ such that the above limit is equal to $$\lim_{y\to y_0}{\int_0^{\infty}{e^{-x}x^{\eta-1}\ln{x}\,dx}}\,.$$ But now I'm not sure what to do. This is similar to a previous question I posted; for that problem we knew the derivative of the original integrand was bounded, so we applied the bounded convergence theorem. Would it be enough to prove that the derivative of my integrand is bounded, and apply BCT?

  • 1
    $\begingroup$ 1. Bound it by something more convenient. Remember that exponentials eventually outgrow polynomials. For y less than 1 you want two different bounds at 0 and at infinity. 2. Have you tried letting the lower bound be positive and then letting it tend to zero? $\endgroup$ Dec 4, 2010 at 19:15
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    $\begingroup$ As the integrand is positive and continuous, the questions of whether it is a Lebesgue integral and whether it is an improper Riemann integral as the same. Then it's just a question of showing the integral converges as $x$ goes to infinity and to zero. Going to zero is the more delicate of the two. Also there are theorems regarding differentiation of integrals under the integral sign. There are conditions but they do apply in this case. $\endgroup$ Dec 4, 2010 at 19:18
  • $\begingroup$ Thanks for your help, Robin and Qiaochu. If either of you post your comment as an answer, I will upvote. $\endgroup$
    – Bey
    Dec 7, 2010 at 5:08

1 Answer 1


A bit late to the game, but here is an answer:

1: To show it is finite: Write $e^{-x}=e^{-x/2}\cdot e^{-x/2}$. For every $y>0$ there exists $N$ such that $e^{-x/2}x^{y-1}<1$ when $x>N$ so that $$\int_N^\infty e^{-x} x^{y-1}dx\leq \int_N^\infty e^{-x/2} <\infty.$$ For $x$ between $0,1$ compare to $\int x^{y-1}dx$ which converges whenever $y-1>-1$, so for all $y$. Lastly bounding $\int_1^N e^{-x}x^{y-1}dx$ I leave to you.

2: For this part, what you have done so far is good. All we need to do is switch the last integral with the limit. To do this, notice that in some neighborhood of radius $\delta$ around $y_0$ we can bound $$\int_0^\infty e^{-x} x^{y-1}dx$$ by similar methods as in 1. Then, applying the Dominated Convergence Theorem to that neighborhood, we can switch the limit and the integral, solving the problem.

Remark: For that last part, notice we cannot bound $$\int_0^\infty e^{-x} x^{y-1}dx$$ uniformely for all $y\in(0,\infty)$ since the function blows up at $0$ and at $\infty$. But fortunately, we can bound it uniformely for all $y$ in any compact subset of $(0,\infty)$, allowing us to use the DCT.


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