Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can someone show me a proof or any clear resource about convergence of gamma function for values of $p$ less than zero.

If possible I need proofs using integration by parts.

My problem evaluating convergence is below.

$\displaystyle\int^\infty_0 e^{-x} \hspace{1 mm}x^p \ dx$ and $p<0$

Why does this integral not converge for $p \le -1$, but converge for $-1< p\le 0$

A proof using series or integrals (like an integral smaller than other convergent integral is convergent) would be appericiated.

share|cite|improve this question
The problem is around the zero, because of the function $x^p$ in $(0,\varepsilon)$ is not integrable for $p\leq -1$. – Kolmo Mar 9 '12 at 12:31

The integrand is non-negative, and the problem is only at $0$. Since $\lim_{x\to 0^+}e^{-x}=1$, we have $3\cdot 2^{-1}\geq e^{-x}\geq 2^{—1}$ for $x\leq x_0$, and so $3\cdot 2^{-1}x^p\geq e^{-x}x^p\geq 2^{-1}x^p\geq 0$. Since for $p\geq -1$ the integral $\int_0^1 x^pdx$ is divergent $\int_0^{+\infty}e^{-x}x^pdx$ is divergent and if $p<-1$ the integral $\int_0^1 x^pdx$ is convergent and so is $\int_0^{+\infty}e^{-x}x^pdx$.

share|cite|improve this answer
I appreciate your reasoning. But I think $\lim_{x\to 0^+}e^{-x}=1$. – Gardel Mar 9 '12 at 12:48
Yes, it's a typo. Thanks for pointing it out. – Davide Giraudo Mar 9 '12 at 12:48

Basically, given $p\leq0$

$$\Gamma(p)=\int\limits_0^1 x^{p-1} e^{-x} dx+\int\limits_1^\infty x^{p-1} e^{-x} dx$$

There is no convergence problems for $\displaystyle \int\limits_1^\infty x^{p-1} e^{-x} dx$, however, for $0<a<1$

$$\int\limits_a^1 x^{p-1} e^{-x} dx\geq e^{-1}\int\limits_a^1 \frac {dx} x =-e^{-1} \log a$$

and the limit for $a \to 0^{-}$ does not exist.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.