# Convergence of $\Gamma(p)$ for $0<p\leq 1$ and divergence for $p \leq0$.

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.

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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 $\int_0^{+\infty}e^{-x}x^pdx$.

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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.

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